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1l6m5k24x | maths | circle | position-of-a-point-with-respect-to-circle | <p>For $$\mathrm{t} \in(0,2 \pi)$$, if $$\mathrm{ABC}$$ is an equilateral triangle with vertices $$\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t)$$ and $$C(a, b)$$ such that its orthocentre lies on a circle with centre $$\left(1, \frac{1}{3}\right)$$, then $$\left(a^{2}-b^{2}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{8}{3}$$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "$$\\frac{77}{9}$$"}, {"identifier": "D", "content": "$$\\frac{80}{9}$$"}] | ["B"] | null | <p>Let $$P(h,k)$$ be the orthocentre of $$\Delta$$ABC</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7raysn2/64b7b0e9-e0f5-48bc-ba4c-fe9b8ae351d3/2e1229e0-2e7f-11ed-8702-156c00ced081/file-1l7raysn3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7raysn2/64b7b0e9-e0f5-48bc-ba4c-fe9b8ae351d3/2e1229e0-2e7f-11ed-8702-156c00ced081/file-1l7raysn3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Circle Question 45 English Explanation"></p>
<p>Then</p>
<p>$$h = {{\sin t + \cos t + a} \over 3},\,k = {{ - \cos t + \sin t + b} \over 3}$$</p>
<p>(Orthocentre coincide with centroid)</p>
<p>$$\therefore$$ $${(3h - a)^2} + {(3k - b)^2} = 2$$</p>
<p>$$\therefore$$ $${\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}$$</p>
<p>$$\because$$ Orthocentre lies on circle with centre $$\left( {1,{1 \over 3}} \right)$$</p>
<p>$$\therefore$$ $$a = 3,\,b = 1$$</p>
<p>$$\therefore$$ $${a^2} - {b^2} = 8$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
lsbkwd8l | maths | circle | position-of-a-point-with-respect-to-circle | Four distinct points $(2 k, 3 k),(1,0),(0,1)$ and $(0,0)$ lie on a circle for $k$ equal to : | [{"identifier": "A", "content": "$\\frac{3}{13}$"}, {"identifier": "B", "content": "$\\frac{2}{13}$"}, {"identifier": "C", "content": "$\\frac{5}{13}$"}, {"identifier": "D", "content": "$\\frac{1}{13}$"}] | ["C"] | null | <p>$$(2 k, 3 k)$$ will lie on circle whose diameter is $$A B$$.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1dia1n/7e33c93b-3bc8-4d95-93f9-8ca9b48aaae2/7f9962b0-d3c9-11ee-a50b-bb659a2e1d74/file-1lt1dia1o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1dia1n/7e33c93b-3bc8-4d95-93f9-8ca9b48aaae2/7f9962b0-d3c9-11ee-a50b-bb659a2e1d74/file-1lt1dia1o.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Morning Shift Mathematics - Circle Question 18 English Explanation"></p>
<p>$$\begin{aligned}
& (x-1)(x)+(y-1)(y)=0 \\
& x^2+y^2-x-y=0 \quad \text{.... (i)}
\end{aligned}$$</p>
<p>Satisfy (2k, 3k) in (i)</p>
<p>$$\begin{aligned}
& (2 k)^2+(3 k)^2-2 k-3 k=0 \\
& 13 k^2-5 k=0 \\
& k=0, k=\frac{5}{13} \\
& \text { hence } k=\frac{5}{13}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lsf0k4jf | maths | circle | position-of-a-point-with-respect-to-circle | <p>Equations of two diameters of a circle are $$2 x-3 y=5$$ and $$3 x-4 y=7$$. The line joining the points $$\left(-\frac{22}{7},-4\right)$$ and $$\left(-\frac{1}{7}, 3\right)$$ intersects the circle at only one point $$P(\alpha, \beta)$$. Then, $$17 \beta-\alpha$$ is equal to _________.</p> | [] | null | 2 | <p>Centre of circle is $$(1,-1)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2z9gaq/def8db7c-9d66-41d8-85bb-325814b0bd97/5ba32a20-d4ab-11ee-bdd1-01c80c3e2d9a/file-1lt2z9gar.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2z9gaq/def8db7c-9d66-41d8-85bb-325814b0bd97/5ba32a20-d4ab-11ee-bdd1-01c80c3e2d9a/file-1lt2z9gar.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Mathematics - Circle Question 14 English Explanation"></p>
<p>Equation of $$A B$$ is $$7 x-3 y+10=0 \ldots$$ (i)</p>
<p>Equation of $$\mathrm{CP}$$ is $$3 x+7 y+4=0 \ldots$$ (ii)</p>
<p>Solving (i) and (ii)</p>
<p>$$\alpha=\frac{-41}{29}, \beta=\frac{1}{29} \quad \therefore 17 \beta-\alpha=2$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
lvb294qv | maths | circle | position-of-a-point-with-respect-to-circle | <p>If $$\mathrm{P}(6,1)$$ be the orthocentre of the triangle whose vertices are $$\mathrm{A}(5,-2), \mathrm{B}(8,3)$$ and $$\mathrm{C}(\mathrm{h}, \mathrm{k})$$, then the point $$\mathrm{C}$$ lies on the circle :</p> | [{"identifier": "A", "content": "$$x^2+y^2-74=0$$\n"}, {"identifier": "B", "content": "$$x^2+y^2-65=0$$\n"}, {"identifier": "C", "content": "$$x^2+y^2-61=0$$\n"}, {"identifier": "D", "content": "$$x^2+y^2-52=0$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwabk4bk/5b6f7bc9-3122-44bf-ae55-efc3ca540e79/9484e510-1419-11ef-b3d9-7392e0033caf/file-1lwabk4bl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwabk4bk/5b6f7bc9-3122-44bf-ae55-efc3ca540e79/9484e510-1419-11ef-b3d9-7392e0033caf/file-1lwabk4bl.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Circle Question 2 English Explanation"></p>
<p>Slope of $$B F=\left(m_{B F}\right)=\frac{3-1}{8-6}=1$$</p>
<p>$$\begin{aligned}
& m_{A C}=\frac{k+2}{h-5}=-1(A C \perp B F) \\
& \Rightarrow k+2=5-h \\
& \Rightarrow k=3-h \quad \ldots \text { (i) } \\
& m_{A D}=\frac{1+2}{6-5}=3 \\
& m_{B C}=\frac{k-3}{h-8}=-\frac{1}{3}(A D \perp B C) \\
& \Rightarrow 3 k-9=8-h \quad \ldots \text { (ii) }
\end{aligned}$$</p>
<p>From (i) & (ii)</p>
<p>$$h=-4, k=7$$</p>
<p>Now, $$h^2+k^2=16+49=65$$</p>
<p>$$h^2+k^2-65=0$$</p>
<p>Locus is $$x^2+y^2-65=0$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
yQICdNcg13IghxZH | maths | circle | radical-axis | If the circles $${x^2}\, + \,{y^2} + \,2ax\, + \,cy\, + a\,\, = 0$$ and $${x^2}\, + \,{y^2} - \,3ax\, + \,dy\, - 1\,\, = 0$$ intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for : | [{"identifier": "A", "content": "exactly one value of a "}, {"identifier": "B", "content": "no value of a"}, {"identifier": "C", "content": "infinitely many values of a "}, {"identifier": "D", "content": "exactly two values of a "}] | ["B"] | null | $${s_1} = {x^2} + {y^2} + 2ax + cy + a = 0$$
<br><br>$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$
<br><br>Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is
<br><br>given by $${s_1} - {s_2} = 0$$
<br><br>$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$
<br><br>Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$
<br><br>$$\therefore$$ The two equations should represent the same line
<br><br>$$ \Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$
<br><br>$$ \Rightarrow a + 1 = - {a^2}$$
<br><br>$${a^2} + a + 1 = 0$$
<br><br>No real value of $$a.$$ | mcq | aieee-2005 |
1lsg53zah | maths | circle | radical-axis | <p>Consider two circles $$C_1: x^2+y^2=25$$ and $$C_2:(x-\alpha)^2+y^2=16$$, where $$\alpha \in(5,9)$$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $$C_1$$ and $$C_2$$ be $$\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right)$$. If the length of common chord of $$C_1$$ and $$C_2$$ is $$\beta$$, then the value of $$(\alpha \beta)^2$$ equals _______.</p> | [] | null | 1575 | <p>$$\begin{gathered}
C_1: x^2+y^2=25, C_2:(x-\alpha)^2+y^2=16 \\
5<\alpha<9
\end{gathered}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxip5f/ce7292b4-b5dd-4ed7-ad89-ffefc757a7b4/91687230-ccf1-11ee-a330-494dca5e9a63/file-6y3zli1lsoxip5g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxip5f/ce7292b4-b5dd-4ed7-ad89-ffefc757a7b4/91687230-ccf1-11ee-a330-494dca5e9a63/file-6y3zli1lsoxip5g.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Circle Question 13 English Explanation"></p>
<p>$$\begin{aligned}
& \theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right) \\
& \sin \theta=\frac{\sqrt{63}}{8}
\end{aligned}$$</p>
<p>Area of $$\triangle \mathrm{OAP}=\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta$$</p>
<p>$$\begin{aligned}
\Rightarrow \quad & \alpha \beta=40 \times \frac{\sqrt{63}}{8} \\
& \alpha \beta=5 \times \sqrt{63} \\
& (\alpha \beta)^2=25 \times 63=1575
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
lv9s1zqi | maths | circle | radical-axis | <p>Let the circle $$C_1: x^2+y^2-2(x+y)+1=0$$ and $$\mathrm{C_2}$$ be a circle having centre at $$(-1,0)$$ and radius 2 . If the line of the common chord of $$\mathrm{C}_1$$ and $$\mathrm{C}_2$$ intersects the $$\mathrm{y}$$-axis at the point $$\mathrm{P}$$, then the square of the distance of P from the centre of $$\mathrm{C_1}$$ is:</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>$$\begin{gathered}
C_1: x^2+y^2-2(x+y)+1=0 \\
C_2:(x+1)^2+y^2=(2)^2 \\
x^2+y^2+2 x-3=0
\end{gathered}$$</p>
<p>Common chord is</p>
<p>$$\begin{aligned}
& C_1-C_2=0 \\
& \Rightarrow 2 x+y-2=0
\end{aligned}$$</p>
<p>also, this line intersects the $$y$$-axis at the point</p>
<p>$$\begin{aligned}
& P(y, 0) . \\
& \Rightarrow y=2
\end{aligned}$$</p>
<p>$$P(2,0)$$</p>
<p>Distance of point $$P$$ from $$(1,1)$$ is</p>
<p>$$\begin{aligned}
& d=\sqrt{(2-1)^2+(0-1)^2} \\
& =\sqrt{1^2+1^2} \\
& d=\sqrt{2} \\
& \Rightarrow d^2=2 \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
JSdvzTMnngQqBri2 | maths | circle | tangent-and-normal | The circle passing through $$(1, -2)$$ and touching the axis of $$x$$ at $$(3, 0)$$ also passes through the point : | [{"identifier": "A", "content": "$$\\left( { - 5,\\,2} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { 2,\\,-5} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { 5,\\,-2} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - 2,\\,5} \\right)$$"}] | ["C"] | null | Since circle touches $$x$$-axis at $$(3,0)$$
<br><br>$$\therefore$$ The equation of circle be
<br><br>$${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265119/exam_images/izz5jb334xfqht2xhhp9.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Circle Question 138 English Explanation">
<br><br>As it passes through $$(1, -2)$$
<br><br>$$\therefore$$ Put $$x=1,$$ $$y=-2$$
<br><br>$$ \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0$$
<br><br>$$ \Rightarrow \lambda = 4$$
<br><br>$$\therefore$$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$
<br><br>Now, from the options $$\left( {5, - 2} \right)$$ satisfies equation of circle. | mcq | jee-main-2013-offline |
bj6fQIovF0KROgnzjJ6iB | maths | circle | tangent-and-normal | Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x − y = 1 and 2x + y = 3 is : | [{"identifier": "A", "content": "4x + y \u2212 3 = 0"}, {"identifier": "B", "content": "x + 4y + 3 = 0"}, {"identifier": "C", "content": "3x \u2212 y \u2212 4 = 0"}, {"identifier": "D", "content": "x \u2212 3y \u2212 4 = 0"}] | ["B"] | null | Point of intersection of lines
<br><br>x $$-$$ y = 1 and 2x + y = 3 is $$\left( {{4 \over 3},{1 \over 3}} \right)$$
<br><br>Slope of OP = $${{{1 \over 3} + 1} \over {{4 \over 3} - 1}}$$ = $${{{4 \over 3}} \over {{1 \over 3}}}$$ = 4
<br><br>Slope of tangent = $$-$$ $${1 \over 4}$$
<br><br>Equation of tangent y + 1 = $$-$$ $${1 \over 4}$$ (x $$-$$ 1)
<br><br>4y + 4 = $$-$$ x + 1
<br><br>x + 4y + 3 = 0
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266529/exam_images/iuvzkrct7gyuczkecdmp.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Circle Question 122 English Explanation"> | mcq | jee-main-2016-online-10th-april-morning-slot |
4hYh8JWttCD0c1UU | maths | circle | tangent-and-normal | The radius of a circle, having minimum area, which touches the curve y = 4 – x<sup>2</sup> and the lines, y = |x| is : | [{"identifier": "A", "content": "$$2\\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "B", "content": "$$4\\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "C", "content": "$$4\\left( {\\sqrt 2 + 1} \\right)$$"}, {"identifier": "D", "content": "$$2\\left( {\\sqrt 2 + 1} \\right)$$"}] | ["B"] | null | Let the radius of circle with least area be r.
<br><br>Then, the coordinate of the center = (0, b)
<br><br>$$ \therefore $$ The equation of circle be
x<sup>2</sup> + (y – b)<sup>2</sup> = r<sup>2</sup>
<br><br>Distance of perpendiculur from (0, 4) to y = x line = r
<br><br>$$ \Rightarrow $$ $$\left| {{{ - b} \over {\sqrt 2 }}} \right| = r$$
<br><br>$$ \Rightarrow $$ b = $${\sqrt 2 r}$$
<br><br>Circle passes through (0, 4),
<br><br>$$ \therefore $$ 0 + (4 – b)<sup>2</sup> = r<sup>2</sup>
<br><br>$$ \Rightarrow $$ 4 - b = r
<br><br>$$ \Rightarrow $$ 4 - $${\sqrt 2 r}$$ = r
<br><br>$$ \Rightarrow $$ r = $${4 \over {\sqrt 2 + 1}}$$ = $$4\left( {\sqrt 2 - 1} \right)$$ | mcq | jee-main-2017-offline |
z7Ciz7cseh37ERJT | maths | circle | tangent-and-normal | If the tangent at (1, 7) to the curve x<sup>2</sup> = y - 6
<br/><br/>touches the circle x<sup>2</sup> + y<sup>2</sup> + 16x + 12y + c = 0, then the value of c is : | [{"identifier": "A", "content": "95"}, {"identifier": "B", "content": "195"}, {"identifier": "C", "content": "185"}, {"identifier": "D", "content": "85"}] | ["A"] | null | <b>NOTE :</b>
<br><br>Equation of tangent at (x<sub>1</sub>, y<sub>1</sub>) to the curve x<sup>2</sup> = 4ay is
<br><br>xx<sub>1</sub> = 4a $$\left( {{{y + {y_1}} \over 2}} \right)$$
<br><br>Now equation of tangent at (1, 7) to x<sup>2</sup> = y $$-$$ 6 is
<br><br>$$x\,.\,1 = 4\,.\,{1 \over 4}\left( {{{y + 7} \over 2}} \right) - 6$$
<br><br>$$ \Rightarrow 2x = y + 7 - 12$$
<br><br>$$ \Rightarrow 2x - y + 5 = 0$$
<br><br>This tangent touches the circle.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267101/exam_images/vuu1dbzfbd8ek3dtzrtw.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Circle Question 132 English Explanation">
<br><br>So, perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.
<br><br>For the circle,
<br><br>$${x^2} + {y^2} + 16x + 12y + C = 0$$
<br><br>center is ($$-$$8, $$-$$6)
<br><br>and radius (r) = $$\sqrt {{8^2} + {6^2} - c} $$
<br><br>$$ = \sqrt {100 - c} $$
<br><br>Distance of the tangent from the center of the circle
<br><br>d = $$\left| {{{2\left( { - 8} \right) - \left( { - 6} \right) + 5} \over {\sqrt {{2^2} + {1^2}} }}} \right|$$
<br><br>$$ = \left| {{{ - 16 + 6 + 5} \over {\sqrt 5 }}} \right|$$
<br><br>And we know d = r
<br><br>$$\therefore\,\,\,$$ $$\left| {{{ - 16 + 11} \over {\sqrt 5 }}} \right| = \sqrt {100 - c} $$
<br><br>$$ \Rightarrow \left| {{{ - 5} \over {\sqrt 5 }}} \right| = \sqrt {100 - c} $$
<br><br>$$ \Rightarrow \left| { - \sqrt 5 } \right| = \sqrt {100 - c} $$
<br><br>$$ \Rightarrow 5 = 100 - c$$
<br><br>$$ \Rightarrow c = 95$$ | mcq | jee-main-2018-offline |
8nYxtTBHvZiNiZEYJGMbO | maths | circle | tangent-and-normal | The tangent to the circle C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x $$-$$ 1 = 0 at the point (2, 1) cuts off a chord of length 4 from a circle C<sub>2</sub> whose center is (3, $$-$$2). The radius of C<sub>2</sub> is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\sqrt 6 $$"}] | ["D"] | null | Here, equation of tangent on C<sub>1</sub> at (2, 1) is :
<br><br>2x + y $$-$$ (x + 2) $$-$$1 = 0
<br><br>Or x + y = 3
<br><br>If it cuts off the chord of the circle C<sub>2</sub> then the equation of the chord is :
<br>x + y = 3
<br><br>$$\therefore\,\,\,$$ distance of the chord from (3, $$-$$ 2) is :
<br><br>d = $$\left| {{{3 - 2 - 3} \over {\sqrt 2 }}} \right|$$ = $$\sqrt 2 $$
<br><br>Also, length of the chord is $$l$$ = 4
<br><br>$$\therefore\,\,\,$$ radius of C<sub>2</sub> = r = $$\sqrt {{{\left( {{l \over 2}} \right)}^2} + {d^2}} $$
<br><br>= $$\sqrt {{{\left( 2 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} = \sqrt 6 $$ | mcq | jee-main-2018-online-15th-april-evening-slot |
xyLfjlL9Szi7FPwhZZtsV | maths | circle | tangent-and-normal | The tangent and the normal lines at the point
( $$\sqrt 3 $$, 1) to the circle x<sup>2</sup>
+ y<sup>2</sup> = 4 and the x-axis form a triangle. The area of this triangle (in
square units) is : | [{"identifier": "A", "content": "$${4 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${1 \\over {3 }}$$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266326/exam_images/zc1u5khmyzw8xvgzk5sb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266898/exam_images/xsjbhlhn0ihvo9jj6lv3.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265355/exam_images/owzi5irsmrxdywbqqopg.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267262/exam_images/flmqizn0e2xqx806nx40.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Circle Question 110 English Explanation"></picture>
<br>Equation of tangent to the circle x<sup>2</sup>
+ y<sup>2</sup> = 4 at point
( $$\sqrt 3 $$, 1) is
<br><br>$$\sqrt 3 $$x + y = 4
<br><br>Slope of this tangent (m) = - $$\sqrt 3 $$
<br><br>$$ \therefore $$ Slope of the normal at point ( $$\sqrt 3 $$, 1) is = $${1 \over {\sqrt 3 }}$$
<br><br>$$ \therefore $$ Equation of normal at point ( $$\sqrt 3 $$, 1),
<br><br>y - 1 = $${1 \over {\sqrt 3 }}\left( {x - \sqrt 3 } \right)$$
<br><br>$$ \Rightarrow $$ x - $$\sqrt 3 $$y = 0
<br><br>So normal passes through the center of the axis.
<br><br>Tangent cut the x-axis at point A when y = 0,
<br><br>So the x coordinate of the point A is
<br><br>$$\sqrt 3 $$x + 0 = 4
<br><br>$$ \Rightarrow $$ x = $${4 \over {\sqrt 3 }}$$
<br><br>$$ \therefore $$ Coordinate of A = $$\left( {{4 \over {\sqrt 3 }},0} \right)$$
<br><br>Area of triangle OAB,
<br><br>$$\Delta $$ = $${1 \over 2}\left| {\matrix{
0 & 0 & 1 \cr
{{4 \over {\sqrt 3 }}} & 0 & 1 \cr
{\sqrt 3 } & 1 & 1 \cr
} } \right|$$
<br><br>= $${1 \over 2}\left( {{4 \over {\sqrt 3 }}} \right)$$
<br><br>= $${{2 \over {\sqrt 3 }}}$$ | mcq | jee-main-2019-online-8th-april-evening-slot |
QDZPAFip0LP89oQqFL18hoxe66ijvwpmdcd | maths | circle | tangent-and-normal | If a tangent to the circle x<sup>2 </sup>+ y<sup>2 </sup> = 1 intersects
the coordinate axes at distinct points P and Q,
then the locus of the mid-point of PQ is : | [{"identifier": "A", "content": "x<sup>2</sup> + y<sup>2</sup> \u2013 4x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "B", "content": "x<sup>2</sup> + y<sup>2</sup> - 2xy = 0"}, {"identifier": "C", "content": "x<sup>2</sup> + y<sup>2</sup> \u2013 2x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "D", "content": "x<sup>2</sup> + y<sup>2</sup> - 16x<sup>2</sup>y<sup>2</sup> = 0"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264900/exam_images/vcvrivi2315vojuhifut.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263280/exam_images/bn8cizbo0h6fmxgzbmhr.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267526/exam_images/mnkser8kjex7nql6v8gm.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264099/exam_images/hkzkp7lu4beshmbhun0w.webp" style="max-width: 100%;max-height:250px;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Morning Slot Mathematics - Circle Question 109 English Explanation"></picture>
Let the point of tangency A(cos $$\theta $$, sin $$\theta $$)
<br><br>Equation of tangent at A,
<br><br>xcos $$\theta $$ + ysin $$\theta $$ = 1
<br><br>$$ \therefore $$ P(sec$$\theta $$, 0) and Q(0, cosec$$\theta $$)
<br><br>Let M(h, k) is the mid-point of PQ.
<br><br>$$ \therefore $$ h = $${{\sec\theta + 0} \over 2}$$
<br><br>$$ \Rightarrow $$ 2h = sec $$\theta $$
<br><br>$$ \Rightarrow $$ cos $$\theta $$ = $${1 \over {2h}}$$ .....(1)
<br><br>$$ \therefore $$ k = $${{\cos ec\theta + 0} \over 2}$$
<br><br>$$ \Rightarrow $$ 2k = cosec $$\theta $$
<br><br>$$ \Rightarrow $$ sin $$\theta $$ = $${1 \over {2k}}$$ .....(2)
<br><br>From (1) and (2),
<br><br>sin<sup>2</sup> $$\theta $$ + cos<sup>2</sup> $$\theta $$ = $${\left( {{1 \over {2h}}} \right)^2} + {\left( {{1 \over {2k}}} \right)^2}$$
<br><br>$$ \Rightarrow $$ $${1 \over {4{h^2}}} + {1 \over {4{k^2}}} = 1$$
<br><br>$$ \Rightarrow $$ $${{{h^2} + {k^2}} \over {4{h^2}{k^2}}} = 1$$
<br><br>$$ \Rightarrow $$ $${h^2} + {k^2} = 4{h^2}{k^2}$$
<br><br>$$ \therefore $$ Locus of the midpoint,
<br><br>x<sup>2</sup> + y<sup>2</sup> – 4x<sup>2</sup>y<sup>2</sup> = 0 | mcq | jee-main-2019-online-9th-april-morning-slot |
nEOUZMGLBdTQbD83ak3rsa0w2w9jwxveluz | maths | circle | tangent-and-normal | The line x = y touches a circle at the point (1,1). If the circle also passes through the point (1, – 3), then its
radius is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "2$$\\sqrt 2 $$"}, {"identifier": "D", "content": "3$$\\sqrt 2 $$"}] | ["C"] | null | Equation of circle = (x – 1)<sup>2</sup> + (y –1)<sup>2</sup> + $$\lambda $$(y – x) = 0<br><br>
Which passes through (1, –3)<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266513/exam_images/n8p3thkkkbmcsid2bddg.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267448/exam_images/ukhawmxx5ef8htaeuuz4.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265000/exam_images/zvqsw5bgwwxgllwyy01j.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267320/exam_images/l6db43bddyk4j1ktraq0.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267431/exam_images/mkuhoiu0fusx3spctxdg.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265059/exam_images/zv2cg2j4jsbottmaknpb.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264427/exam_images/xnlrl8ooeye3vypqphdc.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265199/exam_images/h8caxck7nxqpkaubgyjn.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Circle Question 105 English Explanation"></picture><br><br>
So, 0 + 16 + $$\lambda $$(–3 – 1) = 0<br><br>
16 + $$\lambda $$(–4) = 0<br><br>
$$ \therefore $$ $$\lambda $$ = 4<br><br>
Now equation of circle<br><br>
(x – 1)<sup>2</sup> + (y – 1)<sup>2</sup> + 4y – 4x = 0<br><br>
$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> – 6x + 2y + 2 = 0<br><br>
radius = $$\sqrt {9 + 1 - 2} = 2\sqrt 2 $$ | mcq | jee-main-2019-online-10th-april-morning-slot |
gpNR3C5rVO2OLhmiQq7k9k2k5fnh01t | maths | circle | tangent-and-normal | Let the tangents drawn from the origin to the circle, <br/>x<sup>2</sup>
+ y<sup>2</sup>
- 8x - 4y + 16 = 0 touch it at the
points A and B. The (AB)<sup>2</sup>
is equal to : | [{"identifier": "A", "content": "$${{56} \\over 5}$$"}, {"identifier": "B", "content": "$${{32} \\over 5}$$"}, {"identifier": "C", "content": "$${{52} \\over 5}$$"}, {"identifier": "D", "content": "$${{64} \\over 5}$$"}] | ["D"] | null | Equation of chord of contact is
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264374/exam_images/gls1xhkqkxbk85dqfbxj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Mathematics - Circle Question 101 English Explanation">
<br>x.0 + y.0 – 4(x + 0) –2(y + 0) + 16 = 0
<br><br>$$ \Rightarrow $$ 2x + y – 8 = 0
<br><br>$$ \therefore $$ Length of CM = $$\left[ {{{2.4 + 2 - 8} \over {\sqrt {{2^2} + {1^2}} }}} \right]$$ = $${2 \over {\sqrt 5 }}$$ units
<br><br>$$ \therefore $$ AM = BM = $$\sqrt {4 - {4 \over 5}} $$ = $$\sqrt {{{16} \over 5}} $$
<br><br>$$ \therefore $$ Length of chord of contact (AB) = $${8 \over {\sqrt 5 }}$$
<br><br> $$ \therefore $$ AB<sup>2</sup> = $${{16 \times 16} \over {20}}$$ = $${{64} \over 5}$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
sSpQV8FomKrZLJCJKT7k9k2k5hin6dr | maths | circle | tangent-and-normal | If a line, y = mx + c is a tangent to the circle,
(x – 3)<sup>2</sup> + y<sup>2</sup> = 1 and it is perpendicular to a line L<sub>1</sub>, where L<sub>1</sub> is the tangent to the circle, x<sup>2</sup> + y<sup>2</sup> = 1 at the point $$\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$, then : | [{"identifier": "A", "content": "c<sup>2</sup> + 6c + 7 = 0"}, {"identifier": "B", "content": "c<sup>2</sup> - 7c + 6 = 0"}, {"identifier": "C", "content": "c<sup>2</sup> \u2013 6c + 7 = 0"}, {"identifier": "D", "content": "c<sup>2</sup> + 7c + 6 = 0"}] | ["A"] | null | For circle x<sup>2</sup> + y<sup>2</sup>
= 1 tangnet at point P$$\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$ is
<br><br>T = 0
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt 2 }}x + {1 \over {\sqrt 2 }}y - 1 = 0$$
<br><br>$$ \Rightarrow $$ x + y - $$\sqrt 2 $$ = 0
<br><br>Perpendicular to the line is
<br><br>x - y + c = 0
<br><br>This is tangent to the circle (x – 3)<sup>2</sup>
+ y<sup>2</sup>
= 1
<br><br>Also we know, perpendicular distance from center = radius
<br><br>$$ \therefore $$ $$\left| {{{3 - 0 + c} \over {\sqrt 2 }}} \right|$$ = 1
<br><br>$$ \Rightarrow $$ |3 + c| = $${\sqrt 2 }$$
<br><br>$$ \Rightarrow $$ $${\left( {3 + c} \right)^2} = 2$$
<br><br>$$ \Rightarrow $$ 9 + c<sup>2</sup>
+ 6c = 2
<br><br>$$ \Rightarrow $$ c<sup>2</sup>
+ 6c + 7 = 0
| mcq | jee-main-2020-online-8th-january-evening-slot |
WGkJRKvPHKwf6Krn467k9k2k5isco4j | maths | circle | tangent-and-normal | A circle touches the y-axis at the point (0, 4)
and passes through the point (2, 0). Which of
the following lines is not a tangent to this circle? | [{"identifier": "A", "content": "3x \u2013 4y \u2013 24 = 0"}, {"identifier": "B", "content": "4x + 3y \u2013 8 = 0"}, {"identifier": "C", "content": "3x + 4y \u2013 6 = 0"}, {"identifier": "D", "content": "4x \u2013 3y + 17 = 0"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264590/exam_images/u5rwyx3pmckseb9azywf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Morning Slot Mathematics - Circle Question 99 English Explanation">
Equation of family of circle touching y-axis at
<br><br>(0, 4) is given by (x – 0)<sup>2</sup> + (y – 4)<sup>2</sup> + $$\lambda $$x = 0.
<br><br>$$ \because $$ It passes through (2, 0)
<br><br>$$ \Rightarrow $$ $$\lambda $$ = –10.
<br><br>$$ \Rightarrow $$ Required circle is (x – 0)<sup>2</sup> + (y – 4)<sup>2</sup> – 10x = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> – 10x – 8y + 16 = 0
<br><br>$$ \therefore $$ center of circle (5, 4) and radius = 5
<br><br>By checking all options you can see 4x + 3y – 8 = 0 is not a tangent to the circle.
<br><br>As distance of 4x + 3y – 8 = 0 from (5, 4)
<br><br>= $$\left| {{{24} \over 5}} \right|$$ $$ \ne $$ radius | mcq | jee-main-2020-online-9th-january-morning-slot |
uakaYk4qa6deBC0hCU1klrmo3je | maths | circle | tangent-and-normal | If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x $$-$$ 2)<sup>2</sup> + (y $$-$$ 3)<sup>2</sup> = 25 at the point (5, 7) is A, then 24A is equal to _________. | [] | null | 1225 | This question is bonus if we consider poistive x axis.If we consider only x axis for this question then it is right question. | integer | jee-main-2021-online-24th-february-evening-slot |
1zRTbzNUBHDRmyqbal1kmjamamd | maths | circle | tangent-and-normal | The line 2x $$-$$ y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x $$-$$ 2y = 4. Then, the radius of the circle is : | [{"identifier": "A", "content": "5$$\\sqrt 3 $$"}, {"identifier": "B", "content": "4$$\\sqrt 5 $$"}, {"identifier": "C", "content": "3$$\\sqrt 5 $$"}, {"identifier": "D", "content": "5$$\\sqrt 4 $$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263455/exam_images/vmasvdkamze2tghom94b.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Mathematics - Circle Question 85 English Explanation">
<br>$${m_1} \times {m_2} = - 1$$<br><br>$${{{{a - 4} \over 2} - 5} \over {a - 2}} \times 2 = - 1$$<br><br>$${{a - 14} \over {a - 2}} = - 1$$<br><br>$$a - 14 = 2 - a$$<br><br>$$2a = 16$$<br><br>a = 8<br><br>$$ \therefore $$ Centre (8, 2)<br><br>Radius = $$\sqrt {36 + 9} $$<br><br>$$ = \sqrt {45} $$<br><br>$$ = 3\sqrt 5 $$ | mcq | jee-main-2021-online-17th-march-morning-shift |
dUVWV5PxwRaQBO9JX91kmklu5s8 | maths | circle | tangent-and-normal | Let the tangent to the circle x<sup>2</sup> + y<sup>2</sup> = 25 at the point R(3, 4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r<sup>2</sup> is equal to : | [{"identifier": "A", "content": "$${{585} \\over {66}}$$"}, {"identifier": "B", "content": "$${{625} \\over {72}}$$"}, {"identifier": "C", "content": "$${{529} \\over {64}}$$"}, {"identifier": "D", "content": "$${{125} \\over {72}}$$"}] | ["B"] | null | Given equation of circle<br><br>x<sup>2</sup> + y<sup>2</sup> = 25<br><br>$$ \therefore $$ Tangent equation at (3, 4)<br><br>T : 3x + 4y = 25<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267221/exam_images/arrymdsbetsftxxqter6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - Circle Question 81 English Explanation"><br><br>Incentre of $$\Delta$$OPQ.<br><br>$$I = \left( {{{{{25} \over 4} \times {{25} \over 3}} \over {{{25} \over 3} + {{25} \over 4} + {{125} \over {12}}}},{{{{25} \over 3} \times {{25} \over 4}} \over {{{25} \over 3} + {{25} \over 4} + {{125} \over {12}}}}} \right)$$<br><br>$$ \therefore $$ $$I = \left( {{{625} \over {75 + 100 + 125}},{{625} \over {75 + 100 + 125}}} \right) = \left( {{{25} \over {12}},{{25} \over {12}}} \right)$$<br><br>$$ \because $$ Distance from origin to incenter is r.<br><br>$$ \therefore $$ $${r^2} = {\left( {{{25} \over {12}}} \right)^2} + {\left( {{{25} \over {12}}} \right)^2} = {{625} \over {72}}$$<br><br>Therefore, the correct answer is (B). | mcq | jee-main-2021-online-17th-march-evening-shift |
APA6e6vHNrsxIPXcgo1kmkme0oa | maths | circle | tangent-and-normal | Two tangents are drawn from a point P to the circle x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x $$-$$ 4y + 4 = 0, such that the angle between these tangents is $${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$$, where $${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$$ $$\in$$(0, $$\pi$$). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of $$\Delta$$PAB and $$\Delta$$CAB is : | [{"identifier": "A", "content": "3 : 1"}, {"identifier": "B", "content": "9 : 4"}, {"identifier": "C", "content": "2 : 1"}, {"identifier": "D", "content": "11 : 4"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265875/exam_images/hsjo5h1locnmw5jsj3fw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - Circle Question 82 English Explanation">
<br>Let $$\theta$$ = tan<sup>$$-$$1</sup>$$\left( {{{12} \over 5}} \right)$$<br><br>$$ \Rightarrow $$ tan$$\theta$$ = $${{{12} \over 5}}$$<br><br>$$ \Rightarrow $$ $${{2\tan {\theta \over 2}} \over {1 - {{\tan }^2}{\theta \over 2}}} = {{12} \over 5}$$<br><br>$$ \Rightarrow \tan {\theta \over 2} = {2 \over 3}$$<br><br>$$ \Rightarrow \sin {\theta \over 2} = {2 \over {\sqrt 3 }}$$ and $$\cos {\theta \over 2} = {3 \over {\sqrt {13} }}$$<br><br>In $$\Delta$$CAP,<br><br>$$\tan {\theta \over 2} = {1 \over {AP}}$$<br><br>$$ \Rightarrow AP = {3 \over 2}$$<br><br>In $$\Delta$$APM,
<br><br>$$\sin {\theta \over 2} = {{AM} \over {AP}},\cos {\theta \over 2} = {{PM} \over {AP}}$$<br><br>$$ \Rightarrow AM = {3 \over {\sqrt {13} }}$$<br><br>$$ \Rightarrow PM = {9 \over {2\sqrt {13} }}$$<br><br>$$ \therefore $$ $$AB = {6 \over {\sqrt {13} }}$$<br><br>$$ \therefore $$ Area of $$\Delta PAB = {1 \over 2} \times AB \times PM$$<br><br>$$ = {1 \over 2} \times {6 \over {\sqrt {13} }} \times {9 \over {2\sqrt {13} }} = {{27} \over {16}}$$<br><br>Now, $$\phi = 90^\circ - {\theta \over 2}$$<br><br>In $$\Delta$$CAM,<br><br>$$\cos \phi = {{CM} \over {CA}}$$<br><br>$$ \Rightarrow CM = 1.\cos \left( {{\pi \over 2} - {\theta \over 2}} \right)$$<br><br>$$ = 1.\sin {\theta \over 2} = {2 \over {\sqrt {13} }}$$<br><br>$$ \therefore $$ Area of $$\Delta CAB = {1 \over 2} \times AB \times CM$$<br><br>$$ = {1 \over 2} \times {6 \over {\sqrt {13} }} \times {2 \over {\sqrt {13} }} = {6 \over {13}}$$<br><br>$$ \therefore $$ $${{Area\,of\,\Delta PAB} \over {Area\,of\,\Delta CAB}} = {{27/26} \over {6/13}} = {9 \over 4}$$<br><br>Therefore, the correct answer is (2). | mcq | jee-main-2021-online-17th-march-evening-shift |
1ks093wsl | maths | circle | tangent-and-normal | Two tangents are drawn from the point P($$-$$1, 1) to the circle x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x $$-$$ 6y + 6 = 0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$(3\\sqrt 2 + 2)$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$3(\\sqrt 2 - 1)$$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265883/exam_images/j4eb4u3olf4aobpzkvfj.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263397/exam_images/a0quemgl8enbtkiviozm.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264845/exam_images/vn5mptmseenvrcsn84v1.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Circle Question 73 English Explanation"></picture> <br><br>$$\Delta ABD = {1 \over 2} \times 2 \times 4 = 4$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktkesbfz | maths | circle | tangent-and-normal | Let B be the centre of the circle x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x + 4y + 1 = 0. Let the tangents at two points P and Q on the circle intersect at the point A(3, 1). Then 8.$$\left( {{{area\,\Delta APQ} \over {area\,\Delta BPQ}}} \right)$$ is equal to _____________. | [] | null | 18 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxvqvz06/35771b20-6f71-497a-bc46-b4444dec857f/6a6f4960-6af6-11ec-b350-33e20cd86462/file-1kxvqvz07.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxvqvz06/35771b20-6f71-497a-bc46-b4444dec857f/6a6f4960-6af6-11ec-b350-33e20cd86462/file-1kxvqvz07.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2021 (Online) 31st August Evening Shift Mathematics - Circle Question 64 English Explanation"> </p>
<p>Radius = $$\sqrt {1 + 4 - 1} = 2$$</p>
<p>$$AB = \sqrt {{3^2} + {2^2}} = \sqrt {13} $$</p>
<p>In $$\Delta$$ABP</p>
<p>$$A{P^2} = A{B^2} - B{P^2} = 13 - 4 = 9$$</p>
<p>AP = 3</p>
<p>AQ = AP = 3</p>
<p>Let $$\angle$$ABP = $$\theta$$, $$\angle$$BAP = 90$$-$$ $$\theta$$</p>
<p>In $$\Delta$$ABP, tan$$\theta$$ = 3/2</p>
<p>$$\sin \theta = {3 \over {\sqrt {13} }}$$, $$\cos \theta = {2 \over {\sqrt {13} }}$$</p>
<p>In $$\Delta$$ARP,</p>
<p>$$\cos (90 - \theta ) = {{AR} \over {AP}} \Rightarrow AR = 3\sin \theta $$</p>
<p>In $$\Delta$$BRP,</p>
<p>$$\cos \theta = {{BR} \over {BP}}$$</p>
<p>$$ \Rightarrow BR = 2\cos \theta = {{Area\,(\Delta APQ)} \over {Area\,(\Delta BPQ)}} = {{{1 \over 2} \times PQ \times AR} \over {{1 \over 2} \times PQ \times BR}}$$</p>
<p>$$ = {{AR} \over {RB}} = {{3\sin \theta } \over {2\cos \theta }} = {9 \over 4}$$</p>
<p>$$ \Rightarrow 8\left( {{{Area\,(\Delta APQ)} \over {Area\,(\Delta BPQ)}}} \right) = 18$$</p> | integer | jee-main-2021-online-31st-august-evening-shift |
1l545q6zt | maths | circle | tangent-and-normal | <p>Let the tangent to the circle C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> = 2 at the point M($$-$$1, 1) intersect the circle C<sub>2</sub> : (x $$-$$ 3)<sup>2</sup> + (y $$-$$ 2)<sup>2</sup> = 5, at two distinct points A and B. If the tangents to C<sub>2</sub> at the points A and B intersect at N, then the area of the triangle ANB is equal to :</p> | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 6}$$"}, {"identifier": "D", "content": "$${5 \\over 3}$$"}] | ["C"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5dvedrh/aec1f8d8-ed7e-48d5-8b3d-78754e048fbe/ae421dd0-ff83-11ec-8a90-6db465c3d8db/file-1l5dvedri.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5dvedrh/aec1f8d8-ed7e-48d5-8b3d-78754e048fbe/ae421dd0-ff83-11ec-8a90-6db465c3d8db/file-1l5dvedri.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Circle Question 63 English Explanation"></p>
<p>Equation of tangent at point M is</p>
<p>$$T = 0$$</p>
<p>$$ \Rightarrow x{x_1} + y{y_1} = 2$$</p>
<p>$$ \Rightarrow - x + y = 2$$</p>
<p>$$ \Rightarrow y = x + 2$$</p>
<p>Putting this value to equation of circle C<sub>2</sub>,</p>
<p>$${(x - 3)^2} + {(y - 2)^2} = 5$$</p>
<p>$$ \Rightarrow {(x - 3)^2} + {x^2} = 5$$</p>
<p>$$ \Rightarrow {x^2} - 6x + 9 + {x^2} = 5$$</p>
<p>$$ \Rightarrow 2{x^2} - 6x + 4 = 0$$</p>
<p>$$ \Rightarrow {x^2} - 3x + 2 = 0$$</p>
<p>$$ \Rightarrow (x - 2)(x - 1) = 0$$</p>
<p>$$ \Rightarrow x = 1,2$$</p>
<p>when $$x = 1,$$ $$y = 3$$</p>
<p>and when $$x = 2,$$ $$y = 4$$</p>
<p>$$\therefore$$ Point $$A(1,3)$$ and $$B(2,4)$$</p>
<p>Now, equation of tangent at $$A(1,3)$$ on circle $${(x - 3)^2} + {(y - 2)^2} = 5$$ or $${x^2} + {y^2} - 6x - 4y + 8 = 0$$ is</p>
<p>$$T = 0$$</p>
<p>$$x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + C = 0$$</p>
<p>$$ \Rightarrow x + 3y - 3(x + 1) - 2(y + 3) + 8 = 0$$</p>
<p>$$ \Rightarrow x + 3y - 3x - 3 - 2y - 6 + 8 = 0$$</p>
<p>$$ \Rightarrow - 2x + y - 1 = 0$$</p>
<p>$$ \Rightarrow 2x - y + 1 = 0$$ ....... (1)</p>
<p>Similarly tangent at $$B(2,4)$$ is</p>
<p>$$2x + 4y - 3(x + 2) - 2(y + 4) + 8 = 0$$</p>
<p>$$ \Rightarrow 2x + 4y - 3x - 6 - 2y - 8 + 8 = 0$$</p>
<p>$$ \Rightarrow - x + 2y - 6 = 0$$</p>
<p>$$ \Rightarrow x - 2y + 6 = 0$$ ...... (2)</p>
<p>Solving equation (1) and (2), we get</p>
<p>$$x - 2(2x + 1) + 6 = 0$$</p>
<p>$$ \Rightarrow x - 4x - 2 + 6 = 0$$</p>
<p>$$ \Rightarrow - 3x + 4 = 0$$</p>
<p>$$ \Rightarrow x = {4 \over 3}$$</p>
<p>$$\therefore$$ $$y = 2 \times {4 \over 3} + 1 = {{11} \over 3}$$</p>
<p>$$\therefore$$ Point $$N = \left( {{4 \over 3},{{11} \over 3}} \right)$$</p>
<p>Now area of the triangle ANB</p>
<p>$$ = {1 \over 2}\left| {\matrix{
{{x_1}} & {{y_1}} & 1 \cr
{{x_2}} & {{y_2}} & 1 \cr
{{x_3}} & {{y_3}} & 1 \cr
} } \right|$$</p>
<p>$$ = {1 \over 2}\left| {\matrix{
1 & 3 & 1 \cr
2 & 4 & 1 \cr
{{4 \over 3}} & {{{11} \over 3}} & 1 \cr
} } \right|$$</p>
<p>$$ = {1 \over 2}\left[ {1\left( {4 - {{11} \over 3}} \right) - 3\left( {2 - {4 \over 3}} \right) + 1\left( {{{22} \over 3} - {{16} \over 3}} \right)} \right]$$</p>
<p>$$ = {1 \over 2}\left[ {{1 \over 3} - {6 \over 3} + {6 \over 3}} \right]$$</p>
<p>$$ = {1 \over 6}$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l566usat | maths | circle | tangent-and-normal | <p>If the tangents drawn at the points $$O(0,0)$$ and $$P\left( {1 + \sqrt 5 ,2} \right)$$ on the circle $${x^2} + {y^2} - 2x - 4y = 0$$ intersect at the point Q, then the area of the triangle OPQ is equal to :</p> | [{"identifier": "A", "content": "$${{3 + \\sqrt 5 } \\over 2}$$"}, {"identifier": "B", "content": "$${{4 + 2\\sqrt 5 } \\over 2}$$"}, {"identifier": "C", "content": "$${{5 + 3\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$${{7 + 3\\sqrt 5 } \\over 2}$$"}] | ["C"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5obw2pa/6ae93711-16f3-4dfc-a416-9792742da63d/02d6cce0-0544-11ed-987f-3938cfc0f7f1/file-1l5obw2pb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5obw2pa/6ae93711-16f3-4dfc-a416-9792742da63d/02d6cce0-0544-11ed-987f-3938cfc0f7f1/file-1l5obw2pb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Circle Question 59 English Explanation"> </p>
<p>$$\tan 2\theta = 2 \Rightarrow {{2\tan \theta } \over {1 - {{\tan }^2}\theta }} = 2$$</p>
<p>$$\tan \theta = {{\sqrt 5 - 1} \over 2}$$ (as $$\theta$$ is acute)</p>
<p>Area $$ = {1 \over 2}{L^2}\sin 2\theta = {1 \over 2}.\,{5 \over {{{\tan }^2}\theta }}.\,2\sin \theta \cos \theta $$</p>
<p>$$ = {{5\sin \theta \cos \theta } \over {{{\sin }^2}\theta }}.\,{\cos ^2}\theta $$</p>
<p>$$ = 5\cot \theta .\,{\cos ^2}\theta $$</p>
<p>$$ = 5.\,{2 \over {\sqrt 5 - 1}}.\,{1 \over {1 + {{\left( {{{\sqrt 5 - 1} \over 2}} \right)}^2}}}$$</p>
<p>$$ = {{10} \over {\sqrt 5 - 1}}.\,{4 \over {4 + 6 - 2\sqrt 5 }}$$</p>
<p>$$ = {{40} \over {2\sqrt 5 {{(\sqrt 5 - 1)}^2}}} = {{4\sqrt 5 } \over {6 - 2\sqrt 5 }}$$</p>
<p>$$ = {{4\sqrt 5 (6 + 2\sqrt 5 )} \over {16}}$$</p>
<p>$$ = {{\sqrt 5 (3 + \sqrt 5 )} \over 2}$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l567wcln | maths | circle | tangent-and-normal | <p>Let the lines $$y + 2x = \sqrt {11} + 7\sqrt 7 $$ and $$2y + x = 2\sqrt {11} + 6\sqrt 7 $$ be normal to a circle $$C:{(x - h)^2} + {(y - k)^2} = {r^2}$$. If the line $$\sqrt {11} y - 3x = {{5\sqrt {77} } \over 3} + 11$$ is tangent to the circle C, then the value of $${(5h - 8k)^2} + 5{r^2}$$ is equal to __________.</p> | [] | null | 816 | <p>$${L_1}:y + 2x = \sqrt {11} + 7\sqrt 7 $$</p>
<p>$${L_2}:2y + x = 2\sqrt {11} + 6\sqrt 7 $$</p>
<p>Point of intersection of these two lines is centre of circle i.e. $$\left( {{8 \over 3}\sqrt 7 ,\sqrt {11} + {5 \over 3}\sqrt 7 } \right)$$</p>
<p>$${ \bot ^r}$$ from centre to line $$3x - \sqrt {11} y + \left( {{{5\sqrt {77} } \over 3} + 11} \right) = 0$$ is radius of circle</p>
<p>$$ \Rightarrow r = \left| {{{8\sqrt 7 - 11 - {5 \over 3}\sqrt {77} + {{5\sqrt {77} } \over 3} + 11} \over {\sqrt {20} }}} \right|$$</p>
<p>$$ = \left| {\root 4 \of {{7 \over 5}} } \right| = \root 4 \of {{7 \over 5}} $$ units</p>
<p>So $${(5h - 8K)^2} + 5{r^2}$$</p>
<p>$$ = {\left( {{{40} \over 3}\sqrt 7 - 8\sqrt {11} - {{40} \over 3}\sqrt 7 } \right)^2} + 5.\,16.\,{7 \over 5}$$</p>
<p>$$ = 64 \times 11 + 112 = 816$$.</p> | integer | jee-main-2022-online-28th-june-morning-shift |
1l56u8rwg | maths | circle | tangent-and-normal | <p>Let a circle C of radius 5 lie below the x-axis. The line L<sub>1</sub> : 4x + 3y + 2 = 0 passes through the centre P of the circle C and intersects the line L<sub>2</sub> = 3x $$-$$ 4y $$-$$ 11 = 0 at Q. The line L<sub>2</sub> touches C at the point Q. Then the distance of P from the line 5x $$-$$ 12y + 51 = 0 is ______________.</p> | [] | null | 11 | <p>$${L_1}:4x + 3y + 2 = 0$$</p>
<p>$${L_2}:3x - 4y - 11 = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7n1g0/160bc86a-f674-4729-9933-246027606159/2bb7d100-05c0-11ed-8617-d71e6444d1a0/file-1l5p7n1g1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5p7n1g0/160bc86a-f674-4729-9933-246027606159/2bb7d100-05c0-11ed-8617-d71e6444d1a0/file-1l5p7n1g1.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Evening Shift Mathematics - Circle Question 58 English Explanation"></p>
<p>Since circle C touches the line L<sub>2</sub> at Q intersection point Q of L<sub>1</sub> and L<sub>2</sub>, is (1, $$-$$2)</p>
<p>$$\because$$ P lies of L<sub>1</sub></p>
<p>$$\therefore$$ $$P\left( {x, - {1 \over 3}(2 + 4x)} \right)$$</p>
<p>Now, $$PQ = 5 \Rightarrow {(x - 1)^2} + {\left( {{{4x + 2} \over 3} - 2} \right)^2} = 25$$</p>
<p>$$ \Rightarrow {(x - 1)^2}\left[ {1 + {{16} \over 9}} \right] = 25$$</p>
<p>$$ \Rightarrow {(x - 1)^2} = 9$$</p>
<p>$$ \Rightarrow x = 4,\, - 2$$</p>
<p>$$\because$$ Circle lies below the x-axis</p>
<p>$$\therefore$$ y = $$-$$6</p>
<p>P(4, $$-$$6)</p>
<p>Now distance of P from 5x $$-$$ 12y + 51 = 0</p>
<p>$$ = \left| {{{20 + 72 + 51} \over {13}}} \right| = {{143} \over {13}} = 11$$</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l59kc4xu | maths | circle | tangent-and-normal | <p>A circle touches both the y-axis and the line x + y = 0. Then the locus of its center is :</p> | [{"identifier": "A", "content": "$$y = \\sqrt 2 x$$"}, {"identifier": "B", "content": "$$x = \\sqrt 2 y$$"}, {"identifier": "C", "content": "$${y^2} - {x^2} = 2xy$$"}, {"identifier": "D", "content": "$${x^2} - {y^2} = 2xy$$"}] | ["D"] | null | <p>Let the centre be (h, k)</p>
<p>So, $$\left| h \right| = \left| {{{h + k} \over {\sqrt 2 }}} \right|$$</p>
<p>$$ \Rightarrow 2{h^2} = {h^2} + {k^2} + 2hk$$</p>
<p>Locus will be $${x^2} - {y^2} = 2xy$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5ahluy5 | maths | circle | tangent-and-normal | Let a circle C touch the lines $${L_1}:4x - 3y + {K_1} = 0$$ and $${L_2} = 4x - 3y + {K_2} = 0$$, $${K_1},{K_2} \in R$$. If a line passing through the centre of the circle C intersects L<sub>1</sub> at $$( - 1,2)$$ and L<sub>2</sub> at $$(3, - 6)$$, then the equation of the circle C is : | [{"identifier": "A", "content": "$${(x - 1)^2} + {(y - 2)^2} = 4$$"}, {"identifier": "B", "content": "$${(x + 1)^2} + {(y - 2)^2} = 4$$"}, {"identifier": "C", "content": "$${(x - 1)^2} + {(y + 2)^2} = 16$$"}, {"identifier": "D", "content": "$${(x - 1)^2} + {(y - 2)^2} = 16$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5urhu40/938cdb8a-9891-48ef-8a0d-4f13f0b7d905/bcd66b00-08cd-11ed-a060-0972f3a5152e/file-1l5urhu41.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5urhu40/938cdb8a-9891-48ef-8a0d-4f13f0b7d905/bcd66b00-08cd-11ed-a060-0972f3a5152e/file-1l5urhu41.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th June Morning Shift Mathematics - Circle Question 52 English Explanation"></p>
<p>Co-ordinate of centre $$C \equiv \left( {{{3 + ( - 1)} \over 2},{{ - 6 - 2} \over 2}} \right) \equiv (1, - 2)$$</p>
<p>L<sub>1</sub> is passing through A</p>
<p>$$ \Rightarrow - 4 - 6 + {K_1} = 0$$</p>
<p>$$ \Rightarrow {K_1} = 10$$</p>
<p>L<sub>2</sub> is passing through B</p>
<p>$$ \Rightarrow 12 + 18 + {K_2} = 0$$</p>
<p>$$ \Rightarrow {K_2} = - 30$$</p>
<p>Equation of $${L_1}:4x - 3y + 10 = 0$$</p>
<p>Equation of $${L_2} = 4x - 3y - 30 = 0$$</p>
<p>Diameter of circle $$ = \left| {{{10 + 30} \over {\sqrt {{4^2} + {{( - 3)}^2}} }}} \right| = 8$$</p>
<p>$$\Rightarrow$$ Radius = 4</p>
<p>Equation of circle $${(x - 1)^2} + {(y + 2)^2} = 16$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6kk85jt | maths | circle | tangent-and-normal | <p>A circle $$C_{1}$$ passes through the origin $$\mathrm{O}$$ and has diameter 4 on the positive $$x$$-axis. The line $$y=2 x$$ gives a chord $$\mathrm{OA}$$ of circle $$\mathrm{C}_{1}$$. Let $$\mathrm{C}_{2}$$ be the circle with $$\mathrm{OA}$$ as a diameter. If the tangent to $$\mathrm{C}_{2}$$ at the point $$\mathrm{A}$$ meets the $$x$$-axis at $$\mathrm{P}$$ and $$y$$-axis at $$\mathrm{Q}$$, then $$\mathrm{QA}: \mathrm{AP}$$ is equal to :</p> | [{"identifier": "A", "content": "1 : 4"}, {"identifier": "B", "content": "1 : 5"}, {"identifier": "C", "content": "2 : 5"}, {"identifier": "D", "content": "1 : 3"}] | ["A"] | null | <p>Equation of C<sub>1</sub></p>
<p>$${x^2} + {y^2} - 4x = 0$$</p>
<p>Intersection with</p>
<p>$$y = 2x$$</p>
<p>$${x^2} + 4{x^2} - 4x = 0$$</p>
<p>$$5{x^2} - 4x = 0$$</p>
<p>$$ \Rightarrow x = 0,{4 \over 5}$$</p>
<p>$$y = 0,{8 \over 5}$$</p>
<p>$$A:\left( {{4 \over 5},{8 \over 5}} \right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qace1v/32850382-0912-489c-a1fa-6994a53f61a4/f659d730-2def-11ed-a744-1fb8f3709cfa/file-1l7qace1w.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qace1v/32850382-0912-489c-a1fa-6994a53f61a4/f659d730-2def-11ed-a744-1fb8f3709cfa/file-1l7qace1w.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Evening Shift Mathematics - Circle Question 46 English Explanation"></p>
<p>Tangent of C<sub>2</sub> at $$A\left( {{4 \over 5},{8 \over 5}} \right)$$</p>
<p>$$x + 2y = 4 \Rightarrow P:(4,0),\,Q:(0,2)$$</p>
<p>$$QA:AP = 1:4$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nndhie | maths | circle | tangent-and-normal | <p>Let the tangents at two points $$\mathrm{A}$$ and $$\mathrm{B}$$ on the circle $$x^{2}+\mathrm{y}^{2}-4 x+3=0$$ meet at origin $$\mathrm{O}(0,0)$$. Then the area of the triangle $$\mathrm{OAB}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{3 \\sqrt{3}}{2}$$"}, {"identifier": "B", "content": "$$\\frac{3 \\sqrt{3}}{4}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2 \\sqrt{3}}$$"}, {"identifier": "D", "content": "$$\\frac{3}{4 \\sqrt{3}}$$"}] | ["B"] | null | <p>$${x^2} + {y^2} - 4x + 3 = 0$$</p>
<p>$$ \Rightarrow {(x - 2)^2} + {y^2} = 1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rrgzao/8d049878-91a0-486a-bf74-ffc8968e3205/b9d7eb00-2ebf-11ed-b92e-01f1dabc9173/file-1l7rrgzap.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rrgzao/8d049878-91a0-486a-bf74-ffc8968e3205/b9d7eb00-2ebf-11ed-b92e-01f1dabc9173/file-1l7rrgzap.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Evening Shift Mathematics - Circle Question 44 English Explanation"></p>
<p>$$AO = \sqrt {{{(OC)}^2} - {{(AC)}^2}} $$</p>
<p>$$ = \sqrt {4 - 1} = \sqrt 3 $$</p>
<p>$$\sin \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 6}$$</p>
<p>Also, $$AO = BO$$</p>
<p>Area of $$\Delta OAB = {1 \over 2}\,.\,OA\,.\,OB\sin 60^\circ $$</p>
<p>$$ = {1 \over 2} \times \sqrt 3 \,.\,\sqrt 3 \,.\,{{\sqrt 3 } \over 2} = {{3\sqrt 3 } \over 4}$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldr7omt6 | maths | circle | tangent-and-normal | <p>Let $$y=x+2,4y=3x+6$$ and $$3y=4x+1$$ be three tangent lines to the circle $$(x-h)^2+(y-k)^2=r^2$$. Then $$h+k$$ is equal to :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "5 (1 + $$\\sqrt2$$)"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "5$$\\sqrt2$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leq10yl8/23860a94-17c3-4a42-9ba1-82deb3b85bcd/09ff05c0-b861-11ed-8195-4f3c56fa1eb5/file-1leq10yl9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leq10yl8/23860a94-17c3-4a42-9ba1-82deb3b85bcd/09ff05c0-b861-11ed-8195-4f3c56fa1eb5/file-1leq10yl9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Morning Shift Mathematics - Circle Question 36 English Explanation"></p>
<p>$$(h,k) = \left( {{{5.5 + 5( - 2) + 14\sqrt 2 } \over {10 + 7\sqrt 2 }},{{35 + 21\sqrt 2 } \over {10 + 7\sqrt 2 }}} )\right.$$</p>
<p>$$h + k = {{50 + 35\sqrt 2 } \over {10 + 7\sqrt 2 }} = 5$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldsfvrxm | maths | circle | tangent-and-normal | <p>A circle with centre (2, 3) and radius 4 intersects the line $$x+y=3$$ at the points P and Q. If the tangents at P and Q intersect at the point $$S(\alpha,\beta)$$, then $$4\alpha-7\beta$$ is equal to ___________.</p> | [] | null | 11 | <p>The line $$x + y = 3$$ ..... (i)</p>
<p>is polar of $$S(\alpha ,\beta )$$ w.r.t. circle</p>
<p>$${(x - 2)^2} + {(y - 3)^2} = 16$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} - 4x - 6y - 3 = 0$$</p>
<p>Equation of polar is</p>
<p>$$\alpha x + \beta y - 2(x + \alpha ) - 3(4 + \beta ) - 3 = 0$$</p>
<p>$$(\alpha - 2)x + (\alpha - 3)y - (2\alpha + 3\beta + 3) = 0$$ ..... (ii)</p>
<p>(i) and (ii) represent the same.</p>
<p>$$\therefore$$ $${{\alpha - 2} \over 1} = {{\beta - 3} \over 1} = {{2\alpha + 3\beta + 3} \over 3}$$</p>
<p>$$\alpha - \beta + 1 = 0$$</p>
<p>$$\alpha - 3\beta - 9 = 0$$</p>
<p>$$ \Rightarrow \alpha = - 6,\beta = - 5$$</p>
<p>$$4\alpha - 7\beta = 11$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldu69w0u | maths | circle | tangent-and-normal | <p>Points P($$-$$3, 2), Q(9, 10) and R($$\alpha,4$$) lie on a circle C and PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line $$2x-ky=1$$, then k is equal to ____________.</p> | [] | null | 3 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef6afmm/c04cf9db-da6e-4771-935e-0ffcde0666c2/fff172f0-b268-11ed-9d4d-b96eca78f2e5/file-1lef6afmn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef6afmm/c04cf9db-da6e-4771-935e-0ffcde0666c2/fff172f0-b268-11ed-9d4d-b96eca78f2e5/file-1lef6afmn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Evening Shift Mathematics - Circle Question 33 English Explanation"><br>
Now, $\frac{10-2}{9+3} \times \frac{10-4}{9-\alpha}=-1$
<br><br>
$\Rightarrow \frac{8}{12} \cdot 6=\alpha-9 \Rightarrow \alpha=13$
<br><br>
$\therefore 0=(5,3)$ So,
$$
m_{O Q}=\frac{7}{4}
$$
$$
\begin{aligned}
& m_{O R}=\frac{1}{8}
\end{aligned}
$$
<br><br>
$\therefore Q: y-10=\frac{-4}{7}(x-9)$
<br><br>
$\Rightarrow 4 x+7 y=106$
<br><br>
Tangent at $R: y-4=-8(x-13)$
<br><br>
$$
8 x+y=108\quad...(ii)
$$
<br><br>
By (i) and (ii) $S \equiv\left(\frac{25}{2}, 8\right)$, satisfies with the line<br><br> $\therefore K=3$ | integer | jee-main-2023-online-25th-january-evening-shift |
1lgow4qau | maths | circle | tangent-and-normal | <p>Let the centre of a circle C be $$(\alpha, \beta)$$ and its radius $$r < 8$$. Let $$3 x+4 y=24$$ and $$3 x-4 y=32$$ be two tangents and $$4 x+3 y=1$$ be a normal to C. Then $$(\alpha-\beta+r)$$ is equal to :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh1kl8kh/309857da-6cb3-42cd-b0c0-e753c4a2dc87/993e3110-e652-11ed-b540-cb85a096fb04/file-1lh1kl8ki.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh1kl8kh/309857da-6cb3-42cd-b0c0-e753c4a2dc87/993e3110-e652-11ed-b540-cb85a096fb04/file-1lh1kl8ki.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Evening Shift Mathematics - Circle Question 29 English Explanation">
<br><br>First find point A by solving $4 x+3 y=1$ and $3 x-4 y=32$
<br><br>After solving, point $\mathrm{A}$ is $(4,-5)$
<br><br>Centre $(\alpha, \beta)$ lie on $4 x+3 y=1$
<br><br>$$
4 \alpha+3 \beta=1 \Rightarrow \beta=\frac{1-4 \alpha}{3}
$$
<br><br>Now distance from centre to line $3 x-4 y-32=$ 0 and $3 x+4 y-24=0$ are equal.
<br><br>$$
\left|\frac{3 \alpha-4\left(\frac{1-4 \alpha}{3}\right)-32}{5}\right|=\left|\frac{3 \alpha+4\left(\frac{1-4 \alpha}{3}\right)-24}{5}\right|
$$
<br><br>After solving $\alpha=1$ and $\alpha=\frac{28}{3}$
<br><br>For $\alpha=1$, centre $(1,-1) \Rightarrow$ radius $=5$
<br><br>For $\alpha=\frac{28}{3}$, centre $\left(\frac{28}{3}, \frac{-109}{2}\right)$
<br><br>$$
\Rightarrow \text { radius } \approx 49.78 \text { (rejected })
$$
<br><br>Hence, $\alpha=1, \beta=-1, \mathrm{r}=5$
<br><br>$$
\alpha-\beta+\mathrm{r}=7
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgykzejb | maths | circle | tangent-and-normal | <p>Let O be the origin and OP and OQ be the tangents to the circle $$x^2+y^2-6x+4y+8=0$$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $$\left( {\alpha ,{1 \over 2}} \right)$$, then a value of $$\alpha$$ is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$-\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$\\frac{5}{2}$$"}, {"identifier": "D", "content": "$$\\frac{3}{2}$$"}] | ["C"] | null | Centre $(3,-2)$
<br/><br/>Equation of circumcircle is
<br/><br/>$$
\begin{aligned}
& x(x-3)+y(y+2)=0 \\\\
& \Rightarrow x^2-3 x+y^2+2 y=0
\end{aligned}
$$
<br/><br/>Since $\left(\alpha, \frac{1}{2}\right)$ is on the circle
<br/><br/>$$
\begin{aligned}
& \text { So } \alpha^2-3 \alpha+\frac{1}{4}+1=0 \\\\
& \Rightarrow 4 \alpha^2-12 \alpha+5=0 \\\\
& \Rightarrow \alpha=\frac{12 \pm \sqrt{144-80}}{8} \\\\
& =\frac{12 \pm \sqrt{64}}{8}=\frac{12 \pm 8}{8} \\\\
& \alpha=\frac{20}{8}, \frac{4}{8}=\frac{5}{2}, \frac{1}{2}
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh2y790j | maths | circle | tangent-and-normal | <p>If the tangents at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ on the circle $$x^{2}+y^{2}-2 x+y=5$$ meet at the point $$R\left(\frac{9}{4}, 2\right)$$, then the area of the triangle $$\mathrm{PQR}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{13}{8}$$"}, {"identifier": "B", "content": "$$\\frac{5}{8}$$"}, {"identifier": "C", "content": "$$\\frac{5}{4}$$"}, {"identifier": "D", "content": "$$\\frac{13}{4}$$"}] | ["B"] | null | Equation of circle $x^2+y^2-2 x+y-5=0$
<br><br>On comparing with $x^2+y^2+2 g x+2 f y+c=0$
<br><br>$$
\begin{gathered}
2 g=-2,2 f=1, c=-5 \\\\
g=-1, f=\frac{1}{2}, c=-5
\end{gathered}
$$
<br><br>$\therefore$ Radius of the circle
<br><br>$$
r=\sqrt{(-1)^2+\left(\frac{1}{2}\right)^2+5}=\frac{5}{2}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1loconojy/39d791f1-201c-4315-8547-452eaca7958f/a485a2e0-7704-11ee-b262-010b491aebf0/file-6y3zli1loconojz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1loconojy/39d791f1-201c-4315-8547-452eaca7958f/a485a2e0-7704-11ee-b262-010b491aebf0/file-6y3zli1loconojz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Evening Shift Mathematics - Circle Question 21 English Explanation">
<br><br>Length of tangent, $P R=\sqrt{S_1}$
<br><br>$$
l=\sqrt{\left(\frac{9}{4}\right)^2+(2)^2-2\left(\frac{9}{4}\right)+2-5}=\frac{5}{4}
$$
<br><br>$\begin{aligned} & \therefore \text { Area of } \triangle P Q R=\frac{r l^3}{r^2+l^2} \\\\ &= \frac{\left(\frac{5}{2}\right)\left(\frac{5}{4}\right)^3}{\left(\frac{5}{2}\right)^2+\left(\frac{5}{4}\right)^2}=\frac{\frac{5}{2} \times \frac{125}{64}}{\frac{25}{4}+\frac{25}{16}} \\\\ &=\frac{\frac{5}{2} \times \frac{125}{64} \times 16}{125}=\frac{5}{8}\end{aligned}$ | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lsconj9i | maths | circle | tangent-and-normal | <p>Consider a circle $$(x-\alpha)^2+(y-\beta)^2=50$$, where $$\alpha, \beta>0$$. If the circle touches the line $$y+x=0$$ at the point $$P$$, whose distance from the origin is $$4 \sqrt{2}$$, then $$(\alpha+\beta)^2$$ is equal to __________.</p> | [] | null | 100 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1x38le/83e15979-b7dd-4565-bdb1-2ae72cdc0d26/13ec8220-d416-11ee-b9d5-0585032231f0/file-1lt1x38lf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1x38le/83e15979-b7dd-4565-bdb1-2ae72cdc0d26/13ec8220-d416-11ee-b9d5-0585032231f0/file-1lt1x38lf.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Circle Question 17 English Explanation"></p>
<p>$$\begin{aligned}
& S:(x-\alpha)^2+(y-\beta)^2=50 \\
& C P=r \\
& \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} \\
& \Rightarrow(\alpha+\beta)^2=100
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift |
lv7v3k7d | maths | circle | tangent-and-normal | <p>Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point $$(3,2)$$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point $$(5,5)$$ is :</p> | [{"identifier": "A", "content": "4$$\\sqrt2$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "2$$\\sqrt2$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgjwq15/c6544ab2-5387-4f8a-8398-9aeaf56b974c/efedcb90-1786-11ef-9978-292aa9baaa14/file-1lwgjwq16.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgjwq15/c6544ab2-5387-4f8a-8398-9aeaf56b974c/efedcb90-1786-11ef-9978-292aa9baaa14/file-1lwgjwq16.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Circle Question 5 English Explanation"></p>
<p>Shortest distance of circle $$C$$ form $$(5,5)$$</p>
<p>$$\begin{aligned}
& =\sqrt{9+16}-1 \\
& =5-1=4
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
xG2EPTYG5Tf9yrpg | maths | complex-numbers | algebra-of-complex-numbers | If $$\left| {z - 4} \right| < \left| {z - 2} \right|$$, its solution is given by : | [{"identifier": "A", "content": "$${\\mathop{\\rm Re}\\nolimits} (z) > 0$$ "}, {"identifier": "B", "content": "$${\\mathop{\\rm Re}\\nolimits} (z) < 0$$"}, {"identifier": "C", "content": "$${\\mathop{\\rm Re}\\nolimits} (z) > 3$$"}, {"identifier": "D", "content": "$${\\mathop{\\rm Re}\\nolimits} (z) > 2$$"}] | ["C"] | null | Given $$\left| {z - 4} \right| < \left| {z - 2} \right|$$
<br><br>Let $$\,\,\,z = x + iy$$
<br><br>$$ \Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|$$
<br><br>$$ \Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}$$
<br><br>$$ \Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4$$
<br><br>$$ \Rightarrow 12 < 4x$$
<br><br>$$ \Rightarrow x > 3$$
<br><br>$$ \Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3$$ | mcq | aieee-2002 |
dBRnY1IywoUv56wj | maths | complex-numbers | algebra-of-complex-numbers | If $${z^2} + z + 1 = 0$$, where z is complex number, then value of $${\left( {z + {1 \over z}} \right)^2} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^2} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^2} + .......... + {\left( {{z^6} + {1 \over {{z^6}}}} \right)^2}$$ is : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "12"}] | ["D"] | null | $${z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,$$ or $$\,\,\,{\omega ^2}$$
<br><br>So, $$z + {1 \over z} = \omega + {\omega ^2} = - 1$$
<br><br>$${z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,$$
<br><br>$${z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2$$
<br><br>$${z^4} + {1 \over {{z^4}}} = - 1,$$
<br><br>$${z^5} + {1 \over {{z^5}}} = - 1$$
<br><br>and $$\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2$$
<br><br>$$\therefore$$ The given sum $$ = 1 + 1 + 4 + 1 + 1 + 4 = 12$$ | mcq | aieee-2006 |
LhUeHRBHKbPCLuNY | maths | complex-numbers | algebra-of-complex-numbers | If $$\,\left| {z - {4 \over z}} \right| = 2,$$ then the maximum value of $$\,\left| z \right|$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt 5 + 1$$ "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "$$2 + \\sqrt 2 $$ "}, {"identifier": "D", "content": "$$\\sqrt 3 + 1$$ "}] | ["A"] | null | Given that $$\left| {z - {4 \over z}} \right| = 2$$
<br><br>Now $$\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}$$
<br><br>$$ \Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}$$
<br><br>$$ \Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0$$
<br><br>$$ \Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0$$
<br><br>$$\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0$$
<br><br>$$ \Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)$$
<br><br>$$ \Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1$$ | mcq | aieee-2009 |
cvg6bc045DsjHMd2gYK21 | maths | complex-numbers | algebra-of-complex-numbers | The point represented by 2 + <i>i</i> in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there $$2\sqrt 2 $$ units in the south-westwardsdirection. Then its new position in the Argand plane is at the point represented by : | [{"identifier": "A", "content": "2 + 2i"}, {"identifier": "B", "content": "1 + i"}, {"identifier": "C", "content": "$$-$$1 $$-$$ i"}, {"identifier": "D", "content": "$$-$$2 $$-$$2i"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267287/exam_images/vpkjlli9hudoskpjp7ap.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Mathematics - Complex Numbers Question 136 English Explanation">
<br><br>Here,
<br><br>z $$-$$ (3 + 3i) = $$2\sqrt 2 $$ (cos($$-$$135<sup>o</sup>) + i sin ($$-$$ 135<sup>o</sup>))
<br><br>= $$2\sqrt 2 $$ ($$-$$ $${1 \over {\sqrt 2 }}$$ $$-$$$${i \over {\sqrt 2 }}$$)
<br><br>= $$-$$ 2 $$-$$ 2i
<br><br>$$ \Rightarrow $$ z = 3 + 3 i $$-$$ 2 $$-$$ 2 i = 1 + i
<br><br><u>Note </u>:
<br><br>Polar form of a complex number :
<br><br>z = r (cos$$\theta $$ + i sin$$\theta $$)
<br><br>Here r = modulus of z and $$\theta $$ argument of z. | mcq | jee-main-2016-online-9th-april-morning-slot |
pqnhDHvUErLFbUUw | maths | complex-numbers | algebra-of-complex-numbers | A value of $$\theta \,$$ for which $${{2 + 3i\sin \theta \,} \over {1 - 2i\,\,\sin \,\theta \,}}$$ is purely imaginary, is : | [{"identifier": "A", "content": "$${\\sin ^{ - 1}}\\left( {{{\\sqrt 3 } \\over 4}} \\right)$$ "}, {"identifier": "B", "content": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)\\,$$ "}, {"identifier": "C", "content": "$${\\pi \\over 3}$$ "}, {"identifier": "D", "content": "$${\\pi \\over 6}$$ "}] | ["B"] | null | Rationalizing the given expression
<br><br>$${{\left( {2 + 3i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
<br><br>For the given expression to be purely imaginary, real part of the above expression should be equal to zero.
<br><br>$$ \Rightarrow {{2 - 6{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0$$
<br><br>$$ \Rightarrow {\sin ^2}\theta = {1 \over 3}$$
<br><br>$$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 3 }}$$ | mcq | jee-main-2016-offline |
9MmkhQB3ccxn864L | maths | complex-numbers | algebra-of-complex-numbers | Let $$\omega $$ be a complex number such that 2$$\omega $$ + 1 = z where z = $$\sqrt {-3} $$. If
<br/><br/>$$\left| {\matrix{
1 & 1 & 1 \cr
1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr
1 & {{\omega ^2}} & {{\omega ^7}} \cr
} } \right| = 3k$$,
<br/><br/>then k is equal to : | [{"identifier": "A", "content": "z"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "-z"}] | ["D"] | null | Given 2$$\omega $$ + 1 = z;
<br><br>z = $$\sqrt 3 i$$
<br><br>$$ \Rightarrow $$ $$\omega = {{\sqrt 3 i - 1} \over 2}$$
<br><br>$$ \Rightarrow $$ As $$\omega $$ is complex cube root of unity.
<br><br>$${\omega ^3} = 1$$
<br><br>$$1 + \omega + {\omega ^2} = 0$$
<br><br>$$\left| {\matrix{
1 & 1 & 1 \cr
1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr
1 & {{\omega ^2}} & {{\omega ^7}} \cr
} } \right| = 3k$$
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
1 & 1 & 1 \cr
1 & \omega & {{\omega ^2}} \cr
1 & {{\omega ^2}} & \omega \cr
} } \right| = 3k$$
<br><br>Applying R<sub>1</sub> $$ \to $$ R<sub>1</sub> + R<sub>2</sub> + R<sub>3</sub>
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
3 & 0 & 0 \cr
1 & \omega & {{\omega ^2}} \cr
1 & {{\omega ^2}} & \omega \cr
} } \right| = 3k$$
<br><br>$$ \Rightarrow $$ $$3\left( {{\omega ^2} - {\omega ^4}} \right) = 3k$$
<br><br>$$ \Rightarrow $$ $$\left( {{\omega ^2} - \omega } \right) = k$$
<br><br>$$ \therefore $$ $$k = \left( {{{ - 1 - \sqrt 3 i} \over 2}} \right) - \left( {{{ - 1 + \sqrt 3 i} \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ k = $${ - \sqrt 3 i}$$ = -z | mcq | jee-main-2017-offline |
90f5UU5PdLZXNwMDULEQU | maths | complex-numbers | algebra-of-complex-numbers | The set of all $$\alpha $$ $$ \in $$ <b>R</b>, for which w = $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$ is purely imaginary number, for all z $$ \in $$ <b>C</b> satisfying |z| = 1 and Re z $$ \ne $$ 1, is : | [{"identifier": "A", "content": "an empty set"}, {"identifier": "B", "content": "{0}"}, {"identifier": "C", "content": "$$\\left\\{ {0,{1 \\over 4}, - {1 \\over 4}} \\right\\}$$"}, {"identifier": "D", "content": "equal to <b>R</b> "}] | ["B"] | null | As w = $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$, w is purely imaginary<br><br>
$$ \therefore w$$ + $$\bar w$$ = 0<br><br>
$$ \Rightarrow $$ $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$ + $${{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}$$ = 0<br><br>
$$ \Rightarrow $$ [1 + (1 - 8$$\alpha $$)][1 - $$\bar z$$] + [1 + ( 1 - 8$$\alpha $$)][1 - z] = 0<br><br>
$$ \Rightarrow $$ 2 - (z + $$\bar z$$) + (1 - 8$$\alpha $$)(z + $$\bar z$$) - 2(1 - 8$$\alpha $$) = 0<br><br>
$$ \Rightarrow $$ 2 - (z + $$\bar z$$) + (z + $$\bar z$$) - 8$$\alpha $$(z + $$\bar z$$) - 2 + 16$$\alpha $$ = 0<br><br>
$$ \Rightarrow $$ 16$$\alpha $$ = 8$$\alpha $$(z + $$\bar z)$$<br><br>
z + $$\bar z$$ = 2 or $$\alpha $$ = 0<br><br>
but z + $$\bar z$$ = 2 is not possible as Re(Z) $$ \ne $$ 1<br><br>
$$ \therefore $$ $$\alpha $$ = 0<br><br>
$$ \therefore $$ $$\alpha $$ $$ \in $$ {0} | mcq | jee-main-2018-online-15th-april-morning-slot |
QOeM4OKvohAunsfuxvotj | maths | complex-numbers | algebra-of-complex-numbers | Let
<br/>A = $$\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$$
<br/>. Then the sum of the elements in A is : | [{"identifier": "A", "content": "$${5\\pi \\over 6}$$"}, {"identifier": "B", "content": "$$\\pi $$"}, {"identifier": "C", "content": "$${3\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${{2\\pi } \\over 3}$$"}] | ["D"] | null | Given complex number,
<br><br>$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$
<br><br>$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
<br><br>$$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$
<br><br>$$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
<br><br>As complex number is purely imaginary, So real part of this complex number is zero.
<br><br>$$ \therefore $$ $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$ = 0
<br><br>$$ \Rightarrow $$ $$3 - 4{\sin ^2}\theta = 0$$
<br><br>$$ \Rightarrow $$ $$\sin \theta = \pm {{\sqrt 3 } \over 2}$$
<br><br>as $$\theta $$ $$ \in $$ $$\left( { - {\pi \over 2},\pi } \right)$$
<br><br>$$ \therefore $$ $$\theta $$ $$=$$ $$-$$ $${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$$
<br><br>$$ \therefore $$ Sum of those values of A is
<br><br>$$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$$
<br><br>$$ = {{2\pi } \over 3}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
2kgwaW5saoEQDxFpBDSmf | maths | complex-numbers | algebra-of-complex-numbers | If $${{z - \alpha } \over {z + \alpha }}\left( {\alpha \in R} \right)$$ is a purely imaginary number and | z | = 2, then a value of $$\alpha $$ is : | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $${{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0$$
<br><br>$$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$$
<br><br>$${\left| z \right|^2} = {\alpha ^2},$$ $$a = \pm 2$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
zgUc015yKq58iZfHNP18hoxe66ijvwujxie | maths | complex-numbers | algebra-of-complex-numbers | Let z $$ \in $$ C be such that |z| < 1.
<br/><br/> If $$\omega = {{5 + 3z} \over {5(1 - z)}}$$z, then : | [{"identifier": "A", "content": "4Im( $$\\omega$$) > 5"}, {"identifier": "B", "content": "5Im( $$\\omega$$) < 1"}, {"identifier": "C", "content": "5Re( $$\\omega$$) > 4"}, {"identifier": "D", "content": "5Re( $$\\omega$$) > 1"}] | ["D"] | null | $$\omega = {{5 + 3z} \over {5(1 - z)}}$$z
<br><br>$$ \Rightarrow $$ $$5\omega \left( {1 - z} \right) = 5 + 3z$$
<br><br>$$ \Rightarrow $$ $$5\omega - 5\omega z = 5 + 3z$$
<br><br>$$ \Rightarrow $$ $$5\omega - 5 = 5\omega z + 3z$$
<br><br>$$ \Rightarrow $$ $$z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$$
<br><br>$$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$$
<br><br>$$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {5\left( {\omega + {3 \over 5}} \right)}}$$
<br><br>$$ \Rightarrow $$ $$z = {{\left( {\omega - 1} \right)} \over {\left( {\omega + {3 \over 5}} \right)}}$$
<br><br>Given |z| < 1
<br><br>$$ \Rightarrow $$ $$\left| {{{\omega - 1} \over {\omega + {3 \over 5}}}} \right|$$ < 1
<br><br>$$ \Rightarrow $$ $$\left| {\omega - 1} \right| < \left| {\omega + {3 \over 5}} \right|$$
<br><br>$$ \Rightarrow $$ $$\left| {\omega - 1} \right| < \left| {\omega - \left( { - {3 \over 5}} \right)} \right|$$
<br><br>It says distance of $$\omega $$ from point 1 is less than distance from point $${\left( { - {3 \over 5}} \right)}$$.
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266782/exam_images/fp4f81kxo2vsscgmu4ep.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Complex Numbers Question 123 English Explanation">
<br><br>x coordinate of perpendicular bisector of points $${\left( { - {3 \over 5},0} \right)}$$ and (1, 0),
<br><br>x = $${{1 - {3 \over 5}} \over 2}$$ = $${1 \over 5}$$
<br><br>As $$\omega $$ is closer to point (1, 0) so $$\omega $$ should present in the right side of perpendicular bisector.
<br><br>$$ \therefore $$ $${\mathop{\rm Re}\nolimits} (\omega ) > {1 \over 5}$$
<br><br>$$ \Rightarrow $$ 5Re( $$\omega$$) > 1
<br><br><b>Other Method :</b>
<br><br>$$\omega = {{5 + 3z} \over {5(1 - z)}}$$z
<br><br>$$ \Rightarrow $$ $$5\omega \left( {1 - z} \right) = 5 + 3z$$
<br><br>$$ \Rightarrow $$ $$5\omega - 5\omega z = 5 + 3z$$
<br><br>$$ \Rightarrow $$ $$5\omega - 5 = 5\omega z + 3z$$
<br><br>$$ \Rightarrow $$ $$z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$$
<br><br>$$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$$
<br><br>Given |z| < 1
<br><br>$$ \Rightarrow $$ $$\left| {{{5\left( {\omega - 1} \right)} \over {5\omega + 3}}} \right|$$ < 1
<br><br>$$ \Rightarrow $$ $$5\left| {\omega - 1} \right| < \left| {5\omega + 3} \right|$$
<br><br>$$ \Rightarrow $$ $$25\left( {{\omega ^2} - 2\omega + 1} \right) < 25{\omega ^2} + 30\omega + 9$$
<br><br>$$ \Rightarrow $$ $$25\left( {\omega \overline \omega - \omega - \overline \omega + 1} \right)$$ <
<br><br>$$25\omega \overline \omega + 15\omega + 15\overline \omega + 9$$
<br><br>(Using $${\left| z \right|^2} = z\overline z $$)
<br><br>$$ \Rightarrow $$ $$16 < 40\omega + 40\overline \omega $$
<br><br>$$ \Rightarrow $$ $$\omega + \overline \omega > {2 \over 5}$$
<br><br>$$ \Rightarrow $$ $$2{\mathop{\rm Re}\nolimits} \left( \omega \right) > {2 \over 5}$$
<br><br>$$ \Rightarrow $$ 5Re( $$\omega$$) > 1 | mcq | jee-main-2019-online-9th-april-evening-slot |
gKYuo8JqpDeuA0UDsG3rsa0w2w9jxb3ta1h | maths | complex-numbers | algebra-of-complex-numbers | Let z $$ \in $$ C with Im(z) = 10 and it satisfies $${{2z - n} \over {2z + n}}$$ = 2i - 1 for some natural number n. Then : | [{"identifier": "A", "content": "n = 20 and Re(z) = \u201310"}, {"identifier": "B", "content": "n = 40 and Re(z) = 10"}, {"identifier": "C", "content": "n = 40 and Re(z) = \u201310"}, {"identifier": "D", "content": "n = 20 and Re(z) = 10"}] | ["C"] | null | Let Re (z) = x, then<br><br>
$${{2(x + 10i) - n} \over {2(x + 10i) + n}} = 2i - 1$$<br><br>
$$ \Rightarrow \left( {2x - n} \right) + 20i = - \left( {2x + n} \right) - 40 - 20i + 2ni$$<br><br>
$$ \Rightarrow 2x - n = 2x - n - 40$$<br><br>
& 20 = -20 + 2n $$ \Rightarrow $$ x = -10 & n = 20 | mcq | jee-main-2019-online-12th-april-evening-slot |
3NBa8Gu3laVRzWIBoOjgy2xukfurbxh6 | maths | complex-numbers | algebra-of-complex-numbers | The region represented by<br/> {z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1} is also given by the<br/> inequality :
{z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1} | [{"identifier": "A", "content": "y<sup>2</sup> $$ \\le $$ $$2\\left( {x + {1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "y<sup>2</sup> $$ \\le $$ $${x + {1 \\over 2}}$$"}, {"identifier": "C", "content": "y<sup>2</sup> $$ \\ge $$ 2(x + 1)"}, {"identifier": "D", "content": "y<sup>2</sup> $$ \\ge $$ x + 1"}] | ["A"] | null | Given z = x + iy
<br><br> |z| – Re(z) $$ \le $$ 1
<br><br>$$ \Rightarrow $$ $$\sqrt {{x^2} + {y^2}} $$ - x $$ \le $$ 1
<br><br>$$ \Rightarrow $$ $$\sqrt {{x^2} + {y^2}} $$ $$ \le $$ 1 + x
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> $$ \le $$ 1 + 2x + x<sup>2</sup>
<br><br>$$ \Rightarrow $$ y<sup>2</sup> $$ \le $$ 2x + 1
<br><br>$$ \Rightarrow $$ y<sup>2</sup> $$ \le $$ 2$$\left( {x + {1 \over 2}} \right)$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
KD70hRFZdAQ2A6EDwzjgy2xukg3931gx | maths | complex-numbers | algebra-of-complex-numbers | Let z = x + iy be a non-zero complex number
such that $${z^2} = i{\left| z \right|^2}$$, where i = $$\sqrt { - 1} $$ , then z lies
on the : | [{"identifier": "A", "content": "line, y = \u2013x"}, {"identifier": "B", "content": "real axis"}, {"identifier": "C", "content": "line, y = x"}, {"identifier": "D", "content": "imaginary axis"}] | ["C"] | null | Given z = x + iy
<br><br>and $${z^2} = i{\left| z \right|^2}$$
<br><br>$$ \Rightarrow $$ (x + iy)<sup>2</sup>
= i(x<sup>2</sup> + y<sup>2</sup>)
<br><br>$$ \Rightarrow $$ x<sup>2</sup> - y<sup>2</sup> + 2ixy = i(x<sup>2</sup> + y<sup>2</sup>) + 0
<br><br>Comparing both side we get,
<br><br>x<sup>2</sup> - y<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = y<sup>2</sup>
<br><br>and 2xy = (x<sup>2</sup> + y<sup>2</sup>)
<br><br>$$ \Rightarrow $$ (x - y)<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ x = y
<br><br>$$ \therefore $$ z lies on line x = y | mcq | jee-main-2020-online-6th-september-evening-slot |
1krw3pbyy | maths | complex-numbers | algebra-of-complex-numbers | Let $$S = \left\{ {n \in N\left| {{{\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)}^n}\left( {\matrix{
a & b \cr
c & d \cr
} } \right) = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)\forall a,b,c,d \in R} \right.} \right\}$$, where i = $$\sqrt { - 1} $$. Then the number of 2-digit numbers in the set S is _____________. | [] | null | 11 | Let $$X = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)$$ & $$A = {\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)^n}$$<br><br>$$\Rightarrow$$ AX = IX<br><br>$$\Rightarrow$$ A = I<br><br>$$ \Rightarrow {\left( {\matrix{
0 & i \cr
1 & 0 \cr
} } \right)^n} = I$$<br><br>$$ \Rightarrow {A^8} = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right]$$<br><br>$$\Rightarrow$$ n is multiple of 8<br><br>So, number of 2 digit numbers in the set <br><br>S = 11 (16, 24, 32, .........., 96) | integer | jee-main-2021-online-25th-july-morning-shift |
1kryfpffu | maths | complex-numbers | algebra-of-complex-numbers | If the real part of the complex number $$z = {{3 + 2i\cos \theta } \over {1 - 3i\cos \theta }},\theta \in \left( {0,{\pi \over 2}} \right)$$ is zero, then the value of sin<sup>2</sup>3$$\theta$$ + cos<sup>2</sup>$$\theta$$ is equal to _______________. | [] | null | 1 | Re $$(z) = {{3 - 6{{\cos }^2}\theta } \over {1 + 9{{\cos }^2}\theta }} = 0$$<br><br>$$\Rightarrow$$ $$\theta$$ = $${{\pi \over 4}}$$<br><br>Hence, sin<sup>2</sup>3$$\theta$$ + cos<sup>2</sup>$$\theta$$ = 1. | integer | jee-main-2021-online-27th-july-evening-shift |
1kteil093 | maths | complex-numbers | algebra-of-complex-numbers | If $$S = \left\{ {z \in C:{{z - i} \over {z + 2i}} \in R} \right\}$$, then : | [{"identifier": "A", "content": "S contains exactly two elements"}, {"identifier": "B", "content": "S contains only one element"}, {"identifier": "C", "content": "S is a circle in the complex plane"}, {"identifier": "D", "content": "S is a straight line in the complex plane"}] | ["D"] | null | Given $${{z - i} \over {z + 2i}} \in R$$<br><br>Then $$\arg \left( {{{z - i} \over {z + 2i}}} \right)$$ is 0 or $$\pi $$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266523/exam_images/toje9q4ejzuaxxxnmphc.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Mathematics - Complex Numbers Question 79 English Explanation"><br><br>$$\Rightarrow$$ S is straight line in complex | mcq | jee-main-2021-online-27th-august-morning-shift |
1l544xjnb | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\alpha$$ and $$\beta$$ be the roots of the equation x<sup>2</sup> + (2i $$-$$ 1) = 0. Then, the value of |$$\alpha$$<sup>8</sup> + $$\beta$$<sup>8</sup>| is equal to :</p> | [{"identifier": "A", "content": "50"}, {"identifier": "B", "content": "250"}, {"identifier": "C", "content": "1250"}, {"identifier": "D", "content": "1500"}] | ["A"] | null | <p>Given equation,</p>
<p>$${x^2} + (2i - 1) = 0$$</p>
<p>$$ \Rightarrow {x^2} = 1 - 2i$$</p>
<p>Let $$\alpha$$ and $$\beta$$ are the two roots of the equation.</p>
<p>As, we know roots of a equation satisfy the equation so</p>
<p>$${\alpha ^2} = 1 - 2i$$</p>
<p>and $${\beta ^2} = 1 - 2i$$</p>
<p>$$\therefore$$ $${\alpha ^2} = {\beta ^2} = 1 - 2i$$</p>
<p>$$\therefore$$ $$|{\alpha ^2}| = \sqrt {{1^2} + {{( - 2)}^2}} = \sqrt {15} $$</p>
<p>Now, $$|{\alpha ^8} + {\beta ^8}|$$</p>
<p>$$|{\alpha ^8} + {\alpha ^8}|$$</p>
<p>$$ = 2|{\alpha ^8}|$$</p>
<p>$$ = 2|{\alpha ^2}{|^4}$$</p>
<p>$$ = 2{\left( {\sqrt 5 } \right)^4}$$</p>
<p>$$ = 2 \times 25$$</p>
<p>$$ = 50$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l545wee0 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$S = \{ z \in C:|z - 2| \le 1,\,z(1 + i) + \overline z (1 - i) \le 2\} $$. Let $$|z - 4i|$$ attains minimum and maximum values, respectively, at z<sub>1</sub> $$\in$$ S and z<sub>2</sub> $$\in$$ S. If $$5(|{z_1}{|^2} + |{z_2}{|^2}) = \alpha + \beta \sqrt 5 $$, where $$\alpha$$ and $$\beta$$ are integers, then the value of $$\alpha$$ + $$\beta$$ is equal to ___________.</p> | [] | null | 26 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8b9cup/a3a99286-f2d0-4b94-be08-ef95f78a420f/8454b500-870a-11ed-95c4-a7dd61250e50/file-1lc8b9cuq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8b9cup/a3a99286-f2d0-4b94-be08-ef95f78a420f/8454b500-870a-11ed-95c4-a7dd61250e50/file-1lc8b9cuq.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Complex Numbers Question 73 English Explanation"><br>
$S$ represents the shaded region shown in the diagram.
<br><br>
Clearly $z_{1}$ will be the point of intersection of $P A$ and given circle.
<br><br>
$P A: 2 x+y=4$ and given circle has equation $(x-2)^{2}+y^{2}=1$
<br><br>
On solving we get
<br><br>
$z_{1}=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2}{\sqrt{5}} i \Rightarrow\left|z_{1}\right|^{2}=5-\frac{4}{\sqrt{5}}$
<br><br>
$z_{2}$ will be either $B$ or $C$.
<br><br>
$\because P B=\sqrt{17}$ and $P C=\sqrt{13}$ hence $z_{2}=1$
<br><br>
So $5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=30-4 \sqrt{5}$
<br><br>
Clearly $\alpha=30$ and $\beta=-4 \Rightarrow \alpha+\beta=26$ | integer | jee-main-2022-online-29th-june-morning-shift |
1l58gy62s | maths | complex-numbers | algebra-of-complex-numbers | <p>If $${z^2} + z + 1 = 0$$, $$z \in C$$, then <br/><br/>$$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$$ is equal to _________.</p> | [] | null | 2 | <p>$$\because$$ $${z^2} + z + 1 = 0$$</p>
<p>$$\Rightarrow$$ $$\omega$$ or $$\omega$$<sup>2</sup></p>
<p>$$\because$$ $$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$$</p>
<p>$$ = \left| {\sum\limits_{n = 1}^{15} {{z^{2n}} + \sum\limits_{n = 1}^{15} {{z^{ - 2n}} + 2\,.\,\sum\limits_{n = 1}^{15} {{{( - 1)}^n}} } } } \right|$$</p>
<p>$$ = \left| {0 + 0 - 2} \right|$$</p>
<p>$$ = 2$$</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1l6kiejn2 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let S be the set of all $$(\alpha, \beta), \pi<\alpha, \beta<2 \pi$$, for which the complex number $$\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$$ is purely imaginary and $$\frac{1+i \cos \beta}{1-2 i \cos \beta}$$ is purely real. Let $$Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S$$. Then $$\sum\limits_{(\alpha, \beta) \in S}\left(i Z_{\alpha \beta}+\frac{1}{i \bar{Z}_{\alpha \beta}}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "3 i"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2 $$-$$ i"}] | ["C"] | null | <p>$$\because$$ $${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }}$$ is purely imaginary</p>
<p>$$\therefore$$ $${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }} + {{1 + i\sin \alpha } \over {1 - 2i\sin \alpha }} = 0$$</p>
<p>$$ \Rightarrow 1 - 2{\sin ^2}\alpha = 0$$</p>
<p>$$\therefore$$ $$\alpha = {{5\pi } \over 4},\,{{7\pi } \over 4}$$</p>
<p>and $${{1 + i\cos \beta } \over {1 - 2i\cos \beta }}$$ is purely real</p>
<p>$${{1 + i\cos \beta } \over {1 - 2i\cos \beta }} - {{1 - i\cos \beta } \over {1 + 2i\cos \beta }} = 0$$</p>
<p>$$ \Rightarrow \cos \beta = 0$$</p>
<p>$$\therefore$$ $$\beta = {{3\pi } \over 2}$$</p>
<p>$$\therefore$$ $$S = \left\{ {\left( {{{5\pi } \over 2},{{3\pi } \over 2}} \right),\left( {{{7\pi } \over 4},{{3\pi } \over 2}} \right)} \right\}$$</p>
<p>$${Z_{\alpha \beta }} = 1 - i$$ and $${Z_{\alpha \beta }} = - 1 - i$$</p>
<p>$$\therefore$$ $$\sum\limits_{(\alpha ,\beta ) \in S} {\left( {i{Z_{\alpha \beta }} + {1 \over {i{{\overline Z }_{\alpha \beta }}}}} \right) = i( - 2i) + {1 \over i}\left[ {{1 \over {1 + i}} + {1 \over { - 1 + i}}} \right]} $$</p>
<p>$$ = 2 + {1 \over i}{{2i} \over { - 2}} = 1$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6npfu6z | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\mathrm{z}=a+i b, b \neq 0$$ be complex numbers satisfying $$z^{2}=\bar{z} \cdot 2^{1-z}$$. Then the least value of $$n \in N$$, such that $$z^{n}=(z+1)^{n}$$, is equal to __________.</p> | [] | null | 6 | <p>$$\because$$ $${z^2} = \overline z \,.\,{2^{1 - |z|}}$$ ...... (1)</p>
<p>$$ \Rightarrow |z{|^2} = |\overline z |\,.\,{2^{1 - |z|}}$$</p>
<p>$$ \Rightarrow |z| = {2^{1 - |z|}}$$,</p>
<p>$$\because$$ $$b \ne 0 \Rightarrow |z| \ne 0$$</p>
<p>$$\therefore$$ $$|z| = 1$$ ...... (2)</p>
<p>$$\because$$ $$z = a + ib$$ then $$\sqrt {{a^2} + {b^2}} = 1$$ ...... (3)</p>
<p>Now again from equation (1), equation (2), equation (3) we get :</p>
<p>$${a^2} - {b^2} + i2ab = (a - ib){2^0}$$</p>
<p>$$\therefore$$ $${a^2} - {b^2} = a$$ and $$2ab = - b$$</p>
<p>$$\therefore$$ $$a = - {1 \over 2}$$ and $$b = \, \pm \,{{\sqrt 3 } \over 2}$$</p>
<p>$$z = - {1 \over 2} + {{\sqrt 3 } \over 2}i$$ or $$z = - {1 \over 2} - {{\sqrt 3 } \over 2}i$$</p>
<p>$${z^n} = {(z + 1)^n} \Rightarrow {\left( {{{z + 1} \over z}} \right)^n} = 1$$</p>
<p>$${\left( {1 + {1 \over z}} \right)^n} = 1$$</p>
<p>$$\left( {{{1 + \sqrt 3 i} \over 2}} \right) = 1$$, then minimum value of n is 6.</p> | integer | jee-main-2022-online-28th-july-evening-shift |
1ldsg85m1 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}$$ and $$B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}$$. Then $$\sum\limits_{z \in A \cap B} {({\mathop{\rm Re}\nolimits} z - {\mathop{\rm Im}\nolimits} z)} $$ is equal to ____________.</p> | [] | null | 14 | <p>Let $$z = x + iy$$</p>
<p>and $$\alpha = 8 - 14i$$</p>
<p>$${{\alpha z - \overline \alpha \,\overline z } \over {{z^2} - {{\overline z }^2} - 112i}} = 1$$</p>
<p>$$\therefore$$ $${{(16y - 28x)} \over {4xy - 112i}} = 1$$</p>
<p>$$(16y - 28x + 112)i = 4xy$$</p>
<p>$$\therefore$$ $$z = - 7i$$ or 4</p>
<p>Now, $$z = - 7i$$ satisfy B</p>
<p>$$B:{x^2} + {(y + 3)^2} = 16$$</p>
<p>$$A \cap B = (0, - 7)$$</p>
<p>$${\mathop{\rm Re}\nolimits} z - lm\,z = 7$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldsuuwr5 | maths | complex-numbers | algebra-of-complex-numbers | <p>For two non-zero complex numbers $$z_{1}$$ and $$z_{2}$$, if $$\operatorname{Re}\left(z_{1} z_{2}\right)=0$$ and $$\operatorname{Re}\left(z_{1}+z_{2}\right)=0$$, then which of the following are possible?</p>
<p>A. $$\operatorname{Im}\left(z_{1}\right)>0$$ and $$\operatorname{Im}\left(z_{2}\right) > 0$$</p>
<p>B. $$\operatorname{Im}\left(z_{1}\right) < 0$$ and $$\operatorname{Im}\left(z_{2}\right) > 0$$</p>
<p>C. $$\operatorname{Im}\left(z_{1}\right) > 0$$ and $$\operatorname{Im}\left(z_{2}\right) < 0$$</p>
<p>D. $$\operatorname{Im}\left(z_{1}\right) < 0$$ and $$\operatorname{Im}\left(z_{2}\right) < 0$$</p>
<p>Choose the correct answer from the options given below :</p> | [{"identifier": "A", "content": "A and C"}, {"identifier": "B", "content": "A and B"}, {"identifier": "C", "content": "B and D"}, {"identifier": "D", "content": "B and C"}] | ["D"] | null | <p>Let, $${z_1} = {x_1} + i{y_1}$$</p>
<p>and $${z_2} = {x_2} + i{y_2}$$</p>
<p>$$\therefore$$ $${z_1}{z_2} = {x_1}{x_2} - {y_1}{y_2} + i({x_1}{y_2} + {x_2}{y_1})$$</p>
<p>Given, $${\mathop{\rm Re}\nolimits} ({z_1} + {z_2}) = 0$$</p>
<p>$$ \Rightarrow {x_1} + {x_2} = 0$$ ...... (1)</p>
<p>also given, $${\mathop{\rm Re}\nolimits} ({z_1}{z_2}) = 0$$</p>
<p>$$ \Rightarrow {x_1}{x_2} - {y_1}{y_2} = 0$$</p>
<p>$$ \Rightarrow {x_1}{x_2} = {y_1}{y_2}$$</p>
<p>$$ \Rightarrow {y_1}{y_2} = - x_1^2$$ [$$\because$$ $${x_2} = - {x_1}$$]</p>
<p>So, multiplication of imaginary part's of z<sub>1</sub> and z<sub>2</sub> is negative. It means sign of y<sub>1</sub> and y<sub>2</sub> are opposite of each other.</p>
<p>$$ \therefore $$ B and C are correct.</p> | mcq | jee-main-2023-online-29th-january-morning-shift |
1lgswb5vn | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\mathrm{S}=\left\{z \in \mathbb{C}-\{i, 2 i\}: \frac{z^{2}+8 i z-15}{z^{2}-3 i z-2} \in \mathbb{R}\right\}$$. If $$\alpha-\frac{13}{11} i \in \mathrm{S}, \alpha \in \mathbb{R}-\{0\}$$, then
$$242 \alpha^{2}$$ is equal to _________.</p> | [] | null | 1680 | Put $z=x+i y$
<br/><br/>$$
\begin{aligned}
& \operatorname{lm}\left(\frac{z^2+8 i z-15}{z^2-3 i z-2}\right)=0 \\\\
& \Rightarrow-\left(x^2-y^2-8 y-15\right)(2 x y-3 x)+(2 x y+8 x)\left(x^2-\right. \left.y^2+3 y-2\right)=0 \\\\
& \Rightarrow\left(x^2-y^2\right)(2 x y+8 x-2 x y+3 x)+(8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \\\\
& \Rightarrow 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y-45 x+6 x y^2 -4 x y+24 x y-16 x=0 \\\\
& \Rightarrow 11 x^3+11 x y^2+26 x y-61 x=0 \\\\
& \Rightarrow\left(11 x^2+11 y^2+26 y-61)=0\right. \\\\
& \because \alpha \neq 0, \Rightarrow x=0 \text { (neglected) } \\\\
& \text { Put } y=-\frac{13}{11}, \quad x=\alpha \\\\
& 11 \alpha^2+11 \cdot \frac{13^2}{11^2}-26 \cdot \frac{13}{11}-61=0 \\\\
& \Rightarrow 121 \alpha^2=840 \\\\
& \Rightarrow 242 \alpha^2=1680
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-evening-shift |
1lgvpa470 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$S = \left\{ {z = x + iy:{{2z - 3i} \over {4z + 2i}}\,\mathrm{is\,a\,real\,number}} \right\}$$. Then which of the following is NOT correct?</p> | [{"identifier": "A", "content": "$$y + {x^2} + {y^2} \\ne - {1 \\over 4}$$"}, {"identifier": "B", "content": "$$(x,y) = \\left( {0, - {1 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$x = 0$$"}, {"identifier": "D", "content": "$$y \\in \\left( { - \\infty , - {1 \\over 2}} \\right) \\cup \\left( { - {1 \\over 2},\\infty } \\right)$$"}] | ["B"] | null | Given that $z=x+i y$
<br/><br/>$$
\begin{aligned}
& \text { then } \frac{2 z-3 i}{4 z+2 i}=\frac{2(x+i y)-3 i}{4(x+i y)+2 i} \\\\
& =\frac{2 x+i(2 y-3)}{4 x+i(4 y+2)} \times \frac{4 x-i(4 y+2)}{4 x-i(4 y+2)} \\\\
& =\frac{8 x^2+(2 y-3)(4 y+2)}{(4 x)^2+(4 y+2)^2}+i\left(\frac{4 x(2 y-3)-2 x(4 y+2)}{(4 x)^2+(4 y+2)^2}\right)
\end{aligned}
$$
<br/><br/>Since, $\frac{2 z-3 i}{4 z+2 i}$ is Real $\Rightarrow \operatorname{Img}\left(\frac{2 z-3 i}{4 z+2 i}\right)=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 4 x(2 y-3)-2 x(4 y+2)=0 \\\\
& \Rightarrow 2 x(4 y-6-4 y-2)=0 \\\\
& \Rightarrow 2 x(-8)=0 \Rightarrow x=0
\end{aligned}
$$
<br/><br/>Also, $(4 x)^2+(4 y+2)^2 \neq 0 \Rightarrow y+x^2+y^2 \neq \frac{-1}{4}$
<br/><br/> If $x=0$, then $y \neq-\frac{1}{2}$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxw4now | maths | complex-numbers | algebra-of-complex-numbers | <p>Let the complex number $$z = x + iy$$ be such that $${{2z - 3i} \over {2z + i}}$$ is purely imaginary. If $${x} + {y^2} = 0$$, then $${y^4} + {y^2} - y$$ is equal to :</p> | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["C"] | null | Let, $z=x+i y$
<br/><br/>So, $\frac{2 z-3 i}{2 z+i}$ is purely imaginary
$~~$[Given]
<br/><br/>$$
\begin{aligned}
& \text { Now, }\left(\frac{2 z-3 i}{2 z+i}\right)+\left(\frac{\overline{2 z-3 i}}{2 z+i}\right)=0 \\\\
& \Rightarrow \frac{2 z-3 i}{2 z+i}+\frac{2 \bar{z}+3 i}{2 \bar{z}-i}=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow(2 z-3 i)(2 \bar{z}-i)+(2 \bar{z}+3 i)(2 z+i)=0 \\\\
& \Rightarrow\left\{4|z|^2-6 i \bar{z}-2 i z-3\right\}+\left\{4|z|^2+6 i z+2 i \bar{z}-3\right\}=0 \\\\
& \Rightarrow 8|z|^2-4 i \bar{z}+4 i z-6=0 \\\\
& \Rightarrow 8\left(x^2+y^2\right)-4 i(x-i y)+4 i(x+i y)-6=0 \\\\
& \Rightarrow 8\left(x^2+y^2\right)-4 i x-4 y+4 i x-4 y-6=0 \\\\
& \Rightarrow 8\left(x^2+y^2\right)-8 y-6=0 \\\\
& \Rightarrow 4\left(x^2+y^2\right)-4 y-3=0
\end{aligned}
$$
<br/><br/>Given that, $x+y^2=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow x=-y^2 \\\\
& \Rightarrow 4\left(y^4+y^2\right)-4 y=3 \\\\
& \Rightarrow y^4+y^2-y=\frac{3}{4}
\end{aligned}
$$
<br/><br/>Hence, required answer is $\frac{3}{4}$. | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgyl12p4 | maths | complex-numbers | algebra-of-complex-numbers | <p> Let $$A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.$$ is purely imaginary $$\}$$. Then the sum of the elements in $$\mathrm{A}$$ is :</p> | [{"identifier": "A", "content": "$$3 \\pi$$"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "$$2 \\pi$$"}, {"identifier": "D", "content": "$$4 \\pi$$"}] | ["D"] | null | $$
\begin{aligned}
& \text { Here, } z=\frac{1+2 i \sin \theta}{1-i \sin \theta} \times \frac{1+i \sin \theta}{1+i \sin \theta} \\\\
& \frac{1+i \sin \theta+2 i \sin \theta-2 \sin ^2 \theta}{1-i^2 \sin ^2 \theta} \\\\
& =\frac{\left(1-2 \sin ^2 \theta\right)+i(3 \sin \theta)}{1+\sin ^2 \theta}
\end{aligned}
$$
<br/><br/>$\because z$ is purely imaginary, so $\operatorname{Re} z=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{1-2 \sin ^2 \theta}{1+\sin ^2 \theta}=0 \\\\
& \Rightarrow 2 \sin ^2 \theta=1 \Rightarrow \sin ^2 \theta=\frac{1}{2} \\\\
& \Rightarrow \sin \theta= \pm \frac{1}{\sqrt{2}}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore A=\left[\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right] \because \theta \in(0,2 \pi)\\\\
& \therefore \text { Sum }=\frac{\pi+3 \pi+5 \pi+7 \pi}{4}=\frac{16 \pi}{4}=4 \pi
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh2xxw75 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$a \neq b$$ be two non-zero real numbers. Then the number of elements in the set $$X=\left\{z \in \mathbb{C}: \operatorname{Re}\left(a z^{2}+b z\right)=a\right.$$ and $$\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\}$$ is equal to :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "Infinite"}] | ["D"] | null | Let $z=x+i y$
<br/><br/>$$
\begin{array}{ll}
&\Rightarrow z^2=x^2-y^2+2 i x y \\\\
&\therefore a z^2+b z \\\\
& =a\left(x^2-y^2+2 i x y\right)+b(x+i y) \\\\
& =a\left(x^2-y^2\right)+b x+2 a i x y+b i y
\end{array}
$$
<br/><br/>$\begin{array}{ll}\operatorname{Re}\left(a z^2+b z\right)=b \\\\ \Rightarrow a\left(x^2-y^2\right)+b x=a \\\\ \Rightarrow x^2-y^2+\frac{b}{a} x=1 .........(i)\end{array}$
<br/><br/>$\begin{aligned} & \text { and } b z^2+a z \\\\ & =b\left(x^2-y^2+2 i x y\right)+a(x+i y) \\\\ & =b\left(x^2-y^2\right)+a x+2 b i x y+a i y \\\\ & \operatorname{Re}\left(b z^2+a z\right)=b \\\\ & \Rightarrow b\left(x^2-y^2\right)+a x=b \\\\ & \Rightarrow x^2-y^2+\frac{a}{b} x=1 ..........(ii)\end{aligned}$
<br/><br/>On subtracting Equation (ii) from Equation (i), we get
<br/><br/>$$
\begin{aligned}
& \frac{b}{a} x-\frac{a}{b} x=0 \\\\
& \Rightarrow \left(\frac{b}{a}-\frac{a}{b}\right) x=0 \\\\
& \Rightarrow x=0 \text { or } \frac{b}{a}-\frac{a}{b}=0 \\\\
& \Rightarrow b^2-a^2=0 \\\\
& \Rightarrow a= \pm b \\\\
& \Rightarrow a=-b(\text { since } a \neq b)
\end{aligned}
$$
<br/><br/>From (i), when $x=0$, then
<br/><br/>$$
\begin{aligned}
& 0-y^2=1 \Rightarrow y^2=-1 \\\\
& \Rightarrow y \in \phi \Rightarrow z \in \phi \text { has no solution. }
\end{aligned}
$$
<br/><br/>When, $a=-b$, then $x^2-y^2-x=1$ has infinitely many solutions. | mcq | jee-main-2023-online-6th-april-evening-shift |
lsaognvr | maths | complex-numbers | algebra-of-complex-numbers | Let $\mathrm{S}=|\mathrm{z} \in \mathrm{C}:| z-1 \mid=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \mid$. Let $z_1, z_2 \in \mathrm{S}$ be such that $\left|z_1\right|=\max\limits_{z \in s}|z|$ and $\left|z_2\right|=\min\limits _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["D"] | null | Let $z=x+i y$
<br/><br/>$$
\begin{aligned}
& |z-1|=1 \Rightarrow|x+i y-1|=1 \\\\
& (x-1)^2+y^2=1 .......(1)
\end{aligned}
$$
<br/><br/>$\begin{aligned} & (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \text { (Given) } \\\\ & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} \\\\ & (\sqrt{2}-1) x+y=\sqrt{2} .........(2)\end{aligned}$
<br/><br/>Solving (1) and (2), we get
<br/><br/>$$
\begin{aligned}
& (x-1)^2+(\sqrt{2}-(\sqrt{2}-1) x)^2=1 \\\\
& \left(x^2-2 x+1\right)+2+(\sqrt{2}-1)^2 x^2-2 \sqrt{2}(\sqrt{2}-1) x=0 \\\\
& x^2\left(1+(\sqrt{2}-1)^2\right)+x(-2-2 \sqrt{2}(\sqrt{2}-1))+2=0 \\\\
& x^2(4-2 \sqrt{2})+x(2 \sqrt{2}-6)+2=0 \\\\
& x^2(2-\sqrt{2})+x(\sqrt{2}-3)+1=0
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \Rightarrow x=1 \text { and } x=\frac{1}{2-\sqrt{2}} ..........(3) \\\\ & \text { When } x=1, y=1 \Rightarrow z_2=1+i \\\\ & \text { When } x=\frac{1}{2-\sqrt{2}}, y=\sqrt{2}-\frac{1}{\sqrt{2}} \\\\ & \Rightarrow z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}\end{aligned}$
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
& \left|\sqrt{2} z_1-z_2\right|^2 \\\\
& =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\\\
& =(\sqrt{2})^2 \\\\
& =2
\end{aligned}
$$ | mcq | jee-main-2024-online-1st-february-morning-shift |
lsblk2wv | maths | complex-numbers | algebra-of-complex-numbers | If $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha)^7=A+B \alpha+C \alpha^2, A, B, C \geqslant 0$, then $5(3 A-2 B-C)$ is equal to ____________. | [] | null | 5 | <p>$$x^2+x+1=0 \Rightarrow x=\omega, \omega^2=\alpha$$</p>
<p>Let $$\alpha=\omega$$</p>
<p>Now $$(1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega$$</p>
<p>$$\begin{aligned}
& A=1, B=1, C=0 \\
& \therefore 5(3 A-2 B-C)=5(3-2-0)=5
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lsd4pjmu | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$z_1$$ and $$z_2$$ be two complex numbers such that $$z_1+z_2=5$$ and $$z_1^3+z_2^3=20+15 i$$ Then, $$\left|z_1^4+z_2^4\right|$$ equals -</p> | [{"identifier": "A", "content": "$$15 \\sqrt{15}$$\n"}, {"identifier": "B", "content": "$$30 \\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$25 \\sqrt{3}$$\n"}, {"identifier": "D", "content": "75"}] | ["D"] | null | <p>$$\begin{aligned}
& z_1+z_2=5 \\
& z_1^3+z_2^3=20+15 i \\
& z_1^3+z_2^3=\left(z_1+z_2\right)^3-3 z_1 z_2\left(z_1+z_2\right) \\
& z_1^3+z_2^3=125-3 z_1 \cdot z_2(5) \\
& \Rightarrow 20+15 i=125-15 z_1 z_2 \\
& \Rightarrow 3 z_1 z_2=25-4-3 i \\
& \Rightarrow 3 z_1 z_2=21-3 i \\
& \Rightarrow z_1 \cdot z_2=7-i \\
& \Rightarrow\left(z_1+z_2\right)^2=25 \\
& \Rightarrow z_1^2+z_2^2=25-2(7-i) \\
& \Rightarrow 11+2 i \\
& \left(z_1^2+z_2^2\right)^2=121-4+44 i \\
& \Rightarrow z_1^4+z_2^4+2(7-i)^2=117+44 i \\
& \Rightarrow z_1^4+z_2^4=117+44 i-2(49-1-14 i) \\
& \Rightarrow\left|z_1^4+z_2^4\right|=75
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsf02zvt | maths | complex-numbers | algebra-of-complex-numbers | <p>If $$z=\frac{1}{2}-2 i$$ is such that $$|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$$ and $$\alpha, \beta \in \mathbb{R}$$, then $$\alpha+\beta$$ is equal to</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$-$$4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$-$$1"}] | ["C"] | null | <p>To begin with, let's analyze the given equation:</p>
<p>$$|z+1|=\alpha z+\beta(1+i)$$</p>
<p>First, we compute the modulus of the left side:</p>
<p>Let's take the given value of z,</p>
<p>$$z = \frac{1}{2} - 2i$$</p>
<p>Now, we find $$z + 1$$:</p>
<p>$$z + 1 = \left(\frac{1}{2} - 2i\right) + 1 = \frac{3}{2} - 2i$$</p>
<p>Then, the modulus of $$z + 1$$ is calculated as follows:</p>
<p>$$|z + 1| = \left|\frac{3}{2} - 2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2}$$</p>
<p>$$|z + 1| = \sqrt{\frac{9}{4} + 4}$$</p>
<p>$$|z + 1| = \sqrt{\frac{9}{4} + \frac{16}{4}}$$</p>
<p>$$|z + 1| = \sqrt{\frac{25}{4}}$$</p>
<p>$$|z + 1| = \frac{5}{2}$$</p>
<p>Now we need to equate the modulus to the right-hand side of the equation and solve for $$\alpha$$ and $$\beta$$. Let's rewrite the equation:</p>
<p>$$\frac{5}{2} = \alpha z + \beta(1 + i)$$</p>
<p>Substitute $$z$$ with its value:</p>
<p>$$\frac{5}{2} = \alpha \left(\frac{1}{2} - 2i\right) + \beta(1+i)$$</p>
<p>Rewrite the equation separating real and imaginary parts:</p>
<p>$$\frac{5}{2} = \alpha \left(\frac{1}{2}\right) - 2\alpha i + \beta + \beta i$$</p>
<p>$$\frac{5}{2} = \left(\alpha \frac{1}{2} + \beta\right) + \left(-2\alpha + \beta\right)i$$</p>
<p>For the above equality to hold, both real and imaginary parts must be equal.</p>
<p>Equating real parts:</p>
<p>$$\alpha \frac{1}{2} + \beta = \frac{5}{2}$$</p>
<p>Equating imaginary parts:</p>
<p>$$-2\alpha + \beta = 0$$</p>
<p>We now have a system of two linear equations:</p>
<p>1) $$\frac{\alpha}{2} + \beta = \frac{5}{2}$$</p>
<p>2) $$-2\alpha + \beta = 0$$</p>
<p>Let's solve the system by isolating $$\beta$$ from the second equation and then substituting it into the first one:</p>
<p>$$\beta = 2\alpha$$</p>
<p>Now substitute $$\beta$$ in the first equation:</p>
<p>$$\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$$</p>
<p>$$\alpha \left(\frac{1}{2} + 2\right) = \frac{5}{2}$$</p>
<p>$$\alpha \left(\frac{1}{2} + \frac{4}{2}\right) = \frac{5}{2}$$</p>
<p>$$\alpha \left(\frac{5}{2}\right) = \frac{5}{2}$$</p>
<p>To find the value of $$\alpha$$, we divide both sides by $$\frac{5}{2}$$:</p>
<p>$$\alpha = 1$$</p>
<p>Now, we use the value of $$\alpha$$ to find $$\beta$$:</p>
<p>$$\beta = 2\alpha$$</p>
<p>$$\beta = 2 \cdot 1$$</p>
<p>$$\beta = 2$$</p>
<p>Finally, we add both $$\alpha$$ and $$\beta$$ to find $$\alpha + \beta$$:</p>
<p>$$\alpha + \beta = 1 + 2 = 3$$</p>
<p>The value of $$\alpha + \beta$$ is 3.</p>
<p>So, the correct answer is Option C) 3.</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsf0qr88 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\alpha, \beta$$ be the roots of the equation $$x^2-x+2=0$$ with $$\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$$. Then $$\alpha^6+\alpha^4+\beta^4-5 \alpha^2$$ is equal to ___________.</p> | [] | null | 13 | <p>$$\begin{aligned}
& \alpha^6+\alpha^4+\beta^4-5 \alpha^2 \\
& =\alpha^4(\alpha-2)+\alpha^4-5 \alpha^2+(\beta-2)^2 \\
& =\alpha^5-\alpha^4-5 \alpha^2+\beta^2-4 \beta+4 \\
& =\alpha^3(\alpha-2)-\alpha^4-5 \alpha^2+\beta-2-4 \beta+4 \\
& =-2 \alpha^3-5 \alpha^2-3 \beta+2 \\
& =-2 \alpha(\alpha-2)-5 \alpha^2-3 \beta+2 \\
& =-7 \alpha^2+4 \alpha-3 \beta+2 \\
& =-7(\alpha-2)+4 \alpha-3 \beta+2 \\
& =-3 \alpha-3 \beta+16=-3(1)+16=13
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfl3s5i | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\alpha, \beta$$ be the roots of the equation $$x^2-\sqrt{6} x+3=0$$ such that $$\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$$. Let $$a, b$$ be integers not divisible by 3 and $$n$$ be a natural number such that $$\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+i b), i=\sqrt{-1}$$. Then $$n+a+b$$ is equal to __________.</p> | [] | null | 49 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8r3kr/364277a2-5c20-422c-867f-2e93bd294016/10be8ab0-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8r3ks.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8r3kr/364277a2-5c20-422c-867f-2e93bd294016/10be8ab0-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8r3ks.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Evening Shift Mathematics - Complex Numbers Question 13 English Explanation"></p>
<p>$$\begin{aligned}
& x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) \\
& \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) \\
& \therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) \\
& =\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2} \\
& =3^{49}(-1+i) \\
& =3^n(a+i b) \\
& \therefore n=49, a=-1, b=1 \\
& \therefore n+a+b=49-1+1=49
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift |
1lsgajjz4 | maths | complex-numbers | algebra-of-complex-numbers | <p>If $$z=x+i y, x y \neq 0$$, satisfies the equation $$z^2+i \bar{z}=0$$, then $$\left|z^2\right|$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "$$\\frac{1}{4}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>$$\begin{aligned}
& z^2=-i \bar{z} \\
& \left|z^2\right|=|i \bar{z}| \\
& \left|z^2\right|=|z| \\
& |z|^2-|z|=0 \\
& |z|(|z|-1)=0 \\
& |z|=0 \text { (not acceptable) } \\
& \therefore|z|=1 \\
& \therefore|z|^2=1
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
lv0vxdau | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$\alpha$$ and $$\beta$$ be the sum and the product of all the non-zero solutions of the equation $$(\bar{z})^2+|z|=0, z \in C$$. Then $$4(\alpha^2+\beta^2)$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>$$\begin{aligned}
& (\bar{z})^2+|z|=0 \quad \text{... (1)}\\
& z^2+|\bar{z}|=0 \quad \text{... (2)}
\end{aligned}$$</p>
<p>From equation (1) and (2)</p>
<p>$$\begin{aligned}
& \text { as }|z|=|\bar{z}| \\
& \Rightarrow \quad(\bar{z})^2=z^2 \\
& \Rightarrow \quad z=\bar{z} \text { or } z=-\bar{z} \\
& \Rightarrow \operatorname{Im}(z)=0 \text { or } \operatorname{Re}(z)=0
\end{aligned}$$</p>
<p>Case I : If $$\operatorname{Im}(z)=0$$</p>
<p>$$\Rightarrow z=x$$</p>
<p>Putting value of $$z$$ in equation (1)</p>
<p>$$\begin{aligned}
& x^2+|x|=0 \\
& \Rightarrow x=0 \quad \text{[Rejected]
}
\end{aligned}$$</p>
<p>Case II : If $$\operatorname{Re}(z)=0$$</p>
<p>$$\Rightarrow z=i y$$</p>
<p>Putting value of $$z$$ in equation (1)</p>
<p>$$\begin{aligned}
& -y^2+|y|=0 \\
& y= \pm 1 \text { as } y \neq 0
\end{aligned}$$</p>
<p>Hence, $$z= \pm i$$ are the solution of the given equation</p>
<p>$$\begin{aligned}
& \Rightarrow \alpha=i-i=0 \\
& \text { and } \beta=i(-i)=1 \\
& \Rightarrow \quad 4\left(\alpha^2+\beta^2\right)=4(0+1) \\
& \quad=4
\end{aligned}$$</p>
<p>$$\therefore$$ Option (3) is correct</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv3ve9bf | maths | complex-numbers | algebra-of-complex-numbers | <p>The sum of all possible values of $$\theta \in[-\pi, 2 \pi]$$, for which $$\frac{1+i \cos \theta}{1-2 i \cos \theta}$$ is purely imaginary, is equal to :</p> | [{"identifier": "A", "content": "$$4 \\pi$$\n"}, {"identifier": "B", "content": "$$3 \\pi$$\n"}, {"identifier": "C", "content": "$$2 \\pi$$\n"}, {"identifier": "D", "content": "$$5 \\pi$$"}] | ["B"] | null | <p>$$\frac{1+i \cos \theta}{1-2 i \cos \theta}$$ is purely imaginary</p>
<p>$$n=\frac{1+i \cos \theta}{1-2 i \cos \theta} \times \frac{1+2 i \cos \theta}{1+2 i \cos \theta}=\frac{1+3 i \cos \theta-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}$$</p>
<p>$$n=\frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}+i\left(\frac{3 \cos \theta}{1+4 \cos ^2 \theta}\right)$$</p>
<p>$$n$$ is purely imaginary</p>
<p>$$\Rightarrow \frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}=0$$</p>
<p>$$\Rightarrow \cos ^2 \theta=\frac{1}{2}$$</p>
<p>$$\Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}}$$</p>
<p>$$\theta$$ can be $$\frac{\pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{-3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$$</p>
<p>Sum of all possible values of $$\theta=3 \pi$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv5gsxx4 | maths | complex-numbers | algebra-of-complex-numbers | <p>Let $$z$$ be a complex number such that $$|z+2|=1$$ and $$\operatorname{lm}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$$. Then the value of $$|\operatorname{Re}(\overline{z+2})|$$ is</p> | [{"identifier": "A", "content": "$$\\frac{2 \\sqrt{6}}{5}$$\n"}, {"identifier": "B", "content": "$$\\frac{24}{5}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{6}}{5}$$\n"}, {"identifier": "D", "content": "$$\\frac{1+\\sqrt{6}}{5}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& |z+2|=1 \\
& \operatorname{Im}_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \\
& |\operatorname{Re}(\overline{z+2})|=?
\end{aligned}$$</p>
<p>Let $$z=x+i y$$</p>
<p>$$\begin{aligned}
& \because|z+2|=1 \Rightarrow(x+2)^2+y^2=1 \quad \ldots(1) \\
& I_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \Rightarrow I_m\left(\frac{x+i y+1}{x+i y+2}\right)=\frac{1}{5} \\
& \Rightarrow I_m\left|\frac{[(x+1)+i y][(x+)-i y]}{(x+2)^2+y^2}\right|=\frac{1}{5}
\end{aligned}$$</p>
<p>$$\frac{y(x+2)-y(x+)}{(x+2)^2+y^2}=\frac{1}{5}\quad \text{.... (2)}$$</p>
<p>$$\Rightarrow y=\frac{1}{5}$$</p>
<p>Substituting in equation (1)</p>
<p>$$\begin{aligned}
& (x+2)^2+\frac{1}{25}=1 \\
& (x+2)^2=\frac{24}{25} \\
& \Rightarrow x=-2 \pm \frac{\sqrt{24}}{5} \\
& |\operatorname{Re}(\overline{x+i y+2})| \\
& =x+2= \pm \frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5}
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
zfiW9PrjVIHHJFRs | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | The locus of the centre of a circle which touches the circle $$\left| {z - {z_1}} \right| = a$$ and$$\left| {z - {z_2}} \right| = b\,$$ externally
<br/><br/>($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be : | [{"identifier": "A", "content": "an ellipse "}, {"identifier": "B", "content": "a hyperbola"}, {"identifier": "C", "content": "a circle "}, {"identifier": "D", "content": "none of these"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264708/exam_images/uyfgmdfsahcbye8cxeuf.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Complex Numbers Question 166 English Explanation">
Let the circle be $$\left| {z - {z_3}} \right| = r.$$
<br><br>Then according to given conditions
<br><br>$$\left| {{z_3} - {z_1}} \right| = r + a$$ (Shown in the image)
<br><br>and $$\left| {{z_3} - {z_2}} \right| = r + b.$$ (Shown in the image)
<br><br>Eliminating $$r,$$ we get
<br><br>$$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$$
<br><br>$$\therefore$$ Locus of center $${z_3}$$ is
<br><br>$$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$$ = constant.
<br><br>Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.
<br><br>Here distance of z<sub>1</sub> from z<sub>3</sub> is = $$r + a$$ and distance of z<sub>2</sub> from z<sub>3</sub> is = $$r + b$$
<br><br>Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant
<br><br>$$\therefore$$ Locus of z<sub>3</sub> is a hyperbola. | mcq | aieee-2002 |
EwK1sFRumluyqFqZ | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then : | [{"identifier": "A", "content": "$${a^2} = 4b$$ "}, {"identifier": "B", "content": "$${a^2} = b$$ "}, {"identifier": "C", "content": "$${a^2} = 2b$$ "}, {"identifier": "D", "content": "$${a^2} = 3b$$ "}] | ["D"] | null | Given quadratic equation,
<br>$${Z^2} + aZ + b = 0$$
<br>and two roots are $${Z_1}$$ and $${Z_2}$$.
<br><br>$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$
<br><br>Question says,
<br>There are three complex numbers:
<br>1. Origin (0)
<br>2. $${Z_1}$$
<br>3. $${Z_2}$$
<br>and they form an equilateral triangle. So They are the vertices of the triangle.
<br><br>[ <b>Important Point :</b> If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
<br>$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]
<br><br>In this question,
<br>$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$
<br><br>By putting those values in the equation we get,
<br><br>$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0
<br><br>$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$
<br><br>$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]
<br><br>$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$
<br><br>$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$
<br><br>$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$
<br><br>$$ \Rightarrow $$ $${\left( { - a} \right)^2}$$ = $$3b$$
<br><br>$$ \Rightarrow $$ $${a^2}$$ = $$3b$$
<br><br>So Option (D) is correct.
<br><br>[ <b>Note :</b> This question is asked to check if you know the following formula -
<br><br>"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
<br>$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ] | mcq | aieee-2003 |
dTNQj6NpNJcqoBXF | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | If $$\,\left| {z + 4} \right|\,\, \le \,\,3\,$$, then the maximum value of $$\left| {z + 1} \right|$$ is : | [{"identifier": "A", "content": "6 "}, {"identifier": "B", "content": "0 "}, {"identifier": "C", "content": "4 "}, {"identifier": "D", "content": "10"}] | ["A"] | null | $$z$$ lies on or inside the circle with center $$(-4,0)$$ and radius $$3$$ units.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264288/exam_images/yur8wemdh4kwqpqnxlji.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Complex Numbers Question 154 English Explanation">
<br><br>From the Argand diagram maximum value of $$\left| {z + 1} \right|$$ is $$6$$ | mcq | aieee-2007 |
0Bf9X9z2dXZJuQpg | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let $$\alpha \,,\beta $$ be real and z be a complex number. If $${z^2} + \alpha z + \beta = 0$$ has two distinct roots on the line Re z = 1, then it is necessary that : | [{"identifier": "A", "content": "$$\\beta \\, \\in ( - 1,0)$$ "}, {"identifier": "B", "content": "$$\\left| {\\beta \\,} \\right| = 1$$ "}, {"identifier": "C", "content": "$$\\beta \\, \\in (1,\\infty )$$ "}, {"identifier": "D", "content": "$$\\beta \\, \\in (0,1)$$ "}] | ["C"] | null | As real part of roots is $$1$$
<br><br>Let roots are $$1 + pi,1 + q$$
<br><br>$$\therefore$$ sum of roots $$ = 1 + pi + 1 + qi = - \alpha $$
<br><br>which is real $$ \Rightarrow q = - p\,\,$$
<br><br>or root are $$1+pi$$ and $$1-pi$$
<br><br>product of roots $$ = 1 + {p^2} = \beta \in \left( {1,\infty } \right)$$
<br><br>$$p \ne 0$$ as roots are distinct. | mcq | aieee-2011 |
zrx1He6H0PulM6MQ | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | If $$z \ne 1$$ and $$\,{{{z^2}} \over {z - 1}}\,$$ is real, then the point represented by the complex number z lies : | [{"identifier": "A", "content": "either on the real axis or a circle passing through the origin."}, {"identifier": "B", "content": "on a circle with centre at the origin"}, {"identifier": "C", "content": "either on real axis or on a circle not passing through the origin."}, {"identifier": "D", "content": "on the imaginary axis."}] | ["A"] | null | Let $$z = x + iy$$
<br><br>$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$
<br><br>Now $${{{z^2}} \over {z - 1}}$$ is real
<br><br>$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$
<br><br>$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$
<br><br>$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$
<br><br>$$ \Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$
<br><br>$$ \Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$
<br><br>$$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$
<br><br>$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin. | mcq | aieee-2012 |
qbXThOdgWHTXG1fujwjj9 | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let z$$ \in $$C, the set of complex numbers. Then the equation, 2|z + 3i| $$-$$ |z $$-$$ i| = 0 represents : | [{"identifier": "A", "content": "a circle with radius $${8 \\over 3}.$$"}, {"identifier": "B", "content": "a circle with diameter $${{10} \\over 3}.$$"}, {"identifier": "C", "content": "an ellipse with length of major axis $${{16} \\over 3}.$$"}, {"identifier": "D", "content": "an ellipse with length of minor axis $${{16} \\over 9}.$$"}] | ["A"] | null | Given,
<br><br>2 $$\,\left| \, \right.$$z + 3i$$\,\left| \, \right.$$ = $$\,\left| \, \right.$$z $$-$$i$$\,\left| \, \right.$$
<br><br>Let z = x + iy
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2 $$\,\left| \, \right.$$ x + iy + 3i $$\,\left| \, \right.$$ = $$\,\left| \, \right.$$ x + iy $$-$$ i $$\,\left| \, \right.$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2 $$\,\left| \, \right.$$ x + i (y + 3)$$\,\left| \, \right.$$ = $$\,\left| \, \right.$$ x + i (y $$-$$ 1)$$\,\left| \, \right.$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2 $$\sqrt {{x^2} + {{\left( {y + 3} \right)}^2}} $$ = $$\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 4 (x<sup>2</sup> + y<sup>2</sup> + 6y + 9) = x<sup>2</sup> + y<sup>2</sup> $$-$$ 2y + 1
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 3x<sup>2</sup> + 3y<sup>2</sup> + 26y + 35 = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x<sup>2</sup> + y<sup>2</sup> + $${{26} \over 3}$$ y + $${{35} \over 3}$$ = 0
<br><br>This is a equation of circle with center ($$-$$ $${{13} \over 3}$$, 0)
<br><br>$$\therefore\,\,\,$$ Radius = $$\sqrt {0 + {{169} \over 9} - {{35} \over 3}} $$
<br><br>= $$\sqrt {{{64} \over 9}} $$
<br><br>= $${8 \over 3}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
r2VEe8qSKWNdEek63f6ds | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | The equation
<br/>Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0, z $$ \in $$ <b>C</b>, z $$ \ne $$ i
<br/>represents a part of a circle having radius
equal to :
| [{"identifier": "A", "content": "2 "}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["C"] | null | Let z = x + iy
<br><br>Then,
<br><br>Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0
<br><br>$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$$
<br><br>$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$
<br><br>$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$
<br><br>$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{
i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr
\,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
<br><br>$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{
i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr
\,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
<br><br>$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{
x\left( {y - 1} \right) - xy\, - 2x \hfill \cr
\,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
<br><br>$$ \Rightarrow $$$${{\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} + 1 = 0$$
<br><br>$$ \Rightarrow $$2x<sup>2</sup> + 2y<sup>2</sup> - y - 1 = 0
<br><br>$$ \Rightarrow $$x<sup>2</sup> + y<sup>2</sup> - $$\left( {{1 \over 2}} \right)$$y - $$\left( {{1 \over 2}} \right)$$ = 0
<br><br>$$ \therefore $$ Center of the circle is $$\left( {0,{1 \over 4}} \right)$$
<br><br>$$ \therefore $$ Radius = $$\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}} $$
<br><br>= $$\sqrt {{1 \over {16}} + {1 \over 2}} $$
<br><br>= $$\sqrt {{9 \over {16}}} $$
<br><br>= $${{3 \over 4}}$$ | mcq | jee-main-2017-online-9th-april-morning-slot |
BSbow338TPfoEJmZ5O7Hf | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | All the points in the set<br/>
$$S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )$$ lie on a : | [{"identifier": "A", "content": "straight line whose slope is \u20131"}, {"identifier": "B", "content": "straight line whose slope is 1."}, {"identifier": "C", "content": "circle whose radius is 1."}, {"identifier": "D", "content": "circle whose radius is $$\\sqrt 2$$ ."}] | ["C"] | null | Let h + ik = $${{\alpha + i} \over {\alpha - i}}$$
<br><br>= $${{\left( {\alpha + i} \right)\left( {\alpha + i} \right)} \over {\left( {\alpha - i} \right)\left( {\alpha + i} \right)}}$$
<br><br>= $${{\left( {{\alpha ^2} - 1} \right) + 2i\alpha } \over {{\alpha ^2} + 1}}$$
<br><br>$$ \therefore $$ h = $${{{\alpha ^2} - 1} \over {{\alpha ^2} + 1}}$$ and k = $${{2\alpha } \over {{\alpha ^2} + 1}}$$
<br><br>By squaring and adding we get
<br><br>h<sup>2</sup> + k<sup>2</sup> = 1
<br><br>$$ \therefore $$ This is circle whose radius is 1. | mcq | jee-main-2019-online-9th-april-morning-slot |
tVlWLwS9cygf8yUxA83rsa0w2w9jx5d7bm9 | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | The equation |z – i| = |z – 1|, i = $$\sqrt { - 1} $$, represents : | [{"identifier": "A", "content": "a circle of radius 1"}, {"identifier": "B", "content": "the line through the origin with slope \u2013 1"}, {"identifier": "C", "content": "a circle of radius $${1 \\over 2}$$"}, {"identifier": "D", "content": "the line through the origin with slope 1\n"}] | ["D"] | null | Let the complex number z = x + iy<br><br>
Now given | (x + iy) - 1 | = | (x+iy) - i |<br><br>
$$ \Rightarrow $$ (x - 1)<sup>2</sup> + y<sup>2</sup> = x<sup>2</sup> + (y - 1)<sup>2</sup><br><br>
$$ \Rightarrow $$ x = y<br><br>
$$ \therefore $$ the correct answer is option (D) | mcq | jee-main-2019-online-12th-april-morning-slot |
sDfdRhll5oVpjbpMGu7k9k2k5e2q3wg | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | If $${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$$, where z = x + iy, then the point (x, y) lies on a :
| [{"identifier": "A", "content": "straight line whose slope is $${3 \\over 2}$$"}, {"identifier": "B", "content": "straight line whose slope is $$-{2 \\over 3}$$"}, {"identifier": "C", "content": "circle whose diameter is $${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "circle whose centre is at $$\\left( { - {1 \\over 2}, - {3 \\over 2}} \\right)$$"}] | ["C"] | null | $${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$$
<br><br>Put z = x + iy
<br><br>$$ \therefore $$ $${\mathop{\rm Re}\nolimits} \left( {{{\left( {x + iy} \right) - 1} \over {2\left( {x + iy} \right) + i}}} \right) = 1$$
<br><br>$$ \Rightarrow $$ $${\mathop{\rm Re}\nolimits} \left( {\left( {{{\left( {x - 1} \right) + iy} \over {2x + i\left( {2y + 1} \right)}}} \right)\left( {{{2x - i\left( {2y + 1} \right)} \over {2x - i\left( {2y + 1} \right)}}} \right)} \right) = 1$$
<br><br>$$ \Rightarrow $$ $${\mathop{\rm Re}\nolimits} \left( {{{\left\{ {\left( {x - 1} \right) + iy} \right\}\left\{ {2x - i\left( {2y + 1} \right)} \right\}} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}} \right) = 1$$
<br><br>Real part of this equation is = 1
<br><br>$$ \therefore $$ $${{2x\left( {x - 1} \right) + y\left( {2y + 1} \right)} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ 2x<sup>2</sup> + 2y<sup>2</sup> +2x + 3y + 1 = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> +x + $${3 \over 2}$$y + $${1 \over 2}$$ = 0
<br><br>This is an equation of circle.
<br><br>$$ \therefore $$ Locus is a circle whose
<br><br>center is $$\left( { - {1 \over 2}, - {3 \over 4}} \right)$$ and radius $${{\sqrt 5 } \over 4}$$
<br><br>$$ \therefore $$ Diameter = 2 $$ \times $$ $${{\sqrt 5 } \over 4}$$ = $${{\sqrt 5 } \over 2}$$ | mcq | jee-main-2020-online-7th-january-morning-slot |
TBLslHD22EPp6lLBJjjgy2xukf8z8vmo | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let $$u = {{2z + i} \over {z - ki}}$$, z = x + iy and k > 0. If the curve represented<br/> by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1/2"}, {"identifier": "D", "content": "3/2"}] | ["A"] | null | Given, z = x + iy<br><br>and $$u = {{2z + i} \over {z - ki}}$$<br><br>$$ = {{2(x + iy) + i} \over {(x + iy) - ki}}$$<br><br>$$ = {{2x + i(2y + 1)} \over {x + i(y - k)}} \times {{x - i(y - k)} \over {x - i(y - k)}}$$<br><br>$$ = {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}}$$<br><br>Given, <br><br>$${\mathop{\rm Re}\nolimits} (u) + {\mathop{\rm Im}\nolimits} (u) = 1$$<br><br>$$ \Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1$$<br><br>$$ \Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2}$$<br><br>This curve intersect the y-axis at point P and Q, so at point P and Q x = 0<br><br>Putting x = 0 at the above equation,<br><br>$$ \therefore $$ $$(2y + 1)(y - k) = {(y - k)^2}$$<br><br>$$ \Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky$$<br><br>$$ \Rightarrow {y^2} + y - (k + {k^2}) = 0$$<br><br>Let roots of this quadratic equation y<sub>1</sub> and y<sub>2</sub><br><br>$$ \therefore $$ Point P (0, y<sub>1</sub>) and Q (0, y<sub>2</sub>)<br><br>and $${y_{_1}} + {y_2} = 1$$ , $${y_1}{y_2} = - k - {k^2}$$<br><br>$$ \therefore $$ $${({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2}$$<br><br>$$ = 1 + 4k + 4{k^2}$$<br><br>$$ \Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}} $$<br><br>Given, PQ = 5<br><br>$$ \Rightarrow |{y_1} - {y_2}| = 5$$<br><br>$$ \Rightarrow \sqrt {1 + 4k + 4{k^2}} = 5$$<br><br>$$ \Rightarrow {k^2} + k - 6 = 0$$<br><br>$$ \Rightarrow k = - 3,\,2$$<br><br>So, k = 2 (Given k > 0) | mcq | jee-main-2020-online-4th-september-morning-slot |
yuHYXrHeBUXqF0AGxgjgy2xukfg7cet7 | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | If the four complex numbers $$z,\overline z ,\overline z - 2{\mathop{\rm Re}\nolimits} \left( {\overline z } \right)$$ and $$z-2Re(z)$$ represent the vertices of a square of
side 4 units in the Argand plane, then $$|z|$$ is equal to : | [{"identifier": "A", "content": "4$$\\sqrt 2 $$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "2$$\\sqrt 2 $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265025/exam_images/lalecqmsx1oe6loxq0jd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Morning Slot Mathematics - Complex Numbers Question 107 English Explanation">
Let $$z = x + iy$$<br><br>Length of side = 4<br><br>$$AB = 4$$<br><br>$$|z - \overline z | = 4$$<br><br>$$|2y|\, = 4;$$$$ \Rightarrow $$ $$|y|\, = 2$$<br><br>$$BC = 4$$<br><br>$$ \Rightarrow $$ $$|\overline z - (\overline z - 2{\mathop{\rm Re}\nolimits} (\overline z )|\, = 4$$<br><br>$$ \Rightarrow $$ $$|2x|\, = 4;\,$$$$ \Rightarrow $$ $$|x|\, = 2$$<br><br>$$ \therefore $$ $$|z|\, = \,\sqrt {{x^2} + {y^2}} = \sqrt {4 + 4} = 2\sqrt 2 $$ | mcq | jee-main-2020-online-5th-september-morning-slot |
mhVOjrFE4VXwW6yfRE1kls4xtsd | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let the lines (2 $$-$$ i)z = (2 + i)$$\overline z $$ and (2 $$+$$ i)z + (i $$-$$ 2)$$\overline z $$ $$-$$ 4i = 0, (here i<sup>2</sup> = $$-$$1) be normal to a circle C. If the line iz + $$\overline z $$ + 1 + i = 0 is tangent to this circle C, then its radius is : | [{"identifier": "A", "content": "$${3 \\over {2\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${3 \\over {\\sqrt 2 }}$$"}] | ["A"] | null | $$(2 - i)z = (2 + i)\overline z $$<br><br>$$ \Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)$$<br><br>$$ \Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y$$<br><br>$$ \Rightarrow 2ix - 4iy = 0$$<br><br>$${L_1}:x - 2y = 0$$<br><br>$$ \Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0$$<br><br>$$ \Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$$<br><br>$$ \Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0$$<br><br>$$ \Rightarrow 2ix + 4iy - 4i = 0$$<br><br>$${L_2}:x + 2y - 2 = 0$$<br><br>Solve L<sub>1</sub> and L<sub>2</sub><sub></sub>: x = 1, $$4y = 2,y = {1 \over 2}$$<br><br>$$ \therefore $$ $$x = 1$$<br><br>Centre $$\left( {1,{1 \over 2}} \right)$$<br><br>$${L_3}:iz + \overline z + 1 + i = 0$$<br><br>$$ \Rightarrow i(x + iy) + x - iy + 1 + i = 0$$<br><br>$$ \Rightarrow ix - y + x - iy + 1 + i = 0$$<br><br>$$ \Rightarrow (x - y + 1) + i(x - y + 1) = 0$$<br><br>Radius = distance from $$\left( {1,{1 \over 2}} \right)$$ to $$x - y + 1 = 0$$<br><br>$$ \Rightarrow $$ $$r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}$$<br><br>$$ \Rightarrow $$ $$r = {3 \over {2\sqrt 2 }}$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
t3QY3WP7FQprViAD1n1kluybz0d | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let z be those complex numbers which satisfy<br/><br/>| z + 5 | $$ \le $$ 4 and z(1 + i) + $$\overline z $$(1 $$-$$ i) $$ \ge $$ $$-$$10, i = $$\sqrt { - 1} $$.<br/><br/>If the maximum value of | z + 1 |<sup>2</sup> is $$\alpha$$ + $$\beta$$$$\sqrt 2 $$, then the value of ($$\alpha$$ + $$\beta$$) is ____________. | [] | null | 48 | Let, z = x + iy<br><br>Given, z(1 + i) + $$\overline z $$ (1 $$-$$ i) $$ \ge $$ $$-$$ 10<br><br>$$ \Rightarrow $$ z + $$\overline z $$ + i (z $$-$$ $$\overline z $$) $$ \ge $$ $$-$$ 10<br><br>$$ \Rightarrow $$ 2x + i (2iy) $$ \ge $$ $$-$$ 10<br><br>$$ \Rightarrow $$ x + i<sup>2</sup> y $$ \ge $$ $$-$$ 5<br><br>$$ \Rightarrow $$ x $$-$$ y $$ \ge $$ $$-$$ 5 ...... (1)<br><br>Also given, | z + 5 | $$ \le $$ 4<br><br>$$ \Rightarrow $$ | z $$-$$ ($$-$$5 + 0i) | $$ \le $$ 4 ...... (2)<br><br>It represents a circle whose center at ($$-$$ 5, 0) and radius 4. z is inside of the circle.
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266964/exam_images/iaixfx6nsvqnz71obifn.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Complex Numbers Question 99 English Explanation">
<br><br>From (1) and (2) z is the shaded region of the diagram.<br><br>Now, | z + 1 | = | z $$-$$ ($$-$$1 + 0 i) | = distance of z from ($$-$$1, 0).<br><br>Clearly 'p' is the required position of 'z' when | z + 1 | is maximum.<br><br>$$ \therefore $$ P $$ \equiv $$ ($$-$$5 $$-$$ 4 cos45$$^\circ$$, 0 $$-$$ 4sin45$$^\circ$$) = ($$-$$5$$-$$2$$\sqrt 2 $$, $$-$$2$$\sqrt 2 $$)<br><br>$$ \therefore $$ (PQ)<sup>2</sup>|<sub>max</sub> = 32 + 16$$\sqrt 2 $$<br><br>$$ \Rightarrow $$ $$\alpha$$ = 32<br><br>$$ \Rightarrow $$ $$\beta$$ = 16<br><br>Thus, $$\alpha$$ + $$\beta$$ = 48 | integer | jee-main-2021-online-26th-february-evening-slot |
Y5JJ3LyDE1P4FfE39O1kmjbgw76 | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | The area of the triangle with vertices A(z), B(iz) and C(z + iz) is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${1 \\over 2}$$| z |<sup>2</sup>"}, {"identifier": "C", "content": "$${1 \\over 2}$$| z + iz |<sup>2</sup>"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267682/exam_images/cpesvpbbsxejbpma67r4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Mathematics - Complex Numbers Question 95 English Explanation">
<br>Each side length = |z|
<br><br>Area of $$\Delta$$ = $${1 \over 2}$$ (area of square)<br><br>= $${1 \over 2}$$|z|<sup>2</sup> | mcq | jee-main-2021-online-17th-march-morning-shift |
tYCQViOYWriDal2fUG1kmklebtf | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let S<sub>1</sub>, S<sub>2</sub> and S<sub>3</sub> be three sets defined as<br/><br/>S<sub>1</sub> = {z$$\in$$C : |z $$-$$ 1| $$ \le $$ $$\sqrt 2 $$}<br/><br/>S<sub>2</sub> = {z$$\in$$C : Re((1 $$-$$ i)z) $$ \ge $$ 1}<br/><br/>S<sub>3</sub> = {z$$\in$$C : Im(z) $$ \le $$ 1}<br/><br/>Then the set S<sub>1</sub> $$\cap$$ S<sub>2</sub> $$\cap$$ S<sub>3</sub> : | [{"identifier": "A", "content": "has exactly three elements"}, {"identifier": "B", "content": "is a singleton"}, {"identifier": "C", "content": "has infinitely many elements"}, {"identifier": "D", "content": "has exactly two elements"}] | ["C"] | null | Let, z = x + iy<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266724/exam_images/cnzpzty76azz6qp0mhza.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - Complex Numbers Question 94 English Explanation"><br><br>S<sub>1</sub> $$ \equiv $$ (x $$-$$ 1)<sup>2</sup> + y<sup>2</sup> $$ \le $$ 2 ..... (1)<br><br>S<sub>2</sub> $$ \equiv $$ x + y $$ \ge $$ 1 ..... (2)<br><br>S<sub>3</sub> $$\equiv$$ y $$ \le $$ 1 .... (3)<br><br>$$ \Rightarrow $$ S<sub>1</sub> $$\cap$$ S<sub>2</sub> $$\cap$$ S<sub>3</sub> has infinitely many elements. | mcq | jee-main-2021-online-17th-march-evening-shift |
krFSobiivVYaEzRzw11kmlinglj | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | If the equation $$a|z{|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0$$ represents a circle where a, d are real constants then which of the following condition is correct? | [{"identifier": "A", "content": "|$$\\alpha$$|<sup>2</sup> $$-$$ ad $$\\ne$$ 0"}, {"identifier": "B", "content": "|$$\\alpha$$|<sup>2</sup> $$-$$ ad > 0 and a$$\\in$$R $$-$$ {0}"}, {"identifier": "C", "content": "|$$\\alpha$$|<sup>2</sup> $$-$$ ad $$ \\ge $$ 0 and a$$\\in$$R"}, {"identifier": "D", "content": "$$\\alpha$$ = 0, a, d$$\\in$$R<sup>+</sup>"}] | ["B"] | null | $$a|z{|^2} + \alpha \overline z + \overline \alpha z + d = 0$$<br><br>$$ \Rightarrow $$ $$z\overline z + \left( {{\alpha \over a}} \right)\overline z + \left( {{{\overline \alpha } \over a}} \right)z + {d \over a} = 0$$<br><br>$$ \therefore $$ Centre $$ = - {\alpha \over a}$$<br><br>$$r = \sqrt {{{\left| {{\alpha \over a}} \right|}^2} - {d \over a}} $$<br><br>$$ \Rightarrow {\left| {{\alpha \over a}} \right|^2} \ge {d \over a}$$<br><br>$$ \Rightarrow {\left| \alpha \right|^2} \ge ad$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
Hc7j5aGTrUNOVMU3TI1kmllzn5y | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let z<sub>1</sub>, z<sub>2</sub> be the roots of the equation z<sup>2</sup> + az + 12 = 0 and z<sub>1</sub>, z<sub>2</sub> form an equilateral triangle with origin. Then, the value of |a| is : | [] | null | 6 | For equilateral triangle with vertices z<sub>1</sub>, z<sub>2</sub> and z<sub>3</sub>,<br><br>$$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$$<br><br>Here one vertex z<sub>3</sub> is 0<br><br>$$ \therefore $$ $$z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$$<br><br>Given, z<sub>1</sub>, z<sub>2</sub> are roots of $${z^2} + az + 12 = 0$$<br><br>$$ \therefore $$ $${z_1} + {z_2} = - a$$<br><br>$${z_1}{z_2} = 12$$<br><br>$$ \therefore $$ $$z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$$<br><br>$$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$$<br><br>$$ \Rightarrow {( - a)^2} = 3 \times 12$$<br><br>$$ \Rightarrow {a^2} = 36$$<br><br>$$ \Rightarrow a = \pm 6$$<br><br>$$ \Rightarrow |a|\, = 6$$ | integer | jee-main-2021-online-18th-march-morning-shift |
ACZzZlXbS3L9a3R9ns1kmm2z0b8 | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let a complex number be w = 1 $$-$$ $${\sqrt 3 }$$i. Let another complex number z be such that |zw| = 1 and arg(z) $$-$$ arg(w) = $${\pi \over 2}$$. Then the area of the triangle with vertices origin, z and w is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["D"] | null | <picture><source media="(max-width: 1227px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265980/exam_images/quuj4fvtbuln79e3uwpk.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265539/exam_images/ttoow4xxjqahazhug2ch.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265609/exam_images/nyjekr5wczpp7n4kqfd1.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264025/exam_images/kffie72hsq84xfeart38.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265073/exam_images/bjpn7m5simwe38rccj98.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263484/exam_images/rzzsbpsfml9g59x3uvfs.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Complex Numbers Question 91 English Explanation"></picture>
<br>Given, w = 1 $$-$$ $$\sqrt 3 $$ i<br><br>$$ \Rightarrow |w| = \sqrt {{{(1)}^2} + {{( - \sqrt 3 )}^2}} = 2$$<br><br>Also, given | zw | = 1<br><br>$$ \Rightarrow $$ | z | | w | = 1 (using property)<br><br>$$ \Rightarrow $$ | z | = $${1 \over 2}$$<br><br>Also, arg(z) $$-$$ arg(w) = $${{\pi {} } \over 2}$$<br><br>$$ \therefore $$ Angle between two complex number z and w is $${{\pi {} } \over 2}$$.<br><br>$$ \therefore $$ $$\angle zow = {{\pi {} } \over 2}$$<br><br>$$ \therefore $$ $$\Delta$$zow is a right angle triangle with base $$ow = 2$$ and height $$oz = {1 \over 2}$$<br><br>$$ \therefore $$ Area = $${1 \over 2} \times 2 \times {1 \over 2} = {1 \over 2}$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krxi95ji | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let C be the set of all complex numbers. Let<br/><br/>S<sub>1</sub> = {z$$\in$$C : |z $$-$$ 2| $$\le$$ 1} and <br/><br/>S<sub>2</sub> = {z$$\in$$C : z(1 + i) + $$\overline z $$(1 $$-$$ i) $$\ge$$ 4}.<br/><br/>Then, the maximum value of $${\left| {z - {5 \over 2}} \right|^2}$$ for z$$\in$$S<sub>1</sub> $$\cap$$ S<sub>2</sub> is equal to : | [{"identifier": "A", "content": "$${{3 + 2\\sqrt 2 } \\over 4}$$"}, {"identifier": "B", "content": "$${{5 + 2\\sqrt 2 } \\over 2}$$"}, {"identifier": "C", "content": "$${{3 + 2\\sqrt 2 } \\over 2}$$"}, {"identifier": "D", "content": "$${{5 + 2\\sqrt 2 } \\over 4}$$"}] | ["D"] | null | |t $$-$$ 2| $$\le$$ 1<br><br>Put t = x + iy<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264870/exam_images/xng9d9n757qbnimzbcss.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264992/exam_images/znksuycbiq7wsgsy3hkn.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266723/exam_images/d8quz7o1xhunjaulq2j3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Complex Numbers Question 86 English Explanation"></picture><br><br>(x $$-$$ 2)<sup>2</sup> + y<sup>2</sup> $$\le$$ 1<br><br>Also, t(1 + i) + $$\overline t$$(1 $$-$$ i) $$\ge$$ 4<br><br>x $$-$$ y $$\ge$$ 2<br><br>Let point on circle be A(2 + cos$$\theta$$, sin$$\theta$$)<br><br>$$\theta \in \left[ { - {{3\pi } \over 4},{\pi \over 4}} \right]$$<br><br>$${(AP)^2} = {\left( {2 + \cos \theta - {5 \over 2}} \right)^2} + \sin \theta $$<br><br>$$ = {\cos ^2}\theta - \cos \theta + {1 \over 4} + {\sin ^2}\theta $$<br><br>$$ = {5 \over 4} - \cos \theta $$<br><br>For (AP)<sup>2</sup> maximum $$\theta = - {{3\pi } \over 4}$$<br><br>$${(AP)^2} = {5 \over 4} + {1 \over {\sqrt 2 }} = {{5\sqrt 2 + 4} \over {4\sqrt 2 }}$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1krzqy3td | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | The equation of a circle is Re(z<sup>2</sup>) + 2(Im(z))<sup>2</sup> + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x<sup>2</sup> $$-$$ 6x $$-$$ y + 13 = 0, has y-intercept equal to ______________. | [] | null | 1 | Equation of circle is (x<sup>2</sup> $$-$$ y<sup>2</sup>) + 2y<sup>2</sup> + 2x = 0<br><br>x<sup>2</sup> + y<sup>2</sup> + 2x = 0<br><br>Centre : ($$-$$1, 0)<br><br>Parabola : x<sup>2</sup> $$-$$ 6x $$-$$ y + 13 = 0<br><br>(x $$-$$ 3)<sup>2</sup> = y $$-$$ 4<br><br>Vertex : (3, 4)<br><br>Equation of line $$ \equiv y - 0 = {{4 - 0} \over {3 + 1}}(x + 1)$$<br><br>$$y = x + 1$$<br><br>y-intercept = 1 | integer | jee-main-2021-online-25th-july-evening-shift |
1ks061tqe | maths | complex-numbers | applications-of-complex-numbers-in-coordinate-geometry | Let C be the set of all complex numbers. Let<br/><br/>$${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $$<br/><br/>$${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $$ and <br/><br/>$${S_3} = \{ z \in C||z - \overline z | \ge 8\} $$.<br/><br/>Then the number of elements in $${S_1} \cap {S_2} \cap {S_3}$$ is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "Infinite"}] | ["A"] | null | $${S_1}:|z - 3 - 2i{|^2} = 8$$<br><br>$$|z - 3 - 2i| = 2\sqrt 2 $$<br><br>$${(x - 3)^2} + {(y - 2)^2} = {(2\sqrt 2 )^2}$$<br><br>$${S_2}:x \ge 5$$<br><br>$${S_3}:|z - \overline z | \ge 8$$<br><br>$$|2iy| \ge 8$$<br><br>$$2|y| \ge 8$$<br><br>$$\therefore$$ $$y \ge 4$$, $$y \le - 4$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267632/exam_images/iarlhyp1tybzbzupcjfm.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Complex Numbers Question 84 English Explanation"><br><br>$$n\left( {{S_1} \cap {S_2} \cap {S_3}} \right) = 1$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
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