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lv0vxc6v | maths | circle | basic-theorems-of-a-circle | <p>A square is inscribed in the circle $$x^2+y^2-10 x-6 y+30=0$$. One side of this square is parallel to $$y=x+3$$. If $$\left(x_i, y_i\right)$$ are the vertices of the square, then $$\Sigma\left(x_i^2+y_i^2\right)$$ is equal to:</p> | [{"identifier": "A", "content": "152"}, {"identifier": "B", "content": "148"}, {"identifier": "C", "content": "156"}, {"identifier": "D", "content": "160"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk74s6l/678de8fc-2a12-44e3-8b6b-dd8440b79fd7/1bfb85d0-1988-11ef-a7bd-376696e028ce/file-1lwk74s6m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwk74s6l/678de8fc-2a12-44e3-8b6b-dd8440b79fd7/1bfb85d0-1988-11ef-a7bd-376696e028ce/file-1lwk74s6m.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) 4th April Morning Shift Mathematics - Circle Question 9 English Explanation"></p>
<p>One side of square is $$y=x+k$$ Distance of $$(5,3)$$ to the line $$y=x+k$$ is</p>
<p>$$\begin{aligned}
& \frac{|3-5-k|}{\sqrt{2}}=\sqrt{2} \\
& =|-2-k|=2 \\
& \Rightarrow k=0 \text { or } k=-4
\end{aligned}$$</p>
<p>So lines are $$y=x$$ and $$y=x-4$$</p>
<p>Now, solving these lines with circle</p>
<p>$$\begin{aligned}
y= & x \text { and } x^2+y^2-10 x-6 y+30=0 \\
\Rightarrow & 2 x^2-16 x+30=0 \\
\Rightarrow & x=3, y=3 \\
& x=5, y=5 \\
& y=x-4 \text { and } x^2+y^2-10 x-6 y+30=0 \\
\Rightarrow & x=5, y=1 \\
& x=7, y=3 \\
& \sum_{i=1}^4 x_i^2+y_i^2=9+9+25+25+25+1+49+9 \\
= & 152
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv3ve68b | maths | circle | basic-theorems-of-a-circle | <p>If the image of the point $$(-4,5)$$ in the line $$x+2 y=2$$ lies on the circle $$(x+4)^2+(y-3)^2=r^2$$, then $$r$$ is equal to:</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{x+4}{1}=\frac{y-5}{2}=\frac{-2(4)}{5} \\
& \Rightarrow \quad x=-4-\frac{8}{5}=-\frac{28}{5}, y=5-\frac{16}{5}=\frac{9}{5} \\
& \therefore \quad \text { Image is }\left(\frac{-28}{5}, \frac{9}{5}\right)
\end{aligned}$$</p>
<p>Image lies on circle $$(x+4)^2+(y-3)^2=r^2$$</p>
<p>$$\begin{aligned}
& \left(\frac{-28}{5}+4\right)^2+\left(\frac{9}{5}-3\right)^2=r^2 \\
& \Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \\
& \Rightarrow r=2
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv9s1zub | maths | circle | basic-theorems-of-a-circle | <p>Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :</p> | [{"identifier": "A", "content": "$$\\mathrm{r}=1$$\n"}, {"identifier": "B", "content": "$$2 \\mathrm{r}^2-4 \\mathrm{r}+1=0$$\n"}, {"identifier": "C", "content": "$$2 \\mathrm{r}^2-8 \\mathrm{r}+7=0$$\n"}, {"identifier": "D", "content": "$$\\mathrm{r}^2-8 \\mathrm{r}+8=0$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweo69as/998331bc-73f9-46f0-9260-389c06031bce/0c501540-167e-11ef-9070-f523f4c6bd4b/file-1lweo69at.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweo69as/998331bc-73f9-46f0-9260-389c06031bce/0c501540-167e-11ef-9070-f523f4c6bd4b/file-1lweo69at.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 Evening Shift Mathematics - Circle Question 3 English Explanation"></p>
<p>$$C F=4 \sqrt{2}-2 \sqrt{2}=2 \sqrt{2}=r+r \sqrt{2}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad(2-r) \sqrt{2}=r \\
& \Rightarrow \quad \sqrt{2}=\left(\frac{r}{2-r}\right) \Rightarrow 2=\frac{r^2}{(2-r)^2} \\
& \Rightarrow 2\left(r^2-4 r+4\right)=r^2 \\
& \Rightarrow r^2-8 r+8=0
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvc57bdz | maths | circle | basic-theorems-of-a-circle | <p>A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $$m$$ and $$n$$, respectively, then $$m+n^2$$ is equal to</p> | [{"identifier": "A", "content": "408"}, {"identifier": "B", "content": "414"}, {"identifier": "C", "content": "312"}, {"identifier": "D", "content": "396"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd0dvt3/ce397cb2-87aa-4135-a838-45d211acf529/3c9d7570-1594-11ef-88c8-4b364e13ab15/file-1lwd0dvt4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd0dvt3/ce397cb2-87aa-4135-a838-45d211acf529/3c9d7570-1594-11ef-88c8-4b364e13ab15/file-1lwd0dvt4.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 Morning Shift Mathematics - Circle Question 1 English Explanation"></p>
<p>Inradius of $$\triangle A B C=r=\frac{\Delta}{s}=\frac{\frac{\sqrt{3}}{4} \times(12)^2}{18}$$</p>
<p>$$r=2 \sqrt{3}$$</p>
<p>Side length of square is $$a$$, then $$a^2=2 r^2$$</p>
<p>$$\Rightarrow a^2=24$$</p>
<p>Area of square, $$m=24$$</p>
<p>Perimeter of square, $$n=4 \sqrt{24}$$</p>
<p>$$\begin{aligned}
& \Rightarrow m+n^2=24+384 \\
& =408
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
1l5bb5tdb | maths | circle | chord-of-contact | <p>Let a circle C : (x $$-$$ h)<sup>2</sup> + (y $$-$$ k)<sup>2</sup> = r<sup>2</sup>, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to ___________.</p> | [] | null | 7 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5v64hz0/fe62d372-86ba-43d3-b376-3bcc7cdf6f30/f2e103c0-0906-11ed-a790-b11fa70c8a36/file-1l5v64hz1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5v64hz0/fe62d372-86ba-43d3-b376-3bcc7cdf6f30/f2e103c0-0906-11ed-a790-b11fa70c8a36/file-1l5v64hz1.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Evening Shift Mathematics - Circle Question 51 English Explanation"></p>
<p>Here, $$O{M^2} = O{P^2} - P{M^2}$$</p>
<p>$${\left( {{{|1 + r|} \over {\sqrt 2 }}} \right)^2} = {r^2} - 1$$</p>
<p>$$\therefore$$ $${r^2} - 2r - 3 = 0$$</p>
<p>$$\therefore$$ $$r = 3$$</p>
<p>$$\therefore$$ Equation of circle is</p>
<p>$${(x - 1)^2} + {(y - 3)^2} = {3^2}$$</p>
<p>$$\therefore$$ h = 1, k = 3, r = 3</p>
<p>$$\therefore$$ $$h + k + r = 7$$</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l6rfz5ti | maths | circle | chord-of-contact | <p>Let $$A B$$ be a chord of length 12 of the circle $$(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$$. If tangents drawn to the circle at points $$A$$ and $$B$$ intersect at the point $$P$$, then five times the distance of point $$P$$ from chord $$A B$$ is equal to __________.</p> | [] | null | 72 | Here $A M=B M=6$
<br><br>$$
O M=\sqrt{\left(\frac{13}{2}\right)^{2}-6^{2}}=\frac{5}{2}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7z3hbu0/c0ab874c-f40a-4393-abd9-7c5a221a294e/2d9f8370-32c8-11ed-a433-8d18e842a53d/file-1l7z3hbu1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7z3hbu0/c0ab874c-f40a-4393-abd9-7c5a221a294e/2d9f8370-32c8-11ed-a433-8d18e842a53d/file-1l7z3hbu1.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Circle Question 41 English Explanation">
<br><br>$$
\sin \theta=\frac{12}{13}
$$
<br><br>In $\triangle P A O$ :
<br><br>$$
\begin{aligned}
&\frac{P O}{O A}=\sec \theta \\\\
&P O=\frac{13}{2} \cdot \frac{13}{5}=\frac{169}{10} \\\\
&\therefore P M=\frac{169}{10}-\frac{5}{2}=\frac{144}{10}=\frac{72}{5} \\\\
&\therefore 5 P M=72 .
\end{aligned}
$$ | integer | jee-main-2022-online-29th-july-evening-shift |
1ldsuhn1h | maths | circle | chord-of-contact | <p>Let the tangents at the points $$A(4,-11)$$ and $$B(8,-5)$$ on the circle $$x^{2}+y^{2}-3 x+10 y-15=0$$, intersect at the point $$C$$. Then the radius of the circle, whose centre is $$C$$ and the line joining $$A$$ and $$B$$ is its tangent, is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2\\sqrt{13}}{3}$$"}, {"identifier": "B", "content": "$$\\frac{3\\sqrt{3}}{4}$$"}, {"identifier": "C", "content": "$$\\sqrt{13}$$"}, {"identifier": "D", "content": "$$2\\sqrt{13}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0172l/c56cbb61-8f00-4972-aa26-7f1c65ed7300/4a06dfd0-a637-11ed-8501-8d588d737388/file-1ldt0174e.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ldt0172l/c56cbb61-8f00-4972-aa26-7f1c65ed7300/4a06dfd0-a637-11ed-8501-8d588d737388/file-1ldt0174e.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Circle Question 34 English Explanation"></p>
<p>Equation of AB :</p>
<p>$$y - ( - 5) = {{ - 5 + 11} \over {8 - 4}}(x - 8)$$</p>
<p>$$ \Rightarrow y + 5 = {3 \over 2}(x - 8)$$</p>
<p>$$ \Rightarrow 2y + 10 = 3x - 24$$</p>
<p>$$ \Rightarrow 3x - 2y - 34 = 0$$ .... (1)</p>
<p>Also AB is cord of contact. And we know equation of cord of contact to a circle is $$T = 0$$</p>
<p>$$ \Rightarrow xh + yk - {3 \over 2}(x + h) + 5(y + k) - 15 = 0$$</p>
<p>$$ \Rightarrow x\left( {h - {3 \over 2}} \right) + y(k + 5) + \left( { - {3 \over 2}h + 5k - 15} \right) = 0$$ .... (2)</p>
<p>Comparing equation (1) and (2), we get</p>
<p>$${{h - {3 \over 2}} \over 3} = {{k + 5} \over { - 2}} = {{ - {3 \over 2}h + 5k - 15} \over { - 34}}$$</p>
<p>$$\therefore$$ $$ - 2h + 3 = 3k + 15$$</p>
<p>$$ \Rightarrow 3k + 2h = - 12$$ ..... (3)</p>
<p>and</p>
<p>$$17(k + 5) = - {3 \over 2}h + 5k - 15$$</p>
<p>$$ \Rightarrow 12k + {3 \over 2}h - 100$$</p>
<p>$$ \Rightarrow 3h + 24k = - 200$$ ..... (4)</p>
<p>Solving (3) and (4), we get</p>
<p>$$h = 8$$ and $$k = - {{28} \over 3}$$</p>
<p>$$\therefore$$ Point C is $$\left( {8, - {{28} \over 3}} \right)$$</p>
<p>Now radius of the circle whose centre is at C and tangent is AB is</p>
<p>$$ = \left| {{{3(8) - 2\left( { - {{28} \over 3}} \right) - 34} \over {\sqrt {{3^2} + {2^2}} }}} \right|$$</p>
<p>$$ = \left| {{{26} \over {3\sqrt {13} }}} \right|$$</p>
<p>$$ = {{2\sqrt {13} } \over 3}$$</p> | mcq | jee-main-2023-online-29th-january-morning-shift |
lsapr1j6 | maths | circle | chord-of-contact | Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}$ intersect at two distinct points, is $\mathrm{R}-[\mathrm{a}, \mathrm{b}]$, then the point $(8 \mathrm{a}+12,16 \mathrm{~b}-20)$ lies on the curve : | [{"identifier": "A", "content": "$x^2+2 y^2-5 x+6 y=3$"}, {"identifier": "B", "content": "$5 x^2-y=-11$"}, {"identifier": "C", "content": "$x^2-4 y^2=7$"}, {"identifier": "D", "content": "$6 x^2+y^2=42$"}] | ["D"] | null | $\begin{aligned} & C: x^2+y^2=4 \Rightarrow C(0,0), r_1=2 \\\\ & C^{\prime}: x^2+y^2-4 \lambda x+9=0 \Rightarrow C^{\prime}(2 \lambda, 0), r_2=\sqrt{4 \lambda^2-9} \\\\ & \left|r_1-r_2\right| < C C^{\prime} < \left|r_1+r_2\right| \\\\ & \left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & |2 \lambda|>\left|2-\sqrt{4 \lambda^2-9}\right| \\\\ & \Rightarrow 4 \lambda^2 > 4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9} \\\\ & 4 \sqrt{4 \lambda^2-9}+5>0 \Rightarrow \lambda \in R\end{aligned}$
<br/><br/>$\begin{aligned} & |2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & \Rightarrow 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{\left(4 \lambda^2\right)-9} \\\\ & 5<4 \sqrt{4 \lambda^2-9} \text { and } \lambda^2 \geq \frac{9}{4}\end{aligned}$
<br/><br/>$\begin{aligned} & \frac{25}{16} < 4 \lambda^2-9 \\\\ & \Rightarrow \lambda^2>\frac{169}{64} \\\\ & \lambda \in\left(-\infty, \frac{-13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \\\\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}\right] \\\\ & a=\frac{-13}{8}, b=\frac{13}{8} \\\\ & \Rightarrow(8 a+12,16 b-20)=(-1,6) \\\\ & \Rightarrow 6(-1)^2+(6)^2=42\end{aligned}$ | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lse5ofy4 | maths | circle | chord-of-contact | <p>If one of the diameters of the circle $$x^2+y^2-10 x+4 y+13=0$$ is a chord of another circle $$\mathrm{C}$$, whose center is the point of intersection of the lines $$2 x+3 y=12$$ and $$3 x-2 y=5$$, then the radius of the circle $$\mathrm{C}$$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3$$\\sqrt2$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "$$\\sqrt{20}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjrugw/97f6ed68-0662-482a-8e94-7842336e50a8/07576000-cc2f-11ee-b20d-39b621d226e3/file-6y3zli1lsnjrugx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsnjrugw/97f6ed68-0662-482a-8e94-7842336e50a8/07576000-cc2f-11ee-b20d-39b621d226e3/file-6y3zli1lsnjrugx.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) 31st January Morning Shift Mathematics - Circle Question 15 English Explanation"></p>
<p>$$\begin{aligned}
& 2 x+3 y=12 \\
& 3 x-2 y=5
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 13 x=39 \\
& x=3, y=2
\end{aligned}$$</p>
<p>Center of given circle is $$(5,-2)$$</p>
<p>Radius $$\sqrt{25+4-13}=4$$</p>
<p>$$\begin{aligned}
& \therefore \mathrm{CM}=\sqrt{4+16}=5 \sqrt{2} \\
& \therefore \mathrm{CP}=\sqrt{16+20}=6
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
NnZUsbM8LvOKxeUN | maths | circle | chord-with-a-given-middle-point | If the chord y = mx + 1 of the circle $${x^2}\, + \,{y^2} = 1$$ subtends an angle of measure $${45^ \circ }$$ at the major segment of the circle then value of m is : | [{"identifier": "A", "content": "$$2\\, \\pm \\,\\sqrt 2 \\,\\,$$ "}, {"identifier": "B", "content": "$$ - \\,2\\, \\pm \\,\\sqrt 2 \\,$$ "}, {"identifier": "C", "content": "$$- 1\\, \\pm \\,\\sqrt 2 \\,\\,$$"}, {"identifier": "D", "content": "none of these"}] | ["C"] | null | Equation of circle $${x^2} + {y^2} = 1 = {\left( 1 \right)^2}$$
<br><br>$$ \Rightarrow {x^2} + {y^2} = {\left( {y - mx} \right)^2}$$
<br><br>$$ \Rightarrow {x^2} = {m^2}{x^2} - 2\,\,mxy;$$
<br><br>$$ \Rightarrow {x^2}\left( {1 - {m^2}} \right) + 2mxy = 0.$$
<br><br>Which represents the pair of lines between which the angle is $${45^ \circ }.$$
<br><br>$$\tan 45 = \pm {{2\sqrt {{m^2} - 0} } \over {1 - {m^2}}} = {{ \pm 2m} \over {1 - {m^2}}};$$
<br><br>$$ \Rightarrow 1 - {m^2} = \pm 2m$$
<br><br>$$ \Rightarrow {m^2} \pm 2m - 1 = 0$$
<br><br>$$ \Rightarrow m = {{ - 2 \pm \sqrt {4 + 4} } \over 2}$$
<br><br>$$ = {{ - 2 \pm 2\sqrt 2 } \over 2}$$
<br><br>$$ = - 1 \pm \sqrt 2 .$$ | mcq | aieee-2002 |
XkwXCuYHq4WaJyZN | maths | circle | chord-with-a-given-middle-point | The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is : | [{"identifier": "A", "content": "$${x^2}\\, + \\,{y^2} - \\,2x\\, + \\,2y\\,\\, = \\,62$$ "}, {"identifier": "B", "content": "$${x^2}\\, + \\,{y^2} + \\,2x\\, - \\,2y\\,\\, = \\,62$$ "}, {"identifier": "C", "content": "$${x^2}\\, + \\,{y^2} + \\,2x\\, - \\,2y\\,\\, = \\,47$$"}, {"identifier": "D", "content": "$${x^2}\\, + \\,{y^2} - \\,2x\\, + \\,2y\\,\\, = \\,47$$ "}] | ["D"] | null | $$\pi {r^2} = 154 \Rightarrow r = 7$$
<br><br>For center on solving equation
<br><br>$$2x - 3y = 5\& 3x - 4y = 7$$
<br><br>we get $$x = 1,\,y = - 1$$
<br><br>$$\therefore$$ center $$=(1,-1)$$
<br><br>Equation of circle,
<br><br>$${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$$
<br><br>$${x^2} + {y^2} - 2x + 2y = 47$$ | mcq | aieee-2003 |
gy3TyrWxXDlyzSzo | maths | circle | chord-with-a-given-middle-point | If the lines 2x + 3y + 1 + 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference $$10\,\pi $$, then the equation of the circle is : | [{"identifier": "A", "content": "$${x^2}\\, + \\,{y^2} + \\,2x\\, - \\,2y - \\,23\\,\\, = 0$$ "}, {"identifier": "B", "content": "$${x^2}\\, + \\,{y^2} - \\,2x\\, - \\,2y - \\,23\\,\\, = 0$$ "}, {"identifier": "C", "content": "$${x^2}\\, + \\,{y^2} + \\,2x\\, + \\,2y - \\,23\\,\\, = 0$$ "}, {"identifier": "D", "content": "$${x^2}\\, + \\,{y^2} - \\,2x\\, + \\,2y - \\,23\\,\\, = 0$$ "}] | ["D"] | null | Two diameters are along
<br><br>$$2x+3y+1=0$$ and $$3x-y-4=0$$
<br><br>solving we get center $$(1,-1)$$
<br><br>circumference $$ = 2\pi r = 10\pi $$
<br><br>$$\therefore$$ $$r=5$$.
<br><br>Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$
<br><br>$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$
| mcq | aieee-2004 |
V5nqVAwPhcaJmyA3 | maths | circle | chord-with-a-given-middle-point | Intercept on the line y = x by the circle $${x^2}\, + \,{y^2} - 2x = 0$$ is AB. Equation of the circle on AB as a diameter is : | [{"identifier": "A", "content": "$$\\,{x^2}\\, + \\,{y^2} + \\,x\\, - \\,y\\,\\, = 0$$ "}, {"identifier": "B", "content": "$$\\,{x^2}\\, + \\,{y^2} - \\,x\\, + \\,y\\,\\, = 0$$"}, {"identifier": "C", "content": "$$\\,{x^2}\\, + \\,{y^2} + \\,x\\, + \\,y\\,\\, = 0$$ "}, {"identifier": "D", "content": "$$\\,{x^2}\\, + \\,{y^2} - \\,x\\, - \\,y\\,\\, = 0$$ "}] | ["D"] | null | Solving $$y=x$$ and the circle
<br><br>$${x^2} + {y^2} - 2x = 0,$$ we get
<br><br>$$x = 0,y = 0$$ and $$x=1,$$ $$y=1$$
<br><br>$$\therefore$$ Extremities of diameter of the required circle are
<br><br>$$\left( {0,0} \right)$$ and $$\left( {1,1} \right)$$. Hence, the equation of circle is
<br><br>$$\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0$$
<br><br>$$ \Rightarrow {x^2} + {y^2} - x - y = 0$$ | mcq | aieee-2004 |
wDezU4MM76FhKiUT | maths | circle | chord-with-a-given-middle-point | If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then : | [{"identifier": "A", "content": "$$3{a^2} - 10ab + 3{b^2} = 0$$ "}, {"identifier": "B", "content": "$$3{a^2} - 2ab + 3{b^2} = 0$$ "}, {"identifier": "C", "content": "$$3{a^2} + 10ab + 3{b^2} = 0$$ "}, {"identifier": "D", "content": "$$3{a^2} + 2ab + 3{b^2} = 0$$ "}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265897/exam_images/tcowhr59pgnhld1bijkk.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Circle Question 148 English Explanation">
<br><br>As per question area of one sector $$=3$$ area of another sector
<br><br>$$ \Rightarrow $$ at center by one sector $$ = 3 \times $$ angle at center by another sector
<br><br>Let one angle be $$\theta $$ then other $$=30$$
<br><br>Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$
<br><br>$$\therefore$$ Angle between the diameters represented by combined equation
<br><br>$$a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0$$ is $${45^ \circ }$$
<br><br>$$\therefore$$ Using $$tan$$ $$\theta $$ $$ = {{2\sqrt {{h^2} - ab} } \over {a + b}}$$
<br><br>we get $$\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}$$
<br><br>$$ \Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}$$
<br><br>$$ \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)$$
<br><br>$$ \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab$$
<br><br>$$ \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0$$ | mcq | aieee-2005 |
3U5RrWuDdNdu7DGZ | maths | circle | chord-with-a-given-middle-point | If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi $$ square units, the equation of the circle is : | [{"identifier": "A", "content": "$$\\,{x^2} + {y^2} + 2x\\, - 2y - 47 = 0\\,$$ "}, {"identifier": "B", "content": "$$\\,{x^2} + {y^2} + 2x\\, - 2y - 62 = 0\\,$$"}, {"identifier": "C", "content": "$${x^2} + {y^2} - 2x\\, + 2y - 62 = 0$$ "}, {"identifier": "D", "content": "$${x^2} + {y^2} - 2x\\, + 2y - 47 = 0$$"}] | ["D"] | null | Point of intersection of $$3x - 4y - 7 = 0$$ and
<br><br>$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the
<br><br>circle and radius $$=7$$
<br><br>$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$
<br><br>$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$ | mcq | aieee-2006 |
YnQUn55VJcVWDi9o | maths | circle | chord-with-a-given-middle-point | Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is : | [{"identifier": "A", "content": "$${x^2} + {y^2} = {3 \\over 2}$$ "}, {"identifier": "B", "content": "$${x^2} + {y^2} = 1$$ "}, {"identifier": "C", "content": "$${x^2} + {y^2} = {{27} \\over 4}$$ "}, {"identifier": "D", "content": "$${x^2} + {y^2} = {{9} \\over 4}$$"}] | ["D"] | null | Let $$M\left( {h,k} \right)$$ be the mid point of chord $$AB$$ where
<br><br>$$\angle AOB = {{2\pi } \over 3}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263288/exam_images/uamjsjszwgsvyetzibag.webp" loading="lazy" alt="AIEEE 2006 Mathematics - Circle Question 146 English Explanation">
<br><br>$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$
<br><br>$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$
<br><br>$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$
<br><br>$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is
<br><br>$${x^2} + {y^2} = {9 \over 4}$$ | mcq | aieee-2006 |
wr3znUdfJFsVkS7T | maths | circle | chord-with-a-given-middle-point | If one of the diameters of the circle, given by the equation, $${x^2} + {y^2} - 4x + 6y - 12 = 0,$$ is a chord of a circle $$S$$, whose centre is at $$(-3, 2)$$, then the radius of $$S$$ is : | [{"identifier": "A", "content": "$$5$$ "}, {"identifier": "B", "content": "$$10$$"}, {"identifier": "C", "content": "$$5\\sqrt 2 $$ "}, {"identifier": "D", "content": "$$5\\sqrt 3 $$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266458/exam_images/taibjlarssmymugj1mse.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - Circle Question 133 English Explanation">
<br><br>Center of $$S$$ : $$O(-3, 2)$$ center of given circle $$A(2, -3)$$
<br><br>$$ \Rightarrow OA = 5\sqrt 2 $$
<br><br>Also $$AB=5$$ (as $$AB=r$$ of the given circle)
<br><br>$$ \Rightarrow $$ Using pythagoras theorem in $$\Delta OAB$$
<br><br>$$r = 5\sqrt 3 $$ | mcq | jee-main-2016-offline |
KGtTt7pYyTrsPHIXJGpXP | maths | circle | chord-with-a-given-middle-point | A circle passes through (−2, 4) and touches the y-axis at (0, 2). Which one of the following equations can represent a diameter of this circle? | [{"identifier": "A", "content": "4x + 5y \u2212 6 = 0"}, {"identifier": "B", "content": "2x \u2212 3y + 10 = 0"}, {"identifier": "C", "content": "3x + 4y \u2212 3 = 0"}, {"identifier": "D", "content": "5x + 2y + 4 = 0 "}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264340/exam_images/ylrnxacrnykkxgmlb3iu.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Mathematics - Circle Question 123 English Explanation">
<br>EF = perpendicular bisector of chord AB
<br><br>BG = perpendicular to y-axis
<br><br>Here C = center of the circle
<br><br>mid-point of chord AB, D = ($$-$$ 1, 3)
<br><br>slope of AB = $${{4 - 2} \over { - 2 - 0}}$$ = $$-$$ 1
<br><br>$$ \because $$ EF $$ \bot $$ AB
<br><br>$$ \therefore $$ Slope of EF = 1
<br><br>Equation of EF, y $$-$$ 3 = 1 (x + 1)
<br><br>$$ \Rightarrow $$ y = x + 4 . . . .(i)
<br><br>Equation of BG
<br><br>y = 2 . . . . (ii)
<br><br>From equations (i) and (ii)
<br><br>x = $$-$$ 2, y = 2
<br><br>since C be the point of intersection of EF and BG, therefore center, C = ($$-$$ 2, 2)
<br><br>Now coordinates of center C satiesfy the equation
<br><br>2x $$-$$ 3y + 10 = 0
<br><br>Hence 2x $$-$$ 3y + 10 = 0 is the equation of the diameter | mcq | jee-main-2016-online-9th-april-morning-slot |
Bb1rY9owBIJLvGWfcTGpH | maths | circle | chord-with-a-given-middle-point | If two parallel chords of a circle, having diameter 4units, lie on the opposite sides of the center and subtend angles $${\cos ^{ - 1}}\left( {{1 \over 7}} \right)$$ and sec<sup>$$-$$1</sup> (7) at the center respectivey, then the distance between these chords, is : | [{"identifier": "A", "content": "$${4 \\over {\\sqrt 7 }}$$ "}, {"identifier": "B", "content": "$${8 \\over {\\sqrt 7 }}$$ "}, {"identifier": "C", "content": "$${8 \\over 7}$$ "}, {"identifier": "D", "content": "$${16 \\over 7}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266917/exam_images/e9ndaij1z9oqlqstuhea.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Circle Question 126 English Explanation">
<br><br>Since cos2$$\theta $$ = 1/7 $$ \Rightarrow $$ 2 cos<sup>2</sup> Q $$-$$ 1 = 1/7
<br><br>$$ \Rightarrow $$ 2 cos<sup>2</sup>$$\theta $$ = 8/7
<br><br>$$ \Rightarrow $$ cos<sup>2</sup> $$\theta $$ = 4/7
<br><br>$$ \Rightarrow $$ cos<sup>2</sup>$$\theta $$ = $${4 \over 7}$$
<br><br>$$ \Rightarrow $$ cos<sup>2</sup>$$\theta $$ = $${2 \over {\sqrt 7 }}$$
<br><br>Also, sec<sup>2</sup>$$\phi $$ = 7 = $${1 \over {2{{\cos }^2}\phi - 1}}$$ 7
<br><br>= cos<sup>2</sup>$$\phi $$ $$-$$ 1 = $${1 \over 7}$$
<br><br>= 2 cos<sup>2</sup> $$\phi $$ = $${8 \over 7}$$
<br><br>= cos$$\phi $$ = $${2 \over {\sqrt 7 }}$$
<br><br>P<sub>1</sub>P<sub>2</sub> = r cos$$\theta $$ + r cos$$\phi $$
<br><br>= $${4 \over {\sqrt 7 }} + {4 \over {\sqrt 7 }}$$ = $${8 \over {\sqrt 7 }}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
NN0JSJAcrtZgFuxsgmSCP | maths | circle | chord-with-a-given-middle-point | A line drawn through the point P(4, 7) cuts the circle x<sup>2</sup> + y<sup>2</sup> = 9 at the points A and B. Then PA⋅PB is equal to : | [{"identifier": "A", "content": "53"}, {"identifier": "B", "content": "56"}, {"identifier": "C", "content": "74"}, {"identifier": "D", "content": "65"}] | ["B"] | null | P(4, 7). Here, x = 4, y = 7
<br><br>$$ \therefore $$ PA $$ \times $$ PB = PT<sup>2</sup>
<br><br>Also; PT = $$\sqrt {{x^2} + {y^2} - {{\left( {x - y} \right)}^2}} $$
<br><br>$$ \Rightarrow $$ PT = $$\sqrt {16 + 49 - 9} $$ = $$\sqrt {56} $$
<br><br>$$ \Rightarrow $$ PT<sup>2</sup> = 56
<br><br>$$ \therefore $$ PA $$ \times $$ PB = 56 | mcq | jee-main-2017-online-9th-april-morning-slot |
R6YFDH6A7gVn9tALWCeel | maths | circle | chord-with-a-given-middle-point | If the area of an equilateral triangle inscribed in the circle x<sup>2</sup> + y<sup>2</sup>
+ 10x + 12y + c = 0 is $$27\sqrt 3 $$ sq units then c is equal to : | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "$$-$$ 25"}, {"identifier": "D", "content": "13"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266928/exam_images/dalvkc5ol09xl6tzqcfw.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Circle Question 118 English Explanation">
<br>$$3\left( {{1 \over 2}{r^2}.\sin {{120}^o}} \right) = 27\sqrt 3 $$
<br><br>$${{{r^2}} \over 2}{{\sqrt 3 } \over 2} = {{27\sqrt 3 } \over 3}$$
<br><br>$${r^2} = {{108} \over 3} = 36$$
<br><br>Radius $$ = \sqrt {25 + 36 - C} = \sqrt {36} $$
<br><br>$$C = 25$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
ezwzoSZQrfXnvICEWi3rsa0w2w9jx5zq07t | maths | circle | chord-with-a-given-middle-point | If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90<sup>o</sup>, then
the length (in cm) of their common chord is : | [{"identifier": "A", "content": "$${{13} \\over 5}$$"}, {"identifier": "B", "content": "$${{60} \\over {13}}$$"}, {"identifier": "C", "content": "$${{120} \\over {13}}$$"}, {"identifier": "D", "content": "$${{13} \\over 2}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264886/exam_images/tuxfhzteere5dc5zztu5.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Morning Slot Mathematics - Circle Question 103 English Explanation">
C<sub>1</sub>C<sub>2</sub> = $$\sqrt {{{12}^2} + {5^2}} $$ = 13<br><br>
Area of $$\Delta $$AC<sub>1</sub>C<sub>2</sub> = $${1 \over 2}.12.5 = {1 \over 2}.13{{AB} \over 2} \Rightarrow AB = {{120} \over {13}}units$$
| mcq | jee-main-2019-online-12th-april-morning-slot |
8Hmto3n9u9rvWrSaFljgy2xukfqcq27h | maths | circle | chord-with-a-given-middle-point | If the length of the chord of the circle,
<br/>x<sup>2</sup> + y<sup>2</sup> = r<sup>2</sup>
(r > 0) along the line, y – 2x = 3 is r,
<br/>then r<sup>2</sup>
is equal to :
| [{"identifier": "A", "content": "$${9 \\over 5}$$"}, {"identifier": "B", "content": "$${{24} \\over 5}$$"}, {"identifier": "C", "content": "$${{12} \\over 5}$$"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267098/exam_images/ce8t39fixejz7c7rqvi3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Mathematics - Circle Question 93 English Explanation">
<br><br>In right $$\Delta $$CDB,
<br><br>sin 60<sup>o</sup> = $${{CD} \over r}$$
<br><br>$$ \Rightarrow $$ CD = $$r \times {{\sqrt 3 } \over 2}$$ = $${{\sqrt 3 r} \over 2}$$
<br><br>Now equation of AB is
<br><br>y – 2x – 3 = 0
<br><br>So $${{\sqrt 3 r} \over 2}$$ = $${{\left| {0 + 0 - 3} \right|} \over {\sqrt 5 }}$$
<br><br>$$ \Rightarrow $$ $${{\sqrt 3 r} \over 2}$$ = $${3 \over {\sqrt 5 }}$$
<br><br>$$ \Rightarrow $$ r = $${{2\sqrt 3 } \over 5}$$
<br><br>$$ \Rightarrow $$ r<sup>2</sup> = $${{12} \over 5}$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
J6DqCFwbzQxYPjKuAF1klrjlmf3 | maths | circle | chord-with-a-given-middle-point | If one of the diameters of the circle x<sup>2</sup> + y<sup>2</sup> - 2x - 6y + 6 = 0 is a chord of another circle 'C',
whose center is at (2, 1), then its radius is ________. | [] | null | 3 | Circle x<sup>2</sup> + y<sup>2</sup> - 2x - 6y + 6 = 0 has centre
O<sub>1</sub>(1, 3) and radius r
= 2.
<br><br>Let centre O<sub>2</sub>
(2, 1) of required circle and its
radius being r.
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266179/exam_images/zp0tmjnahzn7ywtwha3e.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Circle Question 92 English Explanation">
<br><br>Distance between (1, 3) and (2, 1) is $$\sqrt 5 $$<br><br>$$ \therefore $$ $${\left( {\sqrt 5 } \right)^2} + {(2)^2} = {r^2}$$<br><br>$$ \Rightarrow r = 3$$ | integer | jee-main-2021-online-24th-february-morning-slot |
9DizyxVsY5PYMF4DAQ1kluxkl2d | maths | circle | chord-with-a-given-middle-point | If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x<sup>2</sup> + y<sup>2</sup> = 1 is a circle of radius r, then r is equal to : | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | Let P(h, k) and point on the circle is (cos$$\theta$$, sin$$\theta$$)<br><br>$$ \therefore $$ $${{3 + \cos \theta } \over 2} = h$$ and $${{2 + \sin \theta } \over 2} = k$$<br><br>cos$$\theta$$ = 2h $$-$$ 3 and sin$$\theta$$ = 2h $$-$$ 2<br><br>Squaring and adding we get<br><br>$${(2h - 3)^2} + {(2h - 2)^2} = 1$$<br><br>$$ \Rightarrow 4{x^2} - 12x + 9 + 4{y^2} - 8y + 4 = 1$$<br><br>$$ \Rightarrow 4{x^2} + 4{y^2} - 12x - 8y + 12 = 0$$<br><br>$$ \Rightarrow {x^2} + {y^2} - 3x - 2y + 3 = 0$$<br><br>Radius = $$\sqrt {{9 \over 4} + 1 - 3} = {1 \over 2}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
1ktbehi2f | maths | circle | chord-with-a-given-middle-point | If a line along a chord of the circle 4x<sup>2</sup> + 4y<sup>2</sup> + 120x + 675 = 0, passes through the point ($$-$$30, 0) and is tangent to the parabola y<sup>2</sup> = 30x, then the length of this chord is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "5$${\\sqrt 3 }$$"}, {"identifier": "D", "content": "3$${\\sqrt 5 }$$"}] | ["D"] | null | Equation of tangent to y<sup>2</sup> = 30 x<br><br>y = mx + $${{30} \over {4m}}$$<br><br>Pass thru ($$-$$30, 0) : a = $$-$$30m + $${{30} \over {4m}}$$ $$\Rightarrow$$ m<sup>2</sup> = 1/4<br><br>$$\Rightarrow$$ m = $${1 \over 2}$$ or m = $$-$$$${1 \over 2}$$<br><br>At m = $${1 \over 2}$$ : y = $${x \over 2}$$ + 15 $$\Rightarrow$$ x $$-$$ 2y + 30 = 0<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265047/exam_images/smft0g8rdwr7refltkmp.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267004/exam_images/fs750d5ymgmfsjcrszl0.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263855/exam_images/wmekhvnakbugxniy2cw2.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Circle Question 70 English Explanation"></picture> <br><br>l<sub>AB</sub> = 2$$\sqrt {{R^2} - {P^2}} $$ = 2$$\sqrt {{{225} \over 4} - {{225} \over 5}} $$<br><br>$$\Rightarrow$$ l<sub>AB</sub> = 30 . $$\sqrt {{1 \over {20}}} $$ = $${{{15} \over {\sqrt 5 }}}$$ = 3$${\sqrt 5 }$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktd2kuim | maths | circle | chord-with-a-given-middle-point | A circle C touches the line x = 2y at the point (2, 1) and intersects the circle <br/><br/>C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> + 2y $$-$$ 5 = 0 at two points P and Q such that PQ is a diameter of C<sub>1</sub>. Then the diameter of C is : | [{"identifier": "A", "content": "$$7\\sqrt 5 $$"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "$$\\sqrt {285} $$"}, {"identifier": "D", "content": "$$4\\sqrt {15} $$"}] | ["A"] | null | (x $$-$$ 2)<sup>2</sup> + (y $$-$$ 1)<sup>2</sup> + $$\lambda$$(x $$-$$ 2y) = 0<br><br>C : x<sup>2</sup> + y<sup>2</sup> + x($$\lambda$$ $$-$$ 4) + y($$-$$2 $$-$$2$$\lambda$$) + 5 = 0<br><br>C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> + 2y $$-$$ 5 = 0<br><br>S<sub>1</sub> $$-$$ S<sub>2</sub> = 0 (Equation of PQ)<br><br>($$\lambda$$ $$-$$ 4)x $$-$$ (2$$\lambda$$ + 4)y + 10 = 0 Passes through (0, $$-$$1)<br><br>$$\Rightarrow$$ $$\lambda$$ = $$-$$7<br><br>C : x<sup>2</sup> + y<sup>2</sup> $$-$$ 11x + 12y + 5 = 0<br><br>= $${{\sqrt {245} } \over 4}$$<br><br>Diameter = $$7\sqrt 5 $$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1l55j0dpj | maths | circle | chord-with-a-given-middle-point | <p>If one of the diameters of the circle $${x^2} + {y^2} - 2\sqrt 2 x - 6\sqrt 2 y + 14 = 0$$ is a chord of the circle $${(x - 2\sqrt 2 )^2} + {(y - 2\sqrt 2 )^2} = {r^2}$$, then the value of r<sup>2</sup> is equal to ____________.</p> | [] | null | 10 | For $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$
<br/><br/>
$$
\text { Radius }=\sqrt{(\sqrt{2})^{2}+(3 \sqrt{2})^{2}-14}=\sqrt{6}
$$
<br/><br/>
$\Rightarrow$ Diameter $=2 \sqrt{6}$
<br/><br/>
If this diameter is chord to
<br/><br/>
$$
\begin{aligned}
&(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\\\
&\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\\\
&\Rightarrow r^{2}=6+4=10 \\\\
&\Rightarrow r^{2}=10
\end{aligned}
$$ | integer | jee-main-2022-online-28th-june-evening-shift |
1l5ajotbj | maths | circle | chord-with-a-given-middle-point | <p>Let the abscissae of the two points P and Q be the roots of $$2{x^2} - rx + p = 0$$ and the ordinates of P and Q be the roots of $${x^2} - sx - q = 0$$. If the equation of the circle described on PQ as diameter is $$2({x^2} + {y^2}) - 11x - 14y - 22 = 0$$, then $$2r + s - 2q + p$$ is equal to __________.</p> | [] | null | 7 | <p>Let $$P({x_1},{y_1})$$ & $$Q({x_2},{y_2})$$</p>
<p>$$\therefore$$ Roots of $$2{x^2} - rx + p = 0$$ are $${x_1},\,{x_2}$$</p>
<p>and roots of $${x^2} - sx - q = 0$$ are $${y_1},\,{y_2}$$.</p>
<p>$$\therefore$$ Equation of circle $$ \equiv (x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$$</p>
<p>$$ \Rightarrow {x^2} - ({x_1} + {x_2})x + {x_1}{x_2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} = 0$$</p>
<p>$$ \Rightarrow {x^2} - {r \over 2}x + {p \over 2} + {y^2} + sy - q = 0$$</p>
<p>$$ \Rightarrow 2{x^2} + 2{y^2} - rx + 2sy + p - 2q = 0$$</p>
<p>Compare with $$2{x^2} + 2{y^2} - 11x - 14y - 22 = 0$$</p>
<p>We get $$r = 11,\,s = 7,\,p - 2q = - 22$$</p>
<p>$$ \Rightarrow 2r + s + p - 2q = 22 + 7 - 22 = 7$$</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l6hyv47p | maths | circle | chord-with-a-given-middle-point | <p>Let the abscissae of the two points $$P$$ and $$Q$$ on a circle be the roots of $$x^{2}-4 x-6=0$$ and the ordinates of $$\mathrm{P}$$ and $$\mathrm{Q}$$ be the roots of $$y^{2}+2 y-7=0$$. If $$\mathrm{PQ}$$ is a diameter of the circle $$x^{2}+y^{2}+2 a x+2 b y+c=0$$, then the value of $$(a+b-c)$$ is _____________.</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "14"}, {"identifier": "D", "content": "16"}] | ["A"] | null | <p>Abscissae of PQ are roots of $${x^2} - 4x - 6 = 0$$</p>
<p>Ordinates of PQ are roots of $${y^2} + 2y - 7 = 0$$</p>
<p>and PQ is diameter</p>
<p>$$\Rightarrow$$ Equation of circle is</p>
<p>$${x^2} + {y^2} - 4x + 2y - 13 = 0$$</p>
<p>But, given $${x^2} + {y^2} + 2ax + 2by + c = 0$$</p>
<p>By comparison $$a = - 2,b = 1,c = - 13$$</p>
<p>$$ \Rightarrow a + b - c = - 2 + 1 + 13 = 12$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
ldoa8b8u | maths | circle | chord-with-a-given-middle-point | The set of all values of $a^{2}$ for which the line $x+y=0$ bisects two distinct chords drawn from a point $\mathrm{P}\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2 x^{2}+2 y^{2}-(1+a) x-(1-a) y=0$, is equal to : | [{"identifier": "A", "content": "$(0,4]$"}, {"identifier": "B", "content": "$(4, \\infty)$"}, {"identifier": "C", "content": "$(2,12]$"}, {"identifier": "D", "content": "$(8, \\infty)$"}] | ["D"] | null | $x^{2}+y^{2}-\frac{(1+a) x}{2}-\frac{(1-a) y}{2}=0$
<br><br>Centre $\left(\frac{1+\mathrm{a}}{4}, \frac{1-\mathrm{a}}{4}\right) \Rightarrow(\mathrm{h}, \mathrm{k})$
<br><br>$\mathrm{P}\left(\frac{1+\mathrm{a}}{2}, \frac{1-\mathrm{a}}{2}\right) \Rightarrow(2 \mathrm{h}, 2 \mathrm{k})$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lefxmkpv/6d2d69d3-c054-4a9a-b9ad-79779d129ce3/e8127240-b2d3-11ed-8169-e1635469e777/file-1lefxmkpw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lefxmkpv/6d2d69d3-c054-4a9a-b9ad-79779d129ce3/e8127240-b2d3-11ed-8169-e1635469e777/file-1lefxmkpw.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Circle Question 39 English Explanation">
<br>Equation of chord $\Rightarrow \mathrm{T}=\mathrm{S}_{1}$
<br><br>$\Rightarrow(\mathrm{x}-\mathrm{y}) \lambda-\frac{2 \mathrm{~h}(\mathrm{x}+\lambda)}{2}-\frac{(2 \mathrm{k})(\mathrm{y}-\lambda)}{2}$
<br><br>$=2 \lambda^{2}-2 \mathrm{~h}(\lambda)+2 \mathrm{k} \lambda$
<br><br>Now, $\lambda(2 \mathrm{~h}, 2 \mathrm{k})$ satisfies the chord
<br><br>$\therefore(2 \mathrm{~h}-2 \mathrm{k}) \lambda-\mathrm{h}(\mathrm{x}+\lambda)-\mathrm{k}(\mathrm{y}-\lambda)$
<br><br>$\Rightarrow 2 \lambda^{2}+4 \mathrm{k} \lambda-4 \mathrm{~h} \lambda+\mathrm{h} \lambda-\mathrm{k} \lambda+\mathrm{hx}+\mathrm{ky}=0$
<br><br>$\Rightarrow 2 \lambda^{2}+\lambda(3 \mathrm{k}-3 \mathrm{~h})+\mathrm{ky}+\mathrm{hx}=0$
<br><br>$\Rightarrow \mathrm{D}>0$
<br><br>$\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8(\mathrm{ky}+\mathrm{hx})>0$
<br><br>$\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8\left(2 \mathrm{k}^{2}+2 \mathrm{~h}^{2}\right)>0$
<br><br>$\Rightarrow-7 \mathrm{k}^{2}-7 \mathrm{~h}^{2}-18 \mathrm{kh}>0$
<br><br>$\Rightarrow 7 \mathrm{k}^{2}+7 \mathrm{~h}^{2}+18 \mathrm{kh}<0$
<br><br>$\Rightarrow 7\left(\frac{1-\mathrm{a}}{4}\right)^{2}+7\left(\frac{1+\mathrm{a}}{4}\right)^{2}+18\left(\frac{1-\mathrm{a}^{2}}{16}\right)<0$
<br><br>$\Rightarrow 7\left[\frac{2\left(1+\mathrm{a}^{2}\right)}{16}\right]+\frac{18\left(1-\mathrm{a}^{2}\right)}{16}<0, \quad \mathrm{a}^{2}=\mathrm{t}$
<br><br>$\Rightarrow \frac{7}{8}(1+\mathrm{t})+\frac{18(1-\mathrm{t})}{16}<0$
<br><br>$\Rightarrow \frac{14+14 \mathrm{t}+18-18 \mathrm{t}}{16}<0$
<br><br>$\Rightarrow 4 \mathrm{t}>32$
<br><br>$$ \Rightarrow $$ $\mathrm{t}>8 $
<br><br>$$ \Rightarrow $$ $ \mathrm{a}^{2}>8$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldwwp6i8 | maths | circle | chord-with-a-given-middle-point | <p>The locus of the mid points of the chords of the circle $${C_1}:{(x - 4)^2} + {(y - 5)^2} = 4$$ which subtend an angle $${\theta _i}$$ at the centre of the circle $$C_1$$, is a circle of radius $$r_i$$. If $${\theta _1} = {\pi \over 3},{\theta _3} = {{2\pi } \over 3}$$ and $$r_1^2 = r_2^2 + r_3^2$$, then $${\theta _2}$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5ga481/2ad2340b-953f-4942-9834-237efb8eb334/3bee0510-ad10-11ed-a86d-8dfe0389db88/file-1le5ga482.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5ga481/2ad2340b-953f-4942-9834-237efb8eb334/3bee0510-ad10-11ed-a86d-8dfe0389db88/file-1le5ga482.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Circle Question 31 English Explanation"></p>
<p>$\therefore \cos \left(\frac{\theta_{1}}{2}\right)=\frac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\frac{\theta_{i}}{2}\right)$ </p>
<p>Given $r_{1}^{2}=r_{2}^{2}+r_{3}^{3}$</p>
<p>$\Rightarrow\left(\cos \left(\frac{\theta_{1}}{2}\right)\right)^{2}=\left(\cos \left(\frac{\theta_{2}}{2}\right)\right)^{2}+\left(\cos \left(\frac{\theta_{3}}{2}\right)\right)^{2}$</p>
<p>$\Rightarrow \frac{3}{4}=\cos ^{2}\left(\frac{\theta_{2}}{2}\right)+\frac{1}{4}$</p>
<p>$\Rightarrow \cos ^{2}\left(\frac{\theta_{2}}{2}\right)=\frac{1}{2}$</p>
<p>$\Rightarrow \frac{\theta_{2}}{2}=\frac{\pi}{4}$</p>
<p>$\Rightarrow \theta_{2}=\frac{\pi}{2}$</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
1lgxvzvzw | maths | circle | chord-with-a-given-middle-point | <p>A line segment AB of length $$\lambda$$ moves such that the points A and B remain on the periphery of a circle of radius $$\lambda$$. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :</p> | [{"identifier": "A", "content": "$${2 \\over 3}\\lambda $$"}, {"identifier": "B", "content": "$${3 \\over 5}\\lambda $$"}, {"identifier": "C", "content": "$${{\\sqrt {19} } \\over 7}\\lambda $$"}, {"identifier": "D", "content": "$${{\\sqrt {19} } \\over 5}\\lambda $$"}] | ["D"] | null | Given, length of $A B=\lambda$
<br><br>So, $A C=\frac{\lambda}{2}$ and $A M=\frac{2 \lambda}{5}$
<br><br>$$
C M=A C-A M=\frac{\lambda}{2}-\frac{2 \lambda}{5}=\frac{\lambda}{10}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lncsenht/b19528f0-d9c0-44df-a496-57a56c3fa6bf/168f1e10-6347-11ee-8e13-d93c7a44ebe2/file-6y3zli1lncsenhu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lncsenht/b19528f0-d9c0-44df-a496-57a56c3fa6bf/168f1e10-6347-11ee-8e13-d93c7a44ebe2/file-6y3zli1lncsenhu.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) 10th April Morning Shift Mathematics - Circle Question 26 English Explanation">
<br><br>Now, from $\triangle O C M$, we get
<br><br>$$
\begin{array}{ll}
&O M^2=C M^2+O C^2 \\\\
&\Rightarrow x^2+y^2=\left(\frac{\lambda}{10}\right)^2+\left(\lambda^2-\frac{\lambda^2}{4}\right) \\\\
&\Rightarrow x^2+y^2=\frac{\lambda^2}{100}+\frac{3 \lambda^2}{4}=\frac{\lambda^2+75 \lambda^2}{100} \\\\
&\Rightarrow x^2+y^2=\frac{76 \lambda^2}{100}=\frac{19 \lambda^2}{25} \\\\
&\therefore x^2+y^2=\left(\frac{\sqrt{19} \lambda}{5}\right)^2
\end{array}
$$
<br><br>So, radius is $\frac{\sqrt{19} \lambda}{5}$. | mcq | jee-main-2023-online-10th-april-morning-shift |
lsamkf38 | maths | circle | chord-with-a-given-middle-point | Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $\mathrm{P}$ and $\mathrm{Q}$. Then, the length of $\mathrm{PQ}$ is : | [{"identifier": "A", "content": "$\\frac{1}{2}$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$\\frac{1}{\\sqrt{2}}$"}, {"identifier": "D", "content": "$\\sqrt{2}$"}] | ["C"] | null | Let mid-point is $(x, y)$
<br/><br/>$$
\begin{aligned}
& x^2+y^2-2 y=0 \\\\
& x x_1+y y_1-\left(y+y_1\right)=x_1^2+y_1^2-2 y_1
\end{aligned}
$$
<br/><br/>It is passing through origin
<br/><br/>$$
\begin{aligned}
& \text { So, } 0+0-\left(0+y_1\right)=x_1^2+y_1^2-2 y_1 \\\\
& \Rightarrow -y_1=x_1^2+y_1^2-2 y_1 \\\\
& \Rightarrow x_1^2+y_1^2-y_1=0
\end{aligned}
$$
<br/><br/>$$
x^2+y^2-y=0
$$ ............ (1)
<br/><br/>$\because$ It intersects the line $x+y=1$
<br/><br/>So put $x=1-y$ in equation (1)
<br/><br/>$$
\begin{aligned}
& (1-y)^2+y^2-y=0 \\\\
& 2 y^2-3 y+1=0
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \Rightarrow (y-1)(2 y-1)=0 \\\\ & \Rightarrow y=1, \frac{1}{2} \\\\ & \therefore P(0,1) \text { and } Q\left(\frac{1}{2}, \frac{1}{2}\right)\end{aligned}$
<br/><br/>So, $P Q=\sqrt{\left(\frac{1}{2}-0\right)^2+\left(\frac{1}{2}-1\right)^2}=\frac{1}{\sqrt{2}}$ | mcq | jee-main-2024-online-1st-february-evening-shift |
lv2eruuk | maths | circle | chord-with-a-given-middle-point | <p>Let $$\mathrm{C}$$ be a circle with radius $$\sqrt{10}$$ units and centre at the origin. Let the line $$x+y=2$$ intersects the circle $$\mathrm{C}$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. Let $$\mathrm{MN}$$ be a chord of $$\mathrm{C}$$ of length 2 unit and slope $$-1$$. Then, a distance (in units) between the chord PQ and the chord $$\mathrm{MN}$$ is</p> | [{"identifier": "A", "content": "$$3-\\sqrt{2}$$\n"}, {"identifier": "B", "content": "$$2-\\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{2}-1$$\n"}, {"identifier": "D", "content": "$$\\sqrt{2}+1$$"}] | ["A"] | null | <p>$$\text { Let the line by } x+y=\lambda$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhfpfz5/cd7503a4-2670-4a67-8200-c99b5e7fc89f/49a14910-1803-11ef-b156-f754785ad3ce/file-1lwhfpfz6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhfpfz5/cd7503a4-2670-4a67-8200-c99b5e7fc89f/49a14910-1803-11ef-b156-f754785ad3ce/file-1lwhfpfz6.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) 4th April Evening Shift Mathematics - Circle Question 8 English Explanation"></p>
<p>$$\begin{aligned}
& \therefore\left|\frac{\lambda}{\sqrt{2}}\right|=3 \\
& \therefore \quad \lambda= \pm 3 \sqrt{2}
\end{aligned}$$</p>
<p>$$\therefore$$ distance between lines $$x+y=2$$ and $$x+y=3 \sqrt{2}$$ is</p>
<p>$$\frac{3 \sqrt{2}-2}{\sqrt{2}}=3-\sqrt{2}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
1l6m5uq0x | maths | circle | director-circle | <p>Let $$C$$ be the centre of the circle $$x^{2}+y^{2}-x+2 y=\frac{11}{4}$$ and $$P$$ be a point on the circle. A line passes through the point $$\mathrm{C}$$, makes an angle of $$\frac{\pi}{4}$$ with the line $$\mathrm{CP}$$ and intersects the circle at the points $$Q$$ and $$R$$. Then the area of the triangle $$P Q R$$ (in unit $$^{2}$$ ) is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "2$$\\sqrt2$$"}, {"identifier": "C", "content": "$$8 \\sin \\left(\\frac{\\pi}{8}\\right)$$"}, {"identifier": "D", "content": "$$8 \\cos \\left(\\frac{\\pi}{8}\\right)$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb097k/86b1d3e8-f7fd-49d7-85c5-28047f9ac8ec/56adfa00-2e7f-11ed-8702-156c00ced081/file-1l7rb097l.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb097k/86b1d3e8-f7fd-49d7-85c5-28047f9ac8ec/56adfa00-2e7f-11ed-8702-156c00ced081/file-1l7rb097l.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 43 English Explanation"></p>
<p>$$QR = 2r = 4$$</p>
<p>$$P = \left( {{1 \over 2} + 2\cos {\pi \over 4}, - 1 + 2\sin {\pi \over 4}} \right)$$</p>
<p>$$ = \left( {{1 \over 2} + \sqrt 2 , - 1 + \sqrt 2 } \right)$$</p>
<p>Area of $$\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2 $$</p>
<p>$$ = 2\sqrt 2 $$ sq. units</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
9LotZu5LbanRuKpm | maths | circle | family-of-circle | The centres of a set of circles, each of radius 3, lie on the circle $${x^2}\, + \,{y^2} = 25$$. The locus of any point in the set is : | [{"identifier": "A", "content": "$$4\\, \\le \\,\\,{x^2}\\, + \\,{y^2}\\, \\le \\,\\,64$$ "}, {"identifier": "B", "content": "$${x^2}\\, + \\,{y^2}\\, \\le \\,\\,25$$ "}, {"identifier": "C", "content": "$${x^2}\\, + \\,{y^2}\\, \\ge \\,\\,25$$ "}, {"identifier": "D", "content": "$$3\\, \\le \\,\\,{x^2}\\, + \\,{y^2}\\, \\le \\,\\,9$$ "}] | ["A"] | null | For any point $$P(x,y)$$ in the given circle,
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267291/exam_images/wa96grbw4mjhrj5s0fuz.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Circle Question 159 English Explanation">
<br><br>we should have
<br><br>$$OA \le OP \le OB$$
<br><br>$$ \Rightarrow \left( {5 - 3} \right) \le \sqrt {{x^2} + {y^2}} \le 5 + 3$$
<br><br>$$ \Rightarrow 4 \le {x^2} + {y^2} \le 64$$ | mcq | aieee-2002 |
vHVMDhgdHeB7Ln2W | maths | circle | family-of-circle | A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is : | [{"identifier": "A", "content": "$${(y\\, - \\,q)^2} = \\,4\\,px$$ "}, {"identifier": "B", "content": "$${(x\\, - \\,q)^2} = \\,4\\,py$$"}, {"identifier": "C", "content": "$${(y\\, - \\,p)^2} = \\,4\\,qx$$"}, {"identifier": "D", "content": "$${(x\\, - \\,p)^2} = \\,4\\,qy$$"}] | ["D"] | null | Let the variable circle be
<br><br>$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$\therefore$$ $${p^2} + {q^2} + 2gp + 2fq + c = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>Circle $$(1)$$ touches $$x$$-axis,
<br><br>$$\therefore$$ $${g^2} - c = 0 \Rightarrow c = {g^2}.\,\,\,$$
<br><br>From $$(2)$$
<br><br>$${p^2} + {q^2} + 2gp + 2fq + {g^2} = 0\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$
<br><br>Let the other end of diameter through $$(p,q)$$ be $$(h,k),$$
<br><br>then, $${{h + p} \over 2} = - g$$ and $${{k + q} \over 2} = - f$$
<br><br>Put in $$(3)$$
<br><br>$${p^2} + {q^2} + 2p\left( { - {{h + p} \over 2}} \right) + 2q\left( { - {{k + q} \over 2}} \right) + {\left( {{{h + p} \over 2}} \right)^2} = 0$$
<br><br>$$ \Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0$$
<br><br>$$\therefore$$ locus of $$\left( {h,k} \right)$$ is $${x^2} + {p^2} - 2xp - 4yq = 0$$
<br><br>$$ \Rightarrow {\left( {x - p} \right)^2} = 4qy$$ | mcq | aieee-2004 |
R6msSj4yWbFMiKDF | maths | circle | family-of-circle | A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is : | [{"identifier": "A", "content": "an ellipse"}, {"identifier": "B", "content": "a circle "}, {"identifier": "C", "content": "a hyperbola "}, {"identifier": "D", "content": "a parabola "}] | ["D"] | null | Equation of circle with center $$(0,3)$$ and radius $$2$$ is
<br><br>$${x^2} + {\left( {y - 3} \right)^2} = 4$$
<br><br>Let locus of the variable circle is $$\left( {\alpha ,\beta } \right)$$
<br><br>As it touches $$x$$-axis.
<br><br>$$\therefore$$ It's equation is $${\left( {x - \alpha } \right)^2} + {\left( {y + \beta } \right)^2} = {\beta ^2}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263352/exam_images/a5hzdfxls1frvo9492ui.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Circle Question 161 English Explanation">
<br><br>Circle touch externally $$ \Rightarrow {c_1}{c_2} = {r_1} + {r_2}$$
<br><br>$$\therefore$$ $$\sqrt {{\alpha ^2} + {{\left( {\beta - 3} \right)}^2}} = 2 + \beta $$
<br><br>$${\alpha ^2} + {\left( {\beta - 3} \right)^2} = {\beta ^2} + 4 + 4\beta $$
<br><br>$$ \Rightarrow {\alpha ^2} = 10\left( {\beta - 1/2} \right)$$
<br><br>$$\therefore$$ Locus is $${x^2} = 10\left( {y - {1 \over 2}} \right)$$ which is parabola. | mcq | aieee-2005 |
uxc61AnvgIDOB2bO | maths | circle | family-of-circle | Consider a family of circles which are passing through the point $$(-1, 1)$$ and are tangent to $$x$$-axis. If $$(h, k)$$ are the coordinate of the centre of the circles, then the set of values of $$k$$ is given by the interval : | [{"identifier": "A", "content": "$$ - {1 \\over 2} \\le k \\le {1 \\over 2}$$ "}, {"identifier": "B", "content": "$$k \\le {1 \\over 2}$$ "}, {"identifier": "C", "content": "$$0 \\le k \\le {1 \\over 2}$$ "}, {"identifier": "D", "content": "$$k \\ge {1 \\over 2}$$ "}] | ["D"] | null | Equation of circle whose center is $$\left( {h,k} \right)$$
<br><br>i.e $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264751/exam_images/dhlchwnp8kfgccfwlmlr.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Circle Question 145 English Explanation">
<br><br>(radius of circle $$=k$$ because circle is tangent to $$x$$-axis)
<br><br>Equation of circle passing through $$\left( { - 1, + 1} \right)$$
<br><br>$$\therefore$$ $${\left( { - 1, - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2}$$
<br><br>$$ \Rightarrow 1 + {h^2} + 2h + 1 + {k^2} - 2k = {k^2}$$
<br><br>$$ \Rightarrow {h^2} + 2h - 2k + 2 = 0$$
<br><br>$$D \ge 0$$
<br><br>$$\therefore$$ $${\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \ge 0$$
<br><br>$$ \Rightarrow 4 - 4\left( { - 2k + 2} \right) \ge 0$$
<br><br>$$ \Rightarrow 1 + 2k - 2 \ge 0$$
<br><br>$$ \Rightarrow k \ge {1 \over 2}$$
| mcq | aieee-2007 |
W5qdG7lPRwsgI7vI | maths | circle | family-of-circle | The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is : | [{"identifier": "A", "content": "$$\\left( {x - 2} \\right){y^2} = 25 - {\\left( {y - 2} \\right)^2}$$ "}, {"identifier": "B", "content": "$$\\left( {y - 2} \\right){y^2} = 25 - {\\left( {y - 2} \\right)^2}$$"}, {"identifier": "C", "content": "$${\\left( {y - 2} \\right)^2}{y^2} = 25 - {\\left( {y - 2} \\right)^2}$$ "}, {"identifier": "D", "content": "$${\\left( {x - 2} \\right)^2}{y^2} = 25 - {\\left( {y - 2} \\right)^2}$$"}] | ["C"] | null | Let the center of the circle be $$(h, 2)$$
<br><br>$$\therefore$$ Equation of circle is
<br><br>$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>Differentiating with respect to $$x,$$ we get
<br><br>$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$
<br><br>$$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$
<br><br>Substituting in equation $$(1)$$ we get
<br><br>$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$
<br><br>$$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$ | mcq | aieee-2008 |
CVivCukoCBuAkUMw | maths | circle | family-of-circle | If $$P$$ and $$Q$$ are the points of intersection of the circles
<br/>$${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q $$ and $$(1, 1)$$ for : | [{"identifier": "A", "content": "all except one value of $$p$$ "}, {"identifier": "B", "content": "all except two values of $$p$$ "}, {"identifier": "C", "content": "exactly one value of $$p$$ "}, {"identifier": "D", "content": "all values of $$p$$"}] | ["A"] | null | The given circles are
<br><br>$${S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$${S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$\therefore$$ Equation of common chord $$PQ$$ is $${S_1} - {S_2} = 0$$
<br><br>$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$
<br><br>$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is
<br><br>$${S_1} + \lambda \,\,L = 0$$
<br><br>$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda $$
<br><br>$$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$
<br><br>As it passes through $$\left( {1,1} \right),$$ therefore
<br><br>$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$
<br><br>$$ \Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},$$
<br><br>which does not exist for $$p=-1$$ | mcq | aieee-2009 |
pdGMim6mel1wVHfj | maths | circle | family-of-circle | The centres of those circles which touch the circle, $${x^2} + {y^2} - 8x - 8y - 4 = 0$$, externally and also touch the $$x$$-axis, lie on : | [{"identifier": "A", "content": "a circle "}, {"identifier": "B", "content": "an ellipse which is not a circle "}, {"identifier": "C", "content": "a hyperbola"}, {"identifier": "D", "content": "a parabola "}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264280/exam_images/cr0skvuxufhkzumdnzhn.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - Circle Question 134 English Explanation">
<br><br>For the given circle,
<br><br>center : $$(4,4)$$
<br><br>radius $$=6$$
<br><br>$$6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} $$
<br><br>$${\left( {h - 4} \right)^2} = 20k + 20$$
<br><br>$$\therefore$$ locus of $$(h, k)$$ is
<br><br>$${\left( {h - 4} \right)^2} = 20\left( {y + 1} \right),$$
<br><br>which is parabola. | mcq | jee-main-2016-offline |
DtjBungYu7lolDmmqPBQA | maths | circle | family-of-circle | A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line, $$y - 4x + 3 = 0,$$ then its radius is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\sqrt 5 $$"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "1"}] | ["A"] | null | Centre $(\alpha, \beta)$ lies on line $y-4 x+3=0$
<br/><br/>Then, $\quad \beta=4 \alpha-3$
<br/><br/>And
<br/><br/>$$
\text { Radius }=\sqrt{(\alpha-2)^2+(\beta-3)^2}=\sqrt{(\alpha-4)^2+(\beta-5)^2}
$$
<br/><br/>$$
\begin{aligned}
& \alpha^2+\beta^2+ 13-4 \alpha-6 \beta=\alpha^2+\beta^2+41-8 \alpha-10 \beta \\\\
& 4 \alpha+4 \beta=28 \Rightarrow \alpha+\beta=7 \\\\
& \Rightarrow \alpha+4 \alpha-3=7 \\\\
& \Rightarrow \alpha=2, \beta=5
\end{aligned}
$$
<br/><br/>Therefore, $\quad$ Radius $=\sqrt{(2-2)^2+(5-3)^2}=2$ | mcq | jee-main-2018-online-15th-april-morning-slot |
GhHPQJZMreFqETJ1UGuYQ | maths | circle | family-of-circle | Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent, then : | [{"identifier": "A", "content": "a, b, c are in A.P."}, {"identifier": "B", "content": "$$\\sqrt a ,\\sqrt b ,\\sqrt c $$ are in A.P"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt b }} + {1 \\over {\\sqrt c }}$$ = $${1 \\over {\\sqrt a }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt b }} = {1 \\over {\\sqrt a }} + {1 \\over {\\sqrt c }}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264861/exam_images/znw3pusc82f7kl1xv763.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Circle Question 121 English Explanation">
<br><br>AB = AC + CB
<br><br>$$\sqrt {{{\left( {b + c} \right)}^2} - {{\left( {c - b} \right)}^2}} $$ = $$\sqrt {{{\left( {b + a} \right)}^2} - {{\left( {b - a} \right)}^2}} $$
<br> + $$\sqrt {{{\left( {c + a} \right)}^2} - {{\left( {c - a} \right)}^2}} $$
<br><br>$$ \Rightarrow $$ $$\sqrt {2bc} $$ = $$\sqrt {2ac} $$ + $$\sqrt {2ab} $$
<br><br>Dividing by $$\sqrt {abc} $$ we get.
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt a }}$$ = $${1 \over {\sqrt b }}$$ + $${1 \over {\sqrt c }}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
KCwGH1e7nkaCmpz92Q3rsa0w2w9jx1ytgcp | maths | circle | family-of-circle | The locus of the centres of the circles, which touch the circle, x<sup>2</sup>
+ y<sup>2</sup>
= 1 externally, also touch the y-axis and
lie in the first quadrant, is : | [{"identifier": "A", "content": "$$x = \\sqrt {1 + 2y} ,y \\ge 0$$"}, {"identifier": "B", "content": "$$y = \\sqrt {1 + 2x} ,x \\ge 0$$"}, {"identifier": "C", "content": "$$y = \\sqrt {1 + 4x} ,x \\ge 0$$"}, {"identifier": "D", "content": "$$x = \\sqrt {1 + 4y} ,y \\ge 0$$"}] | ["B"] | null | Let the centre is (h, k) & radius is h (h, k > 0)<br><br>
OP = h + 1<br><br>
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266464/exam_images/pevci3b1ygpkppdlpbgl.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Evening Slot Mathematics - Circle Question 104 English Explanation"><br>
$$\sqrt {{h^2} + {k^2}} = h + 1$$<br><br>
$$ \Rightarrow {h^2} + {k^2} = {h^2} + 2h + 1$$<br><br>
$$ \Rightarrow {k^2} = 2h + 1$$<br><br>
$$ \therefore $$ Locus is y<sup>2</sup> = 2x + 1
| mcq | jee-main-2019-online-10th-april-evening-slot |
S19NaXGqv0LGnFkd1wovy | maths | circle | family-of-circle | If a circle C passing through the point (4, 0) touches the circle x<sup>2</sup> + y<sup>2</sup> + 4x – 6y = 12 externally at the point (1, – 1), then the radius of C is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2$$\\sqrt {5} $$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$\\sqrt {37} $$"}] | ["A"] | null | x<sup>2</sup> + y<sup>2</sup> + 4x $$-$$ 6y $$-$$ 12 = 0
<br><br>Equation of tangent at (1, $$-$$ 1)
<br><br>x $$-$$ y + 2(x + 1) $$-$$ 3(y $$-$$ 1) $$-$$ 12 = 0
<br><br>3x $$-$$ 4y $$-$$ 7 = 0
<br><br>$$ \therefore $$ Equation of circle is
<br><br>(x<sup>2</sup> + y<sup>2</sup> + 4x $$-$$ 6y $$-$$ 12) + $$\lambda $$ (3x $$-$$ 4y $$-$$ 7) = 0
<br><br>It passes through (4, 0) :
<br><br>(16 + 16 $$-$$ 12) + $$\lambda $$ (12 $$-$$ 7) = 0
<br><br>$$ \Rightarrow $$ 20 + $$\lambda $$(5) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $$-$$ 4
<br><br>$$ \therefore $$ (x<sup>2</sup> + y<sup>2</sup> + 4x $$-$$ 6y $$-$$ 12) $$-$$ 4(3x $$-$$ 4y $$-$$ 7) = 0
<br><br>or x<sup>2</sup> + y<sup>2</sup> $$-$$ 8x + 10y + 16 = 0
<br><br>Radius = $$\sqrt {16 + 25 - 16} = 5$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
aLcjY2vZwkdplfoJYSjgy2xukfakl6kz | maths | circle | family-of-circle | The circle passing through the intersection of the circles, <br/>x<sup>2</sup> + y<sup>2</sup> – 6x = 0 and x<sup>2</sup> + y<sup>2</sup> – 4y = 0, having its centre on<br/> the line, 2x – 3y + 12 = 0, also passes through the point :
| [{"identifier": "A", "content": "(\u20133, 1)"}, {"identifier": "B", "content": "(1, \u20133) "}, {"identifier": "C", "content": "(\u20131, 3) "}, {"identifier": "D", "content": "(\u20133, 6) "}] | ["D"] | null | Let S be the circle passing through point of intersection of S<sub>1</sub> & S<sub>2</sub><br><br>$$ \therefore $$ S = S<sub>1</sub> + $$\lambda $$S<sub>2</sub> = 0<br><br>$$ \Rightarrow $$ $$S:({x^2} + {y^2} - 6x) + \lambda ({x^2} + {y^2} - 4y) = 0$$<br><br>$$ \Rightarrow $$ $$S:{x^2} + {y^2} - \left( {{6 \over {1 + \lambda }}} \right)x - \left( {{{4\lambda } \over {1 + \lambda }}} \right)y = 0$$ ....(1)<br><br>Centre $$\left( {{3 \over {1 + \lambda }},{{2\lambda } \over {1 + \lambda }}} \right)$$ lies on<br><br>$$2x - 3y + 12 = 0 \Rightarrow \lambda = - 3$$<br><br>put in $$(1) \Rightarrow S:{x^2} + {y^2} + 3x - 6y = 0$$<br><br>Now check options point $$( - 3,6)$$<br><br>lies on S. | mcq | jee-main-2020-online-4th-september-evening-slot |
NDpPPI6KPUzk1c1oIm1kmm35609 | maths | circle | family-of-circle | Let S<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> = 9 and S<sub>2</sub> : (x $$-$$ 2)<sup>2</sup> + y<sup>2</sup> = 1. Then the locus of center of a variable circle S which touches S<sub>1</sub> internally and S<sub>2</sub> externally always passes through the points : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 2}, \\pm {{\\sqrt 5 } \\over 2}} \\right)$$"}, {"identifier": "B", "content": "(1, $$\\pm$$ 2)"}, {"identifier": "C", "content": "$$\\left( {2, \\pm {3 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "(0, $$\\pm$$ $$\\sqrt 3 $$)"}] | ["C"] | null | S<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> = 9 ; C<sub>1</sub> (0, 0), r<sub>1</sub> = 3<br><br>S<sub>2</sub> : (x $$-$$ 2)<sup>2</sup> + y<sup>2</sup> = 1 ; C<sub>2</sub> (2, 0), r<sub>2</sub> = 1<br><br>Image<br><br>Let the variable circle S and its radius is r units.<br><br>Here S and S<sub>1</sub> touches internally<br><br>$$ \therefore $$ Distance between center,<br><br>S + S<sub>1</sub> = PC<sub>1</sub> = 3 $$-$$ r<br><br>Here S and S<sub>2</sub> touches externally <br><br>$$ \therefore $$ Distance between center,<br><br> S + S<sub>2</sub> = PC<sub>2</sub> = 1 + r<br><br>$$ \therefore $$ PC<sub>1</sub> + PC<sub>2</sub> = 4 > C<sub>1</sub> C<sub>2</sub><br><br>So locus is ellipse whose focii are C<sub>1</sub> & C<sub>2</sub> and major axis is 2a = 4 and 2ae = C<sub>1</sub>C<sub>2</sub> = 2<br><br>$$ \Rightarrow $$ $$e = {1 \over 2}$$<br><br>$$ \Rightarrow $$ $${b^2} = 4\left( {1 - {1 \over 4}} \right) = 3$$<br><br>Centre of ellipse is midpoint of C<sub>1</sub> & C<sub>2</sub> is (1, 0)<br><br>Equation of ellipse is $${{{{(x - 1)}^2}} \over {{2^2}}} + {{{y^2}} \over {{{\left( {\sqrt 3 } \right)}^2}}} = 1$$<br><br>Now by cross checking the option $$\left( {2, \pm {3 \over 2}} \right)$$ satisfied it. | mcq | jee-main-2021-online-18th-march-evening-shift |
1krrorbwe | maths | circle | family-of-circle | Let r<sub>1</sub> and r<sub>2</sub> be the radii of the largest and smallest circles, respectively, which pass through the point ($$-$$4, 1) and having their centres on the circumference of the circle x<sup>2</sup> + y<sup>2</sup> + 2x + 4y $$-$$ 4 = 0. If $${{{r_1}} \over {{r_2}}} = a + b\sqrt 2 $$, then a + b is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264563/exam_images/bn5acdhhem4hvhtrnsbo.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265467/exam_images/qc4gbepselvkks5k9sik.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267795/exam_images/n7dacsflbuzorb3rmaqd.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265624/exam_images/d4ezp6p9d95v96et0sky.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Circle Question 77 English Explanation"></picture> <br><br>Centre of smallest circle is A<br><br>Centre of largest circle is B<br><br>$${r_2} = |CP - CA| = 3\sqrt 2 - 3$$<br><br>$${r_1} = CP - CB = 3\sqrt 2 + 3$$<br><br>$${{{r_1}} \over {{r_2}}} = {{3\sqrt 2 + 3} \over {3\sqrt 2 - 3}} = {{{{(3\sqrt 2 + 3)}^2}} \over 9} = {(\sqrt 2 + 1)^2} = 3 + 2\sqrt 2 $$<br><br>a = 3, b = 2 | mcq | jee-main-2021-online-20th-july-evening-shift |
1kteov0d4 | maths | circle | family-of-circle | Let the equation x<sup>2</sup> + y<sup>2</sup> + px + (1 $$-$$ p)y + 5 = 0 represent circles of varying radius r $$\in$$ (0, 5]. Then the number of elements in the set S = {q : q = p<sup>2</sup> and q is an integer} is __________. | [] | null | 61 | $$r = \sqrt {{{{p^2}} \over 4} + {{{{(1 - p)}^2}} \over 4} - 5} = {{\sqrt {2{p^2} - 2p - 19} } \over 2}$$<br><br>Since, $$r \in (0,5]$$<br><br>So, $$0 < 2{p^2} - 2p - 19 \le 100$$<br><br>$$ \Rightarrow p \in \left[ {{{1 - \sqrt {239} } \over 2},{{1 - \sqrt {39} } \over 2}} \right) \cup \left( {{{1 + \sqrt {39} } \over 2},{{1 + \sqrt {239} } \over 2}} \right]$$ <br><br>so, number of integral values of p<sup>2</sup> is 61. | integer | jee-main-2021-online-27th-august-morning-shift |
1l589r0yu | maths | circle | family-of-circle | <p>Let C be a circle passing through the points A(2, $$-$$1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle $${(x - 5)^2} + {(y - 1)^2} = {{13} \over 2}$$, then r<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "$${{65} \\over 2}$$"}, {"identifier": "C", "content": "$${{61} \\over 2}$$"}, {"identifier": "D", "content": "30"}] | ["B"] | null | <p>Equation of perpendicular bisector of AB is</p>
<p>$$y - {3 \over 2} = - {1 \over 5}\left( {x - {5 \over 2}} \right) \Rightarrow x + 5y = 10$$</p>
<p>Solving it with equation of given circle,</p>
<p>$${(x - 5)^2}{\left( {{{10 - x} \over 5} - 1} \right)^2} = {{13} \over 2}$$</p>
<p>$$ \Rightarrow {(x - 5)^2}\left( {1 + {1 \over {25}}} \right) = {{13} \over 2}$$</p>
<p>$$ \Rightarrow x - 5 = \pm \,{5 \over 2} \Rightarrow x = {5 \over 2}$$ or $${{15} \over 2}$$</p>
<p>But $$x \ne {5 \over 2}$$ because AB is not the diameter.</p>
<p>So, centre will be $$\left( {{{15} \over 2},{1 \over 2}} \right)$$</p>
<p>Now, $${r^2} = {\left( {{{15} \over 2} - 2} \right)^2} + {\left( {{1 \over 2} + 1} \right)^2}$$</p>
<p>$$ = {{65} \over 2}$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1ldpsfpye | maths | circle | family-of-circle | <p>Let a circle $$C_{1}$$ be obtained on rolling the circle $$x^{2}+y^{2}-4 x-6 y+11=0$$ upwards 4 units on the tangent $$\mathrm{T}$$ to it at the point $$(3,2)$$. Let $$C_{2}$$ be the image of $$C_{1}$$ in $$\mathrm{T}$$. Let $$A$$ and $$B$$ be the centers of circles $$C_{1}$$ and $$C_{2}$$ respectively, and $$M$$ and $$N$$ be respectively the feet of perpendiculars drawn from $$A$$ and $$B$$ on the $$x$$-axis. Then the area of the trapezium AMNB is :</p> | [{"identifier": "A", "content": "$$2\\left( {2 + \\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$4\\left( {1 + \\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$3 + 2\\sqrt 2 $$"}, {"identifier": "D", "content": "$$2\\left( {1 + \\sqrt 2 } \\right)$$"}] | ["B"] | null | Given circle is $x^{2}+y^{2}-4 x-6 y+11=0$, centre $(2,3)$
<br/><br/>Tangent at $(3,2)$ is $x-y=1$
<br/><br/>After rolling up by 4 units centre of $C_{1}$ is
<br/><br/>$A \equiv\left(2+\frac{4}{\sqrt{2}}, 3+\frac{4}{\sqrt{2}}\right)$
<br/><br/>$\Rightarrow A=(2+2 \sqrt{2}, 3+2 \sqrt{2})$
<br/><br/>$B$ is the image of $A$ in $x-y=1$
<br/><br/>$\frac{x-(2+2 \sqrt{2})}{1}=\frac{y-(3+2 \sqrt{2})}{-1}=\frac{-2(-2)}{2}=2$
<br/><br/>$\Rightarrow x=4+2 \sqrt{2}, y=1+2 \sqrt{2}$
<br/><br/>Area of $A M N B$
<br/><br/>$=\frac{1}{2}(4+4 \sqrt{2})(4+2 \sqrt{2}-(2+2 \sqrt{2}))$
<br/><br/>$=4(1+\sqrt{2})$ | mcq | jee-main-2023-online-31st-january-morning-shift |
6ss2xEl1EvF1sipa | maths | circle | intercepts-of-a-circle | The circle $${x^2} + {y^2} = 4x + 8y + 5$$ intersects the line $$3x - 4y = m$$ at two distinct points if : | [{"identifier": "A", "content": "$$ - 35 < m < 15$$ "}, {"identifier": "B", "content": "$$ 15 < m < 65$$"}, {"identifier": "C", "content": "$$ 35 < m < 85$$"}, {"identifier": "D", "content": "$$ - 85 < m < -35$$"}] | ["A"] | null | Circle $${x^2} + {y^2} - 4x - 8y - 5 = 0$$
<br><br>Center $$=(2,4),$$ Radius $$ = \sqrt {4 + 16 + 5} = 5$$
<br><br>If circle is intersecting line $$3x-4y=m,$$ at two distinct points.
<br><br>$$ \Rightarrow $$ length of perpendicular from center to the line $$ < $$ radius
<br><br>$$ \Rightarrow {{\left| {6 - 16 - m} \right|} \over 5} < 5 \Rightarrow \left| {10 + m} \right| < 25$$
<br><br>$$ \Rightarrow - 25 < m + 10 < 25 \Rightarrow - 35 < m < 15$$ | mcq | aieee-2010 |
0D5SbWICmfbrSiS6nhSKX | maths | circle | intercepts-of-a-circle | If a circle C, whose radius is 3, touches externally the circle,
<br/>$${x^2} + {y^2} + 2x - 4y - 4 = 0$$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to : | [{"identifier": "A", "content": "$$2\\sqrt 5 $$ "}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$$\\sqrt 5 $$"}, {"identifier": "D", "content": "$$2\\sqrt 3 $$"}] | ["A"] | null | Given circle is :
<br><br>x<sup>2</sup> + y<sup>2</sup> + 2x $$-$$ 4y $$-$$4 = 0
<br><br>$$\therefore\,\,\,$$ its center is ($$-$$ 1, 2) and radius is 3 units.
<br><br>Let A = (x, y) be the center of the circle C
<br><br>$$ \therefore $$$$\,\,\,$$ $${{x - 1} \over 2}$$ = 2 $$ \Rightarrow $$ x = 5 and $${{y + 2} \over 2}$$ = 2 $$ \Rightarrow $$ y = 2
<br><br>So the center of C is (5, 2) and its radius is 3
<br><br>$$\therefore\,\,\,$$ Equation of center C is :
<br><br>x<sup>2</sup> + y<sup>2</sup> $$-$$ 10x $$-$$ 4y + 20 = 0
<br><br>$$\therefore\,\,\,$$ The length of the intercept it cuts on the x-axis
<br><br>= 2$$\sqrt {{g^2} - c} = 2\sqrt {25 - 20} = 2\sqrt 5 $$ | mcq | jee-main-2018-online-16th-april-morning-slot |
bwfEFliy0nuIS6V14kpQ2 | maths | circle | intercepts-of-a-circle | The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is : | [{"identifier": "A", "content": "$$4\\sqrt 5 $$ "}, {"identifier": "B", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "C", "content": "$$2\\sqrt 5 $$"}, {"identifier": "D", "content": "$${{\\sqrt 5 } \\over 4}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264558/exam_images/tv9y0wlrozxdjluksyez.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Mathematics - Circle Question 117 English Explanation">
<br><br>Equation of circle
<br><br>(x $$-$$ 1) (x $$-$$ 0) + (y $$-$$ 0) (y $$-$$ $${1 \over 2}$$) = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup><sup></sup> $$-$$ x $$-$$ $${y \over 2}$$ = 0
<br><br>Equation of tangent of region is 2x + y = 0
<br><br>$$\ell $$<sub>1</sub> + $$\ell $$<sub>2</sub> = $${2 \over {\sqrt 5 }} + {1 \over {2\sqrt 5 }}$$
<br><br>= $${{4 + 1} \over {2\sqrt 5 }} = {{\sqrt 5 } \over 2}$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
dGyz2Ms65BQBVjVOx30tZ | maths | circle | intercepts-of-a-circle | A square is inscribed in the circle x<sup>2</sup> + y<sup>2</sup>
– 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes.
Then the distance of the vertex of this square which is nearest to the origin is : | [{"identifier": "A", "content": "$$\\sqrt {137} $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt {41} $$"}, {"identifier": "D", "content": "13"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266267/exam_images/raqnosna6rdmei2xv6vs.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Mathematics - Circle Question 116 English Explanation">
<br>R $$ = \sqrt {9 + 16 + 103} = 8\sqrt 2 $$
<br><br>OA $$ = 13$$
<br><br>OB $$ = \sqrt {265} $$
<br><br>OC $$ = \sqrt {137} $$
<br><br>OD $$ = \sqrt {41} $$ | mcq | jee-main-2019-online-11th-january-morning-slot |
MNZ8HDiEA2opWMQFzoaec | maths | circle | intercepts-of-a-circle | If a circle of radius R passes through the origin O and intersects the coordinates axes at A and B, then the
locus of the foot of perpendicular from O on AB is : | [{"identifier": "A", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> = 4R<sup>2</sup>x<sup>2</sup>y<sup>2</sup>"}, {"identifier": "B", "content": "(x<sup>2</sup> + y<sup>2</sup>) (x + y) = R<sup>2</sup>xy"}, {"identifier": "C", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> = 4R<sup></sup>x<sup>2</sup>y<sup>2</sup>"}, {"identifier": "D", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>3</sup> = 4R<sup>2</sup>x<sup>2</sup>y<sup>2</sup> "}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266292/exam_images/mqtptxaki1efkgjktaz3.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Circle Question 112 English Explanation">
<br>Slope of AB = $${{ - h} \over k}$$
<br><br>Equation of AB is hx + ky = h<sup>2</sup> + k<sup>2</sup>
<br><br>A $$\left( {{{{h^2} + {k^2}} \over h},0} \right),B\left( {0,{{{h^2} + {k^2}} \over k}} \right)$$
<br><br>AB = 2R
<br><br>$$ \Rightarrow $$ (h<sup>2</sup> + k<sup>2</sup>)<sup>3</sup> = 4R<sup>2</sup>h<sup>2</sup>k<sup>2</sup>
<br><br>$$ \Rightarrow $$ (x<sup>2</sup> + y<sup>2</sup>)<sup>3</sup> = 4R<sup>2</sup>x<sup>2</sup>y<sup>2</sup> | mcq | jee-main-2019-online-12th-january-evening-slot |
61W16N54NVigiXQdvP8G4 | maths | circle | intercepts-of-a-circle | The sum of the squares of the lengths of the chords
intercepted on the circle, x<sup>2</sup> + y<sup>2</sup> = 16, by the lines,
x + y = n, n $$ \in $$ N, where N is the set of all natural
numbers, is : | [{"identifier": "A", "content": "210"}, {"identifier": "B", "content": "160"}, {"identifier": "C", "content": "320"}, {"identifier": "D", "content": "105"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265623/exam_images/doy3xrxadrncldcwhphr.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265990/exam_images/c61gxkjkwqx5wosotcyk.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265002/exam_images/bk22gu7jwtdummbjsmw8.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Circle Question 111 English Explanation"></picture>
<br>Let the chord x + y = n cuts the circle x<sup>2</sup> +
y<sup>2</sup> = 16 at A and B.
<br><br>Length of perpendicular from
O on AB,
<br><br>OM = $$\left| {{{0 + 0 - n} \over {\sqrt {{1^2} + {1^2}} }}} \right| = {n \over {\sqrt 2 }}$$
<br><br>Radius of the circle = 4
<br><br>From the picture you can see,
<br><br>$${n \over {\sqrt 2 }}$$ < 4
<br><br>$$ \Rightarrow $$ n < 5.65
<br><br>$$ \therefore $$ Possible value of n = 1, 2, 3, 4, 5
<br><br>Length of chord AB = $$2\sqrt {{{\left( 4 \right)}^2} - {{\left( {{n \over {\sqrt 2 }}} \right)}^2}} $$
<br><br>= $$2\sqrt {16 - {{{n^2}} \over 2}} $$
<br><br>= $$\sqrt {64 - 2{n^2}} $$ = $${l}$$
<br><br>For n = 1, $${l^2}$$ = 62
<br><br>For n = 2, $${l^2}$$ = 56
<br><br>For n = 3, $${l^2}$$ = 46
<br><br>For n = 4, $${l^2}$$ = 32
<br><br>For n = 5, $${l^2}$$ = 14
<br><br>$$ \therefore $$ Sum of square of length of chords
= 62 + 56 + 46 + 32 + 14 = 210
| mcq | jee-main-2019-online-8th-april-morning-slot |
zzLYX99HzUqacuDxIm3rsa0w2w9jxaykypb | maths | circle | intercepts-of-a-circle | A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the
point : | [{"identifier": "A", "content": "(1, 5)"}, {"identifier": "B", "content": "( 2, 3)"}, {"identifier": "C", "content": "(3, 5)"}, {"identifier": "D", "content": "(3, 10)"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265069/exam_images/iuwti6tm5cysz6ebxdbi.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Circle Question 102 English Explanation"><br><br>
From the above figure equation of circle is (x - 3)<sup>2</sup> + (y - 5)<sup>5</sup> = 5<sup>2</sup><br><br>
So (3, 10) will satisfy the equation. | mcq | jee-main-2019-online-12th-april-evening-slot |
ZkMgZkefk4fu6UwEbm1kmixaf77 | maths | circle | intercepts-of-a-circle | Let the lengths of intercepts on x-axis and y-axis made by the circle <br/>x<sup>2</sup> + y<sup>2</sup> + ax + 2ay + c = 0, (a < 0) be 2$${\sqrt 2 }$$ and 2$${\sqrt 5 }$$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to : | [{"identifier": "A", "content": "$${\\sqrt {10} }$$"}, {"identifier": "B", "content": "$${\\sqrt {6} }$$"}, {"identifier": "C", "content": "$${\\sqrt {11} }$$"}, {"identifier": "D", "content": "$${\\sqrt {7} }$$"}] | ["B"] | null | $$2\sqrt {{{{a^2}} \over 4} - c} = 2\sqrt 2 $$<br><br>$$\sqrt {{a^2} - 4c} = 2\sqrt 2 $$<br><br>$${a^2} - 4c = 8$$ .... (1)<br><br>$$2\sqrt {{a^2} - c} = 2\sqrt 5 $$<br><br>$${a^2} - c = 5$$ .... (2)<br><br>$$(2) - (1)$$<br><br>$$3c = - 3a \Rightarrow c = - 1$$<br><br>$${a^2} = 4 \Rightarrow a = - 2$$ (Given a < 0)<br><br>Equation of circle
<br><br>$${x^2} + {y^2} - 2x - 4y - 1 = 0$$<br><br>Equation of tangent which is perpendicular to the line x + 2y = 0 is
<br><br>$$2x - y + \lambda = 0$$<br><br>$$ \therefore $$ p = r<br><br>$$\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6 $$<br><br>$$ \Rightarrow \lambda = \pm \sqrt {30} $$<br><br>$$ \therefore $$ Tangent $$2x - y \pm \sqrt {30} = 0$$<br><br>Distance from origin = $${{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6 $$ | mcq | jee-main-2021-online-16th-march-evening-shift |
1kryesdeh | maths | circle | intercepts-of-a-circle | Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $$6\sqrt 5 $$ on the x-axis. Then the radius of the circle C is equal to : | [{"identifier": "A", "content": "$$\\sqrt {53} $$"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "$$\\sqrt {82} $$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266278/exam_images/kb34kqlzzf1i40yfp522.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Circle Question 75 English Explanation"><br><br>$$r = \sqrt {{6^2} + {{\left( {3 + \sqrt 5 } \right)}^2}} $$<br><br>$$ = \sqrt {36 + 45} = 9$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1l6jdc4r8 | maths | circle | intercepts-of-a-circle | <p>If the circle $$x^{2}+y^{2}-2 g x+6 y-19 c=0, g, c \in \mathbb{R}$$ passes through the point $$(6,1)$$ and its centre lies on the line $$x-2 c y=8$$, then the length of intercept made by the circle on $$x$$-axis is :</p> | [{"identifier": "A", "content": "$$\\sqrt{11}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$2 \\sqrt{23}$$"}] | ["D"] | null | <p>Circle : $${x^2} + {y^2} - 2gx + 6y - 19c = 0$$</p>
<p>It passes through $$h(6,1)$$</p>
<p>$$ \Rightarrow 36 + 1 - 12g + 6 - 19c = 0$$</p>
<p>$$ = 12g + 19c = 43$$ ..... (1)</p>
<p>Line $$x - 2cy = 8$$ passes through centre</p>
<p>$$ \Rightarrow g + 6c = 8$$ ...... (2)</p>
<p>From (1) & (2)</p>
<p>$$g = 2,\,c = 1$$</p>
<p>$$C:{x^2} + {y^2} - 4x + 6y - 19 = 0$$</p>
<p>x intercept $$= 2\sqrt {{g^2} - C} $$</p>
<p>$$ = 2\sqrt {4 + 19} $$</p>
<p>$$ = 2\sqrt {23} $$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1lgrgje37 | maths | circle | intercepts-of-a-circle | <p>Two circles in the first quadrant of radii $$r_{1}$$ and $$r_{2}$$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $$x+y=2$$. Then $$r_{1}^{2}+r_{2}^{2}-r_{1} r_{2}$$ is equal to ___________.</p> | [] | null | 7 | $$
\begin{aligned}
& \text { Circle }(x-a)^2+(y-a)^2=a^2 \\\\
& x^2+y^2-2 a x-2 a y+a^2=0 \\\\
& \text { intercept }=2 \\\\
& \Rightarrow 2 \sqrt{a^2-d^2}=2
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1li9wqjod/6e47a7f9-a970-48fa-94e5-b8b0e7d86b93/96fe4cd0-feb4-11ed-9a3f-cb8a261721bd/file-1li9wqjoe.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1li9wqjod/6e47a7f9-a970-48fa-94e5-b8b0e7d86b93/96fe4cd0-feb4-11ed-9a3f-cb8a261721bd/file-1li9wqjoe.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 12th April Morning Shift Mathematics - Circle Question 28 English Explanation">
<br><br>Where $\mathrm{d}=$ perpendicular distance of centre from line $x+y=2$
<br><br>$$
\begin{aligned}
& \Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2 \\\\
& \Rightarrow a^2-\frac{(2 a-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\\\
& \Rightarrow 2 a^2-8 a+6=0 \Rightarrow a^2-4 a+3=0 \\\\
& \therefore r_1+r_2=4 \text { and } r_1 r_2=3 \\\\
& \therefore r_1^2+r_2^2-r_1 r_2=\left(r_1+r_2\right)^2-3 r_1 r_2 \\\\
& =16-9=7
\end{aligned}
$$ | integer | jee-main-2023-online-12th-april-morning-shift |
1lgvq76au | maths | circle | intercepts-of-a-circle | <p>Let A be the point $$(1,2)$$ and B be any point on the curve $$x^{2}+y^{2}=16$$. If the centre of the locus of the point P, which divides the line segment $$\mathrm{AB}$$ in the ratio $$3: 2$$ is the point C$$(\alpha, \beta)$$, then the length of the line segment $$\mathrm{AC}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{3 \\sqrt{5}}{5}$$"}, {"identifier": "B", "content": "$$\\frac{6 \\sqrt{5}}{5}$$"}, {"identifier": "C", "content": "$$\\frac{2 \\sqrt{5}}{5}$$"}, {"identifier": "D", "content": "$$\\frac{4 \\sqrt{5}}{5}$$"}] | ["A"] | null | We have, equation of circle is $x^2+y^2=16$
<br><br>Let any point on the circle $x^2+y^2=4^2$ is $B(4 \cos \theta, 4 \sin \theta)$ and $A(1,2)$
<br><br>Let $\mathrm{P}$ be $(h, k)$ which divides $\mathrm{AB}$ in $3: 2$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk5g4ko/0e3a3eb2-16a6-4d51-952c-ed316089a0d7/92c95a80-6753-11ee-a917-d528e09e481b/file-6y3zli1lnk5g4kp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnk5g4ko/0e3a3eb2-16a6-4d51-952c-ed316089a0d7/92c95a80-6753-11ee-a917-d528e09e481b/file-6y3zli1lnk5g4kp.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) 10th April Evening Shift Mathematics - Circle Question 27 English Explanation">
<br><br>So
<br><br>$$
h=\frac{12 \cos \theta+2}{3+2} \text { and } k=\frac{12 \sin \theta+2 \times 2}{3+2}
$$
<br><br>$\Rightarrow \cos \theta=\frac{5 h-2}{12}$ and $\sin \theta=\frac{5 k-4}{12}$
<br><br>As, $\cos ^2 \theta+\sin ^2 \theta=1,\left(\frac{5 h-2}{12}\right)^2+\left(\frac{5 k-4}{12}\right)^2=1$
<br><br>$$
\Rightarrow\left(h-\frac{2}{5}\right)^2+\left(k-\frac{4}{5}\right)^2=\frac{12^2}{5^2}
$$
<br><br>$\therefore$ Locus of point $P$ is $\left(x-\frac{2}{5}\right)^2+\left(y-\frac{4}{5}\right)^2=\frac{12^2}{5^2}$
<br><br>which is equation of circle with centre $\left(\frac{2}{5}, \frac{4}{5}\right)$
<br><br>Hence, $A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5}$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lh23y6we | maths | circle | intercepts-of-a-circle | <p> A circle passing through the point $$P(\alpha, \beta)$$ in the first quadrant touches the two coordinate axes at the points $$A$$ and $$B$$. The point $$P$$ is above the line $$A B$$. The point $$Q$$ on the line segment $$A B$$ is the foot of perpendicular from $$P$$ on $$A B$$. If $$P Q$$ is equal to 11 units, then the value of $$\alpha \beta$$ is
___________.</p> | [] | null | 121 | Let equation of circle is $(x-a)^2+(y-a)^2=a^2$
<br><br>Since, (i) passes through $P(\alpha, \beta)$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo4iz7t1/8abb90cb-5a21-43f7-858e-e4e5d897d907/29530750-7288-11ee-955e-53a531d428f9/file-6y3zli1lo4iz7t2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lo4iz7t1/8abb90cb-5a21-43f7-858e-e4e5d897d907/29530750-7288-11ee-955e-53a531d428f9/file-6y3zli1lo4iz7t2.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 Morning Shift Mathematics - Circle Question 22 English Explanation">
<br><br>$\therefore (\alpha-a)^2+(\beta-a)^2=a^2$
<br><br>$$
\begin{array}{lr}
&\Rightarrow \alpha^2+\beta^2-2 \alpha a-2 \beta a+a^2=0 .........(i)
\end{array}
$$
<br><br>Equation of $A B=\frac{x}{a}+\frac{y}{a}=1$
<br><br>$$\Rightarrow x+y=a$$ .........(ii)
<br><br>Let $Q(p, q)$ be the foot of the perpendicular from $P$ to line (iii)
<br><br>$$
\begin{aligned}
& \therefore \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-\alpha)}{(1)^2+(1)^2} \\\\
& \Rightarrow \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-a)}{2} \\\\
& \Rightarrow p-\alpha=\frac{-(\alpha+\beta-a)}{2} \\\\
& \text { and } q-\beta=\frac{-(\alpha+\beta-a)}{2}
\end{aligned}
$$
<br><br>Now,
<br><br>$$
\begin{aligned}
& P Q^2 =(p-\alpha)^2+(q-\beta)^2 \\\\
& =\frac{1}{4}(\alpha+\beta-a)^2+\frac{1}{4}(\alpha+\beta-a)^2
\end{aligned}
$$
<br><br>$$
\Rightarrow (11)^2=\frac{1}{2}(\alpha+\beta-\alpha)^2
$$
<br><br>$$
\begin{array}{ll}
&\Rightarrow \alpha^2+\beta^2+a^2+2 \alpha \beta-2 \beta a-2 \alpha a=242 \\\\
&\Rightarrow 2 \alpha \beta=242 \text { [Using Eq. (ii)] }\\\\
&\Rightarrow \alpha \beta=121
\end{array}
$$ | integer | jee-main-2023-online-6th-april-morning-shift |
jaoe38c1lsd4sm2c | maths | circle | intercepts-of-a-circle | <p>Let a variable line passing through the centre of the circle $$x^2+y^2-16 x-4 y=0$$, meet the positive co-ordinate axes at the points $$A$$ and $$B$$. Then the minimum value of $$O A+O B$$, where $$O$$ is the origin, is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "18"}] | ["D"] | null | <p>$$\begin{aligned}
& (y-2)=m(x-8) \\
& \Rightarrow x \text {-intercept } \\
& \Rightarrow\left(\frac{-2}{m}+8\right) \\
& \Rightarrow y \text {-intercept } \\
& \Rightarrow(-8 \mathrm{~m}+2) \\
& \Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{~m}}+8-8 \mathrm{~m}+2 \\
& \mathrm{f}^{\prime}(\mathrm{m})=\frac{2}{\mathrm{~m}^2}-8=0 \\
& \Rightarrow \mathrm{m}^2=\frac{1}{4} \\
& \Rightarrow \mathrm{m}=\frac{-1}{2} \\
& \Rightarrow \mathrm{f}\left(\frac{-1}{2}\right)=18 \\
& \Rightarrow \text { Minimum }=18
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
TKL3WEcvBv7bL9pT | maths | circle | number-of-common-tangents-and-position-of-two-circle | If the two circles $${(x - 1)^2}\, + \,{(y - 3)^2} = \,{r^2}$$ and $$\,{x^2}\, + \,{y^2} - \,8x\, + \,2y\, + \,\,8\,\, = 0$$ intersect in two distinct point, then : | [{"identifier": "A", "content": "$$r > 2$$"}, {"identifier": "B", "content": "$$2 < r < 8$$"}, {"identifier": "C", "content": "$$r < 2$$"}, {"identifier": "D", "content": "$$r = 2.$$"}] | ["B"] | null | $$\left| {{r_1} - {r_2}} \right| < {C_1}{C_2}$$ for intersection
<br><br>$$ \Rightarrow r - 3 < 5 \Rightarrow r < 8\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>and $${r_1} + {r_2} > {C_1}{C_2},\,$$
<br><br>$$r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)$$
<br><br>From $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$2 < r < 8.$$ | mcq | aieee-2003 |
pRzLsQDdQrCvWwSH | maths | circle | number-of-common-tangents-and-position-of-two-circle | The two circles x<sup>2</sup> + y<sup>2</sup> = ax, and x<sup>2</sup> + y<sup>2</sup> = c<sup>2</sup> (c > 0) touch each other if : | [{"identifier": "A", "content": "| a | = c"}, {"identifier": "B", "content": "a = 2c"}, {"identifier": "C", "content": "| a | = 2c"}, {"identifier": "D", "content": "2 | a | = c"}] | ["A"] | null | As center of one circle is $$\left( {0,0} \right)$$ and other circle passes through $$(0,0),$$ therefore
<br><br>Also $${C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)$$
<br><br>$${r_1} = {a \over 2}{r_2} = C$$
<br><br>$${C_1}{C_2} = {r_1} - {r_2} = {a \over 2}$$
<br><br>$$ \Rightarrow C - {a \over 2} = {a \over 2}$$
<br><br>$$ \Rightarrow C = a$$
<br><br>If the two circles touch each other, then they must touch each other internally. | mcq | aieee-2011 |
Dym616D1FHZIYakE | maths | circle | number-of-common-tangents-and-position-of-two-circle | Let $$C$$ be the circle with centre at $$(1, 1)$$ and radius $$=$$ $$1$$. If $$T$$ is the circle centred at $$(0, y)$$, passing through origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over {\\sqrt 2 }}$$ "}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 2}$$ "}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263502/exam_images/az5z8h8zms4j96efvsxd.webp" loading="lazy" alt="JEE Main 2014 (Offline) Mathematics - Circle Question 137 English Explanation">
<br><br>Equation of circle $$C \equiv {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 1$$
<br><br>Radius of $$T = \left| y \right|$$
<br><br>$$T$$ touches $$C$$ externally
<br><br>therefore,
<br><br>Distance between the centers $$=$$ sum of their radii
<br><br>$$ \Rightarrow \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} = 1 + \left| y \right|$$
<br><br>$$ \Rightarrow {\left( {0 - 1} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {1 + \left| y \right|} \right)^2}$$
<br><br>$$ \Rightarrow 1 + {y^2} + 1 - 2y = 1 + {y^2} + 2\left| y \right|$$
<br><br>$$2\left| y \right| = 1 - 2y$$
<br><br>If $$y>0$$ then $$2y=1-2y$$ $$ \Rightarrow y = {1 \over 4}$$
<br><br>$$y<0$$ then $$-2y=1-2y$$ $$ \Rightarrow 0 = 1$$ (not possible)
<br><br>$$\therefore$$ $$y = {1 \over 4}$$ | mcq | jee-main-2014-offline |
0uEqnZaNFKvdeUZb | maths | circle | number-of-common-tangents-and-position-of-two-circle | The number of common tangents to the circles $${x^2} + {y^2} - 4x - 6x - 12 = 0$$ and $${x^2} + {y^2} + 6x + 18y + 26 = 0,$$ is : | [{"identifier": "A", "content": "$$3$$"}, {"identifier": "B", "content": "$$4$$"}, {"identifier": "C", "content": "$$1$$"}, {"identifier": "D", "content": "$$2$$"}] | ["A"] | null | $${x^2} + {y^2} - 4x - 6y - 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Center, $${c_1} = \left( {2,\,3} \right)$$ and Radius, $${r_1} = 5$$ units
<br><br>$${x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Center, $${c_2} = \left( { - 3, - 9} \right)$$ and Radius, $${r_2} = 8$$ units
<br><br>$${C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,$$ units
<br><br>$${r_1} + {r_2} = 5 + 8 = 13$$
<br><br>$$\therefore$$ $${C_1}{C_2} = {r_1} + {r_2}$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91nnzlt/1c575b95-e553-4210-a91e-c044bf073271/e61a5710-47fc-11ed-8757-0f869593f41f/file-1l91nnzlu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91nnzlt/1c575b95-e553-4210-a91e-c044bf073271/e61a5710-47fc-11ed-8757-0f869593f41f/file-1l91nnzlu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2015 (Offline) Mathematics - Circle Question 135 English Explanation">
<br>Therefore there are three common tangents. | mcq | jee-main-2015-offline |
kF4KgNZxNSIv6qYCVW3rsa0w2w9jwxu8ooq | maths | circle | number-of-common-tangents-and-position-of-two-circle | If the circles x<sup>2</sup>
+ y<sup>2</sup>
+ 5Kx + 2y + K = 0 and 2(x<sup>2</sup>
+ y<sup>2</sup>) + 2Kx + 3y –1 = 0, (K$$ \in $$R), intersect at the points
P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for : | [{"identifier": "A", "content": "exactly two values of K"}, {"identifier": "B", "content": "no value of K"}, {"identifier": "C", "content": "exactly one value of K"}, {"identifier": "D", "content": "infinitely many values of K"}] | ["B"] | null | S<sub>1</sub> $$ \equiv $$ x<sup>2</sup> + y<sup>2</sup> + 5Kx + 2y + K = 0<br><br>
$${S_2} \equiv {x^2} + {y^2} + Kx + {3 \over 2}y - {1 \over 2} = 0$$<br><br>
Equation of common chord is<br><br>
S<sub>1</sub> – S<sub>2</sub> = 0<br><br>
$$ \Rightarrow 4Kx + {y \over 2} + K + {1 \over 2} = 0$$ ....(1)<br><br>
4x + 5y – K = 0 …(2) (given)<br><br>
On comparing (1) and (2)<br><br>
$${{4K} \over 4} = {1 \over {10}} = {{2K + 1} \over { - 2K}}$$<br><br>
$$ \Rightarrow K = {1 \over {10}}$$ and $$ - 2K = 20K + 10$$<br><br>
$$ \Rightarrow $$ 22K = –10<br><br>
$$\therefore K = {{ - 5} \over {11}}$$<br><br>
So No value of <b>K</b> exists. | mcq | jee-main-2019-online-10th-april-morning-slot |
p8FE11vq7emHA0SgNw18hoxe66ijvwub61o | maths | circle | number-of-common-tangents-and-position-of-two-circle | The common tangent to the circles x <sup>2</sup> + y<sup>2</sup> = 4 and
x<sup>2</sup> + y<sup>2</sup> + 6x + 8y – 24 = 0 also passes through the
point : | [{"identifier": "A", "content": "(6, \u20132)"}, {"identifier": "B", "content": "(4, \u20132)"}, {"identifier": "C", "content": "(\u20134, 6)"}, {"identifier": "D", "content": "(\u20136, 4)"}] | ["A"] | null | For this circle
x <sup>2</sup> + y<sup>2</sup> = 4
<br><br>Center C<sub>1</sub> = (0, 0) and radius r<sub>1</sub> = 2
<br><br>For this circle
x<sup>2</sup> + y<sup>2</sup> + 6x + 8y – 24 = 0
<br><br>Center C<sub>2</sub> = (-3, -4) and radius r<sub>2</sub> = $$\sqrt {9 + 16 + 24} $$ = 7
<br><br>So distance between center,
<br><br>C<sub>1</sub>C<sub>2</sub> = 5
<br><br>r<sub>1</sub> + r<sub>2</sub> = 7
<br><br>|r<sub>1</sub> - r<sub>2</sub>| = 5
<br><br>As C<sub>1</sub>C<sub>2</sub> = |r<sub>1</sub> - r<sub>2</sub>|
<br><br>$$ \therefore $$ Circles touches internally.
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263600/exam_images/nze8tx1zaegnzuv6v50m.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266752/exam_images/eqogrowu23mfgo9kij83.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264280/exam_images/rjykkmfni3s0h2hccm3q.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Circle Question 108 English Explanation"></picture>
<br><br>Here Equation of common tangent is same as common chord.
<br><br>$$ \therefore $$ S<sub>1</sub> - S<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$ (x <sup>2</sup> + y<sup>2</sup> - 4) - (x<sup>2</sup> + y<sup>2</sup> + 6x + 8y - 24) = 0
<br><br>$$ \Rightarrow $$ - 6x - 8y + 20 = 0
<br><br>$$ \Rightarrow $$ 3x + 4y - 10 = 0
<br><br>By checking all the options you can see that,
<br><br>(6, –2) point satisfy the equation 3x + 4y - 10 = 0. | mcq | jee-main-2019-online-9th-april-evening-slot |
RQXU1QZmaI8cAm09q9pZQ | maths | circle | number-of-common-tangents-and-position-of-two-circle | Let C<sub>1</sub> and C<sub>2</sub> be the centres of the circles x<sup>2</sup> + y<sup>2</sup> – 2x – 2y – 2 = 0 and x<sup>2</sup> + y<sup>2</sup> – 6x – 6y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC<sub>1</sub>QC<sub>2</sub> is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263648/exam_images/kjqkelq5u313u7up3q3r.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Circle Question 113 English Explanation">
<br>Area = 2 $$ \times $$ $${1 \over 2}$$.4 = 2 | mcq | jee-main-2019-online-12th-january-morning-slot |
spzMVkSBmK5b0Xg5TlTdc | maths | circle | number-of-common-tangents-and-position-of-two-circle | If a variable line, 3x + 4y – $$\lambda $$ = 0 is such that the two circles x<sup>2</sup> + y<sup>2</sup> – 2x – 2y + 1 = 0 and x<sup>2</sup> + y<sup>2</sup> – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of $$\lambda $$ is the interval : | [{"identifier": "A", "content": "(23, 31)"}, {"identifier": "B", "content": "(2, 17)"}, {"identifier": "C", "content": "[13, 23] "}, {"identifier": "D", "content": "[12, 21] "}] | ["D"] | null | Centre of circles are opposite side of line
<br><br>(3 + 4 $$-$$ $$\lambda $$) (27 + 4 $$-$$ $$\lambda $$) < 0
<br><br>($$\lambda $$ $$-$$ 7) ($$\lambda $$ $$-$$ 31) < 0
<br><br>$$\lambda $$ $$ \in $$ (7, 31)
<br><br>distance from S<sub>1</sub>
<br><br>$$\left| {{{3 + 4 - \lambda } \over 5}} \right| \ge 1 \Rightarrow \lambda \in ( - \infty ,2] \cup [(12,\infty ]$$
<br><br>distance from S<sub>2</sub>
<br><br>$$\left| {{{27 + 4 - \lambda } \over 5}} \right| \ge 2 \Rightarrow \lambda \in ( - \infty ,21] \cup [41,\infty )$$
<br><br>so $$\lambda \in \left[ {12,21} \right]$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
zbBZfrh5YlxIoiF5Ce8if | maths | circle | number-of-common-tangents-and-position-of-two-circle | Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is : | [{"identifier": "A", "content": "$$2\\sqrt 2 $$"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263285/exam_images/yhmndiwp2ujk1aofixgb.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Mathematics - Circle Question 115 English Explanation">
<br><br>In $$\Delta $$APO
<br><br>$${\left( {{{\sqrt 2 r} \over 2}} \right)^2} + {1^2} = {r^2}$$
<br><br>$$ \Rightarrow $$ $$r = \sqrt 2 $$
<br><br>So distance between centres $$ = \sqrt 2 r = 2$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
BRTMkdqUdllpf8r6uf8jG | maths | circle | number-of-common-tangents-and-position-of-two-circle | If the circles
<br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 16x $$-$$ 20y + 164 = r<sup>2</sup>
<br/><br/>and (x $$-$$ 4)<sup>2</sup> + (y $$-$$ 7)<sup>2</sup> = 36
<br/><br/>intersect at two distinct points, then : | [{"identifier": "A", "content": "r > 11"}, {"identifier": "B", "content": "0 < r < 1"}, {"identifier": "C", "content": "r = 11"}, {"identifier": "D", "content": "1 < r < 11"}] | ["D"] | null | Circles are x<sup>2</sup> + y<sup>2</sup> $$-$$ 16x $$-$$ 20y + 164 = r<sup>2</sup> $$ \Rightarrow $$ c<sub>1</sub> (8, 10)
<br><br>and (x $$-$$ 4)<sup>2</sup> + (y $$-$$ 7)<sup>2</sup> = 36
<br><br>they intersect at two distinct points
<br><br>$$\left| {{r_1} - {r_2}} \right| < {c_1}{c_2} < {r_1} + {r_2}\left\{ {{c_1}{c_2} = \sqrt {16 + 9} = 5} \right\}$$
<br><br>Now $$\left| {r - 6} \right| < 5 < r + 6$$
<br><br>$$\left| {r - 6} \right| < 5$$
<br><br>$$ \Rightarrow $$ $$ - 5 < r - 6 < 5$$
<br><br>$$ \Rightarrow $$ $$1 < r < 11\,\,\,\,\,\,\,\,\,...(i)$$
<br><br>$$5 < r + 6$$
<br><br>$$ - 1 < r\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
<br><br>from (i) and (ii)
<br><br>r $$ \in $$ (1, 11) | mcq | jee-main-2019-online-9th-january-evening-slot |
9GoPmJBIeKYfTPS79ujgy2xukewts732 | maths | circle | number-of-common-tangents-and-position-of-two-circle | The number of integral values of k for which
the line, 3x + 4y = k intersects the circle,
<br/>x<sup>2</sup>
+ y<sup>2</sup>
– 2x – 4y + 4 = 0 at two distinct points is
______. | [] | null | 9 | Circle x<sup>2</sup>
+ y<sup>2</sup>
– 2x – 4y + 4 = 0
<br><br>$$ \Rightarrow $$ (x – 1)<sup>2</sup>
+ (y – 2)<sup>2</sup>
= 1
<br><br>Centre: (1, 2), radius = 1
<br><br>Line 3x + 4y – k = 0 intersects the circle at two distinct points.
<br><br>$$ \Rightarrow $$ distance of centre from the line < radius
<br><br>$$ \Rightarrow $$ $$\left| {{{3 \times 1 + 4 \times 2 - k} \over {\sqrt {{3^2} + {4^2}} }}} \right| < 1$$
<br><br>$$ \Rightarrow $$ |11 - k| < 5
<br><br>$$ \Rightarrow $$ 6 < k < 5
<br><br>$$ \Rightarrow $$ k $$ \in $$ {7, 8, 9, ……15} since k $$ \in $$ I
<br><br>$$ \therefore $$ Total 9 integral value of k. | integer | jee-main-2020-online-2nd-september-morning-slot |
ppjkwHdAOgYLtIrc8d1kmjb64v2 | maths | circle | number-of-common-tangents-and-position-of-two-circle | Choose the incorrect statement about the two circles whose equations are given below :<br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 10x $$-$$ 10y + 41 = 0 and <br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 16x $$-$$ 10y + 80 = 0 | [{"identifier": "A", "content": "Distance between two centres is the average of radii of both the circles."}, {"identifier": "B", "content": "Both circles pass through the centre of each other."}, {"identifier": "C", "content": "Circles have two intersection points."}, {"identifier": "D", "content": "Both circle's centers lie inside region of one another."}] | ["D"] | null | S<sub>1</sub> $$ \equiv $$ x<sup>2</sup> + y<sup>2</sup> $$-$$ 10x $$-$$ 10y + 41 = 0<br><br>Centre C<sub>1</sub> $$ \equiv $$ (5, 5), radius r<sub>1</sub> = 3<br><br>S<sub>2</sub> $$ \equiv $$ x<sup>2</sup> + y<sup>2</sup> $$-$$ 16x $$-$$ 10y + 80 = 0<br><br>Centre C<sub>2</sub> $$ \equiv $$ (8, 5), radius r<sub>2</sub> = 3<br><br>Distance between centres = 3<br><br>Hence both circles pass through the centre of each other, have two intersection point and distance between two centres in average of radii of both the circles.<br/><br/>
Hence, option (d) is the incorrect statement. | mcq | jee-main-2021-online-17th-march-morning-shift |
jj6PYUtpBmn4DW8BTT1kmjco2b6 | maths | circle | number-of-common-tangents-and-position-of-two-circle | The minimum distance between any two points P<sub>1</sub> and P<sub>2</sub> while considering point P<sub>1</sub> on one circle and point P<sub>2</sub> on the other circle for the given circles' equations<br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 10x $$-$$ 10y + 41 = 0<br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 24x $$-$$ 10y + 160 = 0 is ___________. | [] | null | 1 | $${S_1}:{(x - 5)^2} + {(y - 5)^2} = 9$$
<br><br>Centre (5, 5), r<sub>1</sub> = 3<br><br>$${S_2}:{(x - 12)^2} + {(y - 5)^2} = 9$$
<br><br>Centre (12, 5), r<sub>2</sub> = 3<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264982/exam_images/rpihjtwuabjqwdlwqq9s.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 84 English Explanation"><br><br>So (P<sub>1</sub>P<sub>2</sub>)<sub>min</sub> = 1 | integer | jee-main-2021-online-17th-march-morning-shift |
vSsFERngWYuSBpA34g1kmli5r94 | maths | circle | number-of-common-tangents-and-position-of-two-circle | Choose the correct statement about two circles whose equations are given below :<br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 10x $$-$$ 10y + 41 = 0<br/><br/>x<sup>2</sup> + y<sup>2</sup> $$-$$ 22x $$-$$ 10y + 137 = 0 | [{"identifier": "A", "content": "circles have same centre"}, {"identifier": "B", "content": "circles have no meeting point"}, {"identifier": "C", "content": "circles have only one meeting point"}, {"identifier": "D", "content": "circles have two meeting points"}] | ["C"] | null | Let $${S_1}:{x^2} + {y^2} - 10x - 10y + 41 = 0$$<br><br>$$ \Rightarrow {(x - 5)^2} + {(y - 5)^2} = 9$$<br><br>Centre $$({C_1}) = (5,5)$$<br><br>Radius r<sub>1</sub> = 3<br><br>$${S_2}:{x^2} + {y^2} - 22x - 10y + 137 = 0$$<br><br>$$ \Rightarrow {(x - 11)^2} + {(y - 5)^2} = 9$$<br><br>Centre $$({C_2}) = (11,5)$$<br><br>Radius r<sub>2</sub> = 3<br><br>distance $$({C_1}{C_2}) = \sqrt {{{(5 - 11)}^2} + {{(5 - 5)}^2}} $$<br><br>distance $$({C_1}{C_2}) = 6$$<br><br>$$ \because $$ $${r_1} + {r_2} = 3 + 3 = 6$$<br><br>$$ \therefore $$ circles touch externally<br><br>Hence, circle have only one meeting point. | mcq | jee-main-2021-online-18th-march-morning-shift |
1l6f3r23l | maths | circle | number-of-common-tangents-and-position-of-two-circle | <p>If the circles $${x^2} + {y^2} + 6x + 8y + 16 = 0$$ and $${x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6 $$, $$k > 0$$, touch internally at the point $$P(\alpha ,\beta )$$, then $${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2}$$ is equal to ________________.</p> | [] | null | 25 | <p>The circle $${x^2} + {y^2} + 6x + 8y + 16 = 0$$ has centre $$( - 3, - 4)$$ and radius 3 units.</p>
<p>The circle $${x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6 ,\,k > 0$$ has centre $$\left( {\sqrt 3 - 3,\,\sqrt 6 - 4} \right)$$ and radius $$\sqrt {k + 34} $$</p>
<p>$$\because$$ These two circles touch internally hence</p>
<p>$$\sqrt {3 + 6} = \left| {\sqrt {k + 34} - 3} \right|$$</p>
<p>Here, $$k = 2$$ is only possible ($$\because$$ $$k > 0$$)</p>
<p>Equation of common tangent to two circles is $$2\sqrt 3 x + 2\sqrt 6 y + 16 + 6\sqrt 3 + 8\sqrt 6 + k = 0$$</p>
<p>$$\because$$ $$k = 2$$ then equation is</p>
<p>$$x + \sqrt 2 y + 3 + 4\sqrt 2 + 3\sqrt 3 = 0$$ ...... (i)</p>
<p>$$\because$$ ($$\alpha$$, $$\beta$$) are foot of perpendicular from $$( - 3, - 4)$$</p>
<p>To line (i) then</p>
<p>$${{\alpha + 3} \over 1} = {{\beta + 4} \over {\sqrt 2 }} = {{ - \left( { - 3 - 4\sqrt 2 + 3 + 4\sqrt 2 + 3\sqrt 3 } \right)} \over {1 + 2}}$$</p>
<p>$$\therefore$$ $$\alpha + 3 = {{\beta + 4} \over {\sqrt 2 }} = - \sqrt 3 $$</p>
<p>$$ \Rightarrow {\left( {\alpha + \sqrt 3 } \right)^2} = 9$$ and $${\left( {\beta + \sqrt 6 } \right)^2} = 16$$</p>
<p>$$\therefore$$ $${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2} = 25$$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
lgnx2ht3 | maths | circle | number-of-common-tangents-and-position-of-two-circle | The number of common tangents, to the circles <br/><br/>$x^{2}+y^{2}-18 x-15 y+131=0$ <br/><br/>and $x^{2}+y^{2}-6 x-6 y-7=0$, is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>We are given two circles:</p>
<p>(1) $x^2+y^2-18 x-15 y+131=0$</p>
<p>(2) $x^2+y^2-6 x-6 y-7=0$</p>
<p>First, let's find the centers and radii of the circles.</p>
<p>For circle (1):</p>
<p>Completing the square for the equation:</p>
<p>$(x^2-18x+{81})+(y^2-15y+\frac{225}{4})=-131+{81}+\frac{225}{4}$</p>
<p>$(x-9)^2+(y-\frac{15}{2})^2=\frac{25}{4}$</p>
<p>Center 1: $C_1(9, \frac{15}{2})$</p>
<p>Radius 1: $r_1 = \sqrt{\frac{25}{4}}=\frac{5}{2}$</p>
<p>For circle (2):</p>
<p>Completing the square for the equation:</p>
<p>$(x^2-6x+9)+(y^2-6y+9)=7+9+9$</p>
<p>$(x-3)^2+(y-3)^2=25$</p>
<p>Center 2: $C_2(3, 3)$</p>
<p>Radius 2: $r_2 = 5$</p>
<p>Now, let's find the distance between the centers :</p>
<p>$d = \sqrt{(9-3)^2 + (\frac{15}{2}-3)^2} = \sqrt{6^2 + \frac{9}{2}^2} = \sqrt{36 + \frac{81}{4}} = \frac{15}{2}$</p>
<p>Next, let's analyze the relative positions of the circles using the distance between centers and the sum and difference of the radii :</p>
<ol>
<li>If $d > r_1 + r_2$, the circles are separate, and there are 4 common tangents.</li>
<li>If $d = r_1 + r_2$, the circles are externally tangent, and there are 3 common tangents.</li>
<li>If $0 < d < |r_1 - r_2|$, one circle is inside the other, and there are no common tangents.</li>
<li>If $d = |r_1 - r_2|$, the circles are internally tangent, and there is 1 common tangent.</li>
<li>If $d < |r_1 - r_2|$, one circle is completely inside the other, and there are no common tangents.</li>
</ol>
<p>In this case :</p>
<p>$d = \frac{15}{2}$</p>
<p>$r_1 = \frac{5}{2}$</p>
<p>$r_2 = 5$</p>
<p>Now, let's check the conditions :</p>
<p>$r_1 + r_2 = \frac{5}{2} + 5$ = $\frac{15}{2}$ </p>
<p>Since $d = \frac{15}{2} = r_1 + r_2$, the circles touch each other externally, and there are 3 common tangents.</p>
| mcq | jee-main-2023-online-15th-april-morning-shift |
1lsg8yeao | maths | circle | number-of-common-tangents-and-position-of-two-circle | <p>If the circles $$(x+1)^2+(y+2)^2=r^2$$ and $$x^2+y^2-4 x-4 y+4=0$$ intersect at exactly two distinct points, then</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}<\\mathrm{r}<7$$\n"}, {"identifier": "B", "content": "$$3<\\mathrm{r}<7$$\n"}, {"identifier": "C", "content": "$$5<\\mathrm{r}<9$$\n"}, {"identifier": "D", "content": "$$0<\\mathrm{r}<7$$"}] | ["B"] | null | <p>If two circles intersect at two distinct points</p>
<p>$$\begin{aligned}
& \Rightarrow\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2 \\
& |\mathrm{r}-2|<\sqrt{9+16}<\mathrm{r}+2 \\
& |\mathrm{r}-2|<5 \text { and } \mathrm{r}+2>5 \\
& -5<\mathrm{r}-2<5 \quad \mathrm{r}>3 ~\text{......... 2}
\end{aligned}$$</p>
<p>$$-3<\mathrm{r}<7\quad$$ .... (1)</p>
<p>From (1) and (2)</p>
<p>$$3<\text { r }<7$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luy9clii | maths | circle | number-of-common-tangents-and-position-of-two-circle | <p>Let the centre of a circle, passing through the points $$(0,0),(1,0)$$ and touching the circle $$x^2+y^2=9$$, be $$(h, k)$$. Then for all possible values of the coordinates of the centre $$(h, k), 4\left(h^2+k^2\right)$$ is equal to __________.</p> | [] | null | 9 | <p>Circle will touch internally</p>
<p>$$\begin{aligned}
& C_1 C_2=\left|r_1-r_2\right| \\
& =\sqrt{h^2+k^2}=3-\sqrt{h^2+k^2} \\
& \Rightarrow 2 \sqrt{h^2+k^2}=3 \\
& \Rightarrow h^2+k^2=\frac{9}{4} \\
& \therefore 4\left(h^2+k^2\right)=9
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift |
lv5grw3t | maths | circle | number-of-common-tangents-and-position-of-two-circle | <p>Let the circles $$C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$$ and $$C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$$ touch each other externally at the point $$(6,6)$$. If the point $$(6,6)$$ divides the line segment joining the centres of the circles $$C_1$$ and $$C_2$$ internally in the ratio $$2: 1$$, then $$(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$$ equals</p> | [{"identifier": "A", "content": "130"}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "145"}, {"identifier": "D", "content": "125"}] | ["A"] | null | <p>$$\begin{aligned}
& C_1 \rightarrow(x-\alpha)^2+(y-\beta)^2=r_1^2 \\
& C_2 \rightarrow(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2
\end{aligned}$$</p>
<p>Point $$P$$ divide the line segment internally $$C_1 C_2$$ in the ratio 2 : 1</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8ocyvq/170f2048-377a-4fac-b220-54e6cf0f1120/13082c50-1332-11ef-9f8d-838c388c326d/file-1lw8ocyvr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8ocyvq/170f2048-377a-4fac-b220-54e6cf0f1120/13082c50-1332-11ef-9f8d-838c388c326d/file-1lw8ocyvr.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) 8th April Morning Shift Mathematics - Circle Question 6 English Explanation"></p>
<p>$$\begin{aligned}
& \frac{\alpha \times 1+8 \times 2}{1+2}=6, \alpha=2 \\
& \frac{\beta \times 1+\frac{15}{\alpha} \times 2}{1+2}=6, \beta=3 \\
& r_1=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{25}=5 \\
& r_2=\sqrt{(8-6)^2+\left(\frac{15}{\alpha}-6\right)^2}=\frac{5}{2} \\
& \alpha+\beta+4\left(r_1^2+r_2^2\right)=2+3+4\left(5^2+\left(\frac{5}{2}\right)^2\right) \\
& =5+4\left(\frac{125}{4}\right) \\
& =130 \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
k3MJZWxo319I8veN | maths | circle | orthogonality-of-two-circles | If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = 4$$ orthogonally, then the locus of its centre is : | [{"identifier": "A", "content": "$$2ax\\, - 2by\\, - ({a^2}\\, + \\,{b^2} + 4) = 0$$ "}, {"identifier": "B", "content": "$$2ax\\, + 2by\\, - ({a^2}\\, + \\,{b^2} + 4) = 0$$ "}, {"identifier": "C", "content": "$$2ax\\, - 2by\\, + ({a^2}\\, + \\,{b^2} + 4) = 0$$ "}, {"identifier": "D", "content": "$$2ax\\, + 2by\\, + ({a^2}\\, + \\,{b^2} + 4) = 0$$ "}] | ["B"] | null | Let the variable circle is
<br><br>$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>It passes through $$(a,b)$$
<br><br>$$\therefore$$ $${a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$(1)$$ cuts $${x^2} + {y^2} = 4$$ orthogonally
<br><br>$$\therefore$$ $$2\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4$$
<br><br>$$\therefore$$ from $$(2)$$ $$\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0$$
<br><br>$$\therefore$$ Locus of center $$\left( { - g, - f} \right)$$ is
<br><br>$${a^2} + {b^2} - 2ax - 2by + 4 = 0$$
<br><br>or $$2ax + 2by = {a^2} + {b^2} + 4$$ | mcq | aieee-2004 |
4lEUAEKUXUmcORcA | maths | circle | orthogonality-of-two-circles | If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = {p^2}$$ orthogonally, then the equation of the locus of its centre is : | [{"identifier": "A", "content": "$${x^2}\\, + \\,{y^2} - \\,3ax\\, - \\,4\\,by\\,\\, + \\,({a^2}\\, + \\,{b^2} - {p^2}) = 0$$ "}, {"identifier": "B", "content": "$$2ax\\, + \\,\\,2\\,by\\,\\, - \\,({a^2}\\, - \\,{b^2} + {p^2}) = 0$$ "}, {"identifier": "C", "content": "$${x^2}\\, + \\,{y^2} - \\,2ax\\, - \\,\\,3\\,by\\,\\, + \\,({a^2}\\, - \\,{b^2} - {p^2}) = 0$$ "}, {"identifier": "D", "content": "$$2ax\\, + \\,\\,2\\,by\\,\\, - \\,({a^2}\\, + \\,{b^2} + {p^2}) = 0$$ "}] | ["D"] | null | Let the center be $$\left( {\alpha ,\beta } \right)$$
<br><br>As It cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally
<br><br>$$\therefore$$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,$$ we get
<br><br>$$2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0$$
<br><br>$$ = {c_1} - {p^2} \Rightarrow {c_1} = {p^2}$$
<br><br>Let equation of circle is
<br><br>$${x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0$$
<br><br>It passes through
<br><br>$$\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0$$
<br><br>$$\therefore$$ Locus of $$\left( {\alpha ,\beta } \right)$$ is
<br><br>$$\therefore$$ $$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.$$ | mcq | aieee-2005 |
gIWxzoGle7kZLLADRm7k9k2k5kia3nk | maths | circle | orthogonality-of-two-circles | If the curves, x<sup>2</sup> – 6x + y<sup>2</sup> + 8 = 0 and
<br/>x<sup>2</sup> – 8y + y<sup>2</sup> + 16 – k = 0, (k > 0) touch each other
at a point, then the largest value of k is ______. | [] | null | 36 | C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> – 6x + + 8 = 0
<br><br>C<sub>1</sub>(3, 0) and r<sub>1</sub> = 1
<br><br>C<sub>2</sub> : x<sup>2</sup> + y<sup>2</sup> – 8y + 16 – k = 0
<br><br>C<sub>2</sub>(0, 4) and r<sub>2</sub> = $$\sqrt k $$
<br><br>Two circles touch each other
<br><br>$$ \therefore $$ C<sub>1</sub>C<sub>2</sub> = | r<sub>1</sub> $$ \pm $$ r<sub>2</sub> |
<br><br>$$ \Rightarrow $$ 5 = | 1 $$ \pm $$ $$\sqrt k $$ |
<br><br>$$ \therefore $$ 1 + $$\sqrt k $$ = 5 or $$\sqrt k $$ - 1 = 5
<br><br>$$ \Rightarrow $$ k = 16 or k = 36
<br><br>So largest value of k = 36. | integer | jee-main-2020-online-9th-january-evening-slot |
1l6p3thnw | maths | circle | orthogonality-of-two-circles | <p>Let the mirror image of a circle $$c_{1}: x^{2}+y^{2}-2 x-6 y+\alpha=0$$ in line $$y=x+1$$ be $$c_{2}: 5 x^{2}+5 y^{2}+10 g x+10 f y+38=0$$. If $$\mathrm{r}$$ is the radius of circle $$\mathrm{c}_{2}$$, then $$\alpha+6 \mathrm{r}^{2}$$ is equal to ________.</p> | [] | null | 12 | <p>$${c_1}:{x^2} + {y^2} - 2x - 6y + \alpha = 0$$</p>
<p>Then centre $$ = (1,3)$$ and radius $$(r) = \sqrt {10 - \alpha } $$</p>
<p>Image of $$(1,3)$$ w.r.t. line $$x - y + 1 = 0$$ is $$(2,2)$$</p>
<p>$${c_2}:5{x^2} + 5{y^2} + 10gx + 10fy + 38 = 0$$</p>
<p>or $${x^2} + {y^2} + 2gx + 2fy + {{38} \over 5} = 0$$</p>
<p>Then $$( - g, - f) = (2,2)$$</p>
<p>$$\therefore$$ $$g = f = - 2$$ .......... (i)</p>
<p>Radius of $${c_2} = r = \sqrt {4 + 4 - {{38} \over 5}} = \sqrt {10 - \alpha } $$</p>
<p>$$ \Rightarrow {2 \over 5} = 10 - \alpha $$</p>
<p>$$\therefore$$ $$\alpha = {{48} \over 5}$$ and $$r = \sqrt {{2 \over 5}} $$</p>
<p>$$\therefore$$ $$\alpha + 6{r^2} = {{48} \over 5} + {{12} \over 5} = 12$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
wwSAsTzXYKfdbOPl | maths | circle | pair-of-tangents | The centre of the circle passing through (0, 0) and (1, 0) and touching the circle $${x^2}\, + \,{y^2} = 9$$ is : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 2},\\,{1 \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{1 \\over 2},\\, - \\,\\sqrt 2 } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {{3 \\over 2},\\,{1 \\over 2}} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},\\,{3 \\over 2}} \\right)$$ "}] | ["B"] | null | Let the required circle be
<br><br>$${x^2} + {y^2} + 2gx + 2fy + c = 0$$
<br><br>Since it passes through $$\left( {0,0} \right)$$ and $$\left( {1,0} \right)$$
<br><br>$$ \Rightarrow c = 0$$ and $$g = - {1 \over 2}$$
<br><br>Points $$\left( {0,0} \right)$$ and $$\left( {1,0} \right)$$ lie inside the circle $${x^2} + {y^2} = 9,$$
<br><br>so two circles touch internally
<br><br>$$ \Rightarrow c{}_1{c_2} = {r_1} - {r_2}$$
<br><br>$$\therefore$$ $$\sqrt {{g^2} + {f^2}} = 3 - \sqrt {{g^2} + {f^2}} $$
<br><br>$$ \Rightarrow \sqrt {{g^2} + {f^2}} = {3 \over 2}$$
<br><br>$$ \Rightarrow {f^2} = {9 \over 4} - {1 \over 4} = 2$$
<br><br>$$\therefore$$ $$f = \pm \sqrt 2 .$$
<br><br>Hence, the centers of required circle are
<br><br>$$\left( {{1 \over 2}.\sqrt 2 } \right)$$ or $$\left( {{1 \over 2}, - \sqrt 2 } \right)$$ | mcq | aieee-2002 |
1ktgpx85g | maths | circle | pair-of-tangents | Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, and C<sub>1</sub>($$\alpha$$, $$\beta$$) and C<sub>2</sub>($$\gamma$$, $$\delta$$), C<sub>1</sub> $$\ne$$ C<sub>2</sub> are their centres, then |($$\alpha$$ + $$\beta$$) ($$\gamma$$ + $$\delta$$)| is equal to ___________. | [] | null | 40 | Slope of line joining centres of circles = $${4 \over 3} = \tan \theta $$<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kx5g1hrm/3c7cc7eb-253c-4f70-8baa-06a68070a4ec/9eab2120-5c7f-11ec-bf3c-e32253ee0710/file-1kx5g1hrn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kx5g1hrm/3c7cc7eb-253c-4f70-8baa-06a68070a4ec/9eab2120-5c7f-11ec-bf3c-e32253ee0710/file-1kx5g1hrn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Circle Question 66 English Explanation"><br>$$ \Rightarrow \cos \theta = {3 \over 5},\sin \theta = {4 \over 5}$$<br><br>Now using parametric form <br><br>$${{x - 1} \over {\cos \theta }} = {{y - 2} \over {\sin \theta }} = \pm \,5$$<br><br> (x, y) = (1 + 5cos$$\theta$$, 2 + 5sin$$\theta$$)<br><br>($$\alpha$$, $$\beta$$) = (4, 6)<br><br> (x, y) = ($$\gamma$$, $$\delta$$) = (1 $$-$$ 5cos$$\theta$$, 2 $$-$$ 5sin$$\theta$$)<br><br>($$\gamma$$, s) = ($$-$$2, $$-$$2)<br><br>$$\Rightarrow$$ |($$\alpha$$ + $$\beta$$) ($$\gamma$$ + $$\delta$$)| = | 10x $$-$$ 4 | = 40 | integer | jee-main-2021-online-27th-august-evening-shift |
1l5w01mde | maths | circle | pair-of-tangents | <p>Consider three circles:</p>
<p>$${C_1}:{x^2} + {y^2} = {r^2}$$</p>
<p>$${C_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$$</p>
<p>$${C_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$$</p>
<p>If a line L : y = mx + c be a common tangent to C<sub>1</sub>, C<sub>2</sub> and C<sub>3</sub> such that C<sub>1</sub> and C<sub>3</sub> lie on one side of line L while C<sub>2</sub> lies on other side, then the value of $$20({r^2} + c)$$ is equal to :</p> | [{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykyyx8/21715b9a-e296-480d-b13d-bb0f765bfb24/5c010fd0-0ae7-11ed-a51c-73986e88f75f/file-1l5ykyyx9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5ykyyx8/21715b9a-e296-480d-b13d-bb0f765bfb24/5c010fd0-0ae7-11ed-a51c-73986e88f75f/file-1l5ykyyx9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Circle Question 50 English Explanation"></p>
<p>$${c_1}:{x^2} + {y^2} = {r^2}$$ ; center = (0, 0) and radius = r</p>
<p>$${c_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$$ ; center = (1, 1) and radius = r</p>
<p>$${c_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$$ ; center = (2, 1) and radius = r</p>
<p>Distance of $$y = mx + c$$ line from center (0, 0) is,</p>
<p>$$\left| {{{0 + 0 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$$ ..... (1)</p>
<p>Distance of $$y = mx + c$$ line from center (1, 1) is,</p>
<p>$$\left| {{{m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$$ ..... (2)</p>
<p>Distance of $$y = mx + c$$ line from center (2, 1) is,</p>
<p>$$\left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$$ .... (3)</p>
<p>From (1) and (2), we get</p>
<p>$$\left| {{c \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right|$$</p>
<p>$$ \Rightarrow m - 1 + c = \pm \,c$$ ..... (4)</p>
<p>taking positive sign,</p>
<p>$$m - 1 + c = c$$</p>
<p>$$ \Rightarrow m - 1 = 0$$</p>
<p>$$ \Rightarrow m = 1$$</p>
<p>From (2) and (3), we get</p>
<p>$$\left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right|$$</p>
<p>$$ \Rightarrow (m - 1 + c) = \, \pm \,(2m - 1 + c)$$ ...... (5)</p>
<p>taking positive sign,</p>
<p>$$m - 1 + c = 2m - 1 + c$$</p>
<p>$$ \Rightarrow m = 0$$</p>
<p>By taking positive sign we get two different value of m so it is not acceptable.</p>
<p>From equation (4), taking negative sign,</p>
<p>$$m - 1 + c = - c$$</p>
<p>$$ \Rightarrow m + 2c - 1 = 0$$ ..... (6)</p>
<p>From equation (5), taking negative sign</p>
<p>$$m - 1 + c = - (2m - 1 + c)$$</p>
<p>$$ \Rightarrow 3m + 2c - 2 = 0$$ ..... (7)</p>
<p>Solving equation (6) and (7), we get</p>
<p>$$3m + 1 - m - 2 = 0$$</p>
<p>$$ \Rightarrow 2m = 1$$</p>
<p>$$ \Rightarrow m = {1 \over 2}$$</p>
<p>$$\therefore$$ $$2c = 1 - {1 \over 2}$$</p>
<p>$$ \Rightarrow c = {1 \over 4}$$</p>
<p>Putting value of $$m = {1 \over 2}$$ and $$c = {1 \over 4}$$ in equation (1), we get</p>
<p>$$r = \left| {{{{1 \over 4}} \over {\sqrt {1 + {1 \over 4}} }}} \right|$$</p>
<p>$$ = \left| {{1 \over 4} \times {2 \over {\sqrt 5 }}} \right|$$</p>
<p>$$ = {1 \over {2\sqrt 5 }}$$</p>
<p>$$\therefore$$ $$20({r^2} + c)$$</p>
<p>$$ = 20\left( {{1 \over {4 \times 5}} + {1 \over 4}} \right)$$</p>
<p>$$ = 20\left( {{{1 + 5} \over {20}}} \right)$$</p>
<p>$$ = 6$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
JQOYhImF3TPwcQ2r | maths | circle | position-of-a-point-with-respect-to-circle | The length of the diameter of the circle which touches the $$x$$-axis at the point $$(1, 0)$$ and passes through the point $$(2, 3)$$ is : | [{"identifier": "A", "content": "$${{10} \\over 3}$$ "}, {"identifier": "B", "content": "$${{3} \\over 5}$$"}, {"identifier": "C", "content": "$${{6} \\over 5}$$"}, {"identifier": "D", "content": "$${{5} \\over 3}$$"}] | ["A"] | null | Let center of the circle be $$\left( {1,h} \right)$$
<br><br>$$\left[ {\,\,} \right.$$ as circle touches $$x$$-axis at $$\left. {\left( {1,0} \right)\,\,} \right]$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265134/exam_images/q6ltvxbm00bb6jj7jhhr.webp" loading="lazy" alt="AIEEE 2012 Mathematics - Circle Question 139 English Explanation">
<br><br>Let the circle passes through the point $$B(2,3)$$
<br><br>$$\therefore$$ $$CA=CB$$ (radius)
<br><br>$$ \Rightarrow C{A^2} = C{B^2}$$
<br><br>$$ \Rightarrow {\left( {1 - 1} \right)^2} + \left( {h - 0} \right){}^2 = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2}$$
<br><br>$$ \Rightarrow {h^2} = 1 + {h^2} + 9 - 6h$$
<br><br>$$ \Rightarrow h = {{10} \over 6} = {5 \over 3}$$
<br><br>Thus, diameter is $$2h = {{10} \over 3}.$$ | mcq | aieee-2012 |
RKxAtAiPnPuYZZifgEV7R | maths | circle | position-of-a-point-with-respect-to-circle | If a point P has co-ordinates (0, $$-$$2) and Q is any point on the circle, x<sup>2</sup> + y<sup>2</sup> $$-$$ 5x $$-$$ y + 5 = 0, then the maximum value of (PQ)<sup>2</sup> is : | [{"identifier": "A", "content": "$${{25 + \\sqrt 6 } \\over 2}$$ "}, {"identifier": "B", "content": "14 + $$5\\sqrt 3 $$"}, {"identifier": "C", "content": "$${{47 + 10\\sqrt 6 } \\over 2}$$"}, {"identifier": "D", "content": "8 + 5$$\\sqrt 3 $$"}] | ["B"] | null | Given that x<sup>2</sup> + y<sup>2</sup> $$-$$ 5x $$-$$ y + 5 = 0
<br><br>$$ \Rightarrow $$ (x $$-$$ 5/2)<sup>2</sup> $$-$$ $${{25} \over 4}$$ + (y $$-$$ 1/2)<sup>2</sup> $$-$$ 1/4 = 0
<br><br>$$ \Rightarrow $$ (x $$-$$ 5/2)<sup>2</sup> + (y $$-$$ 1/2)<sup>2</sup> = 3/2
<br><br>on circle [ Q $$ \equiv $$ (5/2 + $$\sqrt {3/2} $$ cos Q, $${1 \over 2}$$ + $$\sqrt {3/2} $$ sin Q)]
<br><br>$$ \Rightarrow $$ PQ<sup>2</sup> = $${\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2}$$ + $${\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}$$
<br><br>$$ \Rightarrow $$ PQ<sup>2</sup> = $${{25} \over 2} + {3 \over 2} + 5\sqrt {3/2} $$ (cos Q + sinQ)
<br><br>= 14 + 5$$\sqrt {3/2} $$ (cosQ + sinQ)
<br><br>$$ \therefore $$ Maximum value of PQ<sup>2</sup>
<br><br>= 14 + 5$$\sqrt {3/2} $$ $$ \times $$ $$\sqrt 2 $$ = 14 + 5$$\sqrt 3 $$ | mcq | jee-main-2017-online-8th-april-morning-slot |
pEYxJxihaVS8EGPeEO1kluvq0eh | maths | circle | position-of-a-point-with-respect-to-circle | Let A(1, 4) and B(1, $$-$$5) be two points. Let P be a point on the circle <br/>(x $$-$$ 1)<sup>2</sup> + (y $$-$$ 1)<sup>2</sup> = 1 such that (PA)<sup>2</sup> + (PB)<sup>2</sup> have maximum value, then the points, P, A and B lie on : | [{"identifier": "A", "content": "a straight line"}, {"identifier": "B", "content": "an ellipse"}, {"identifier": "C", "content": "a parabola"}, {"identifier": "D", "content": "a hyperbola"}] | ["A"] | null | P be a point on $${(x - 1)^2} + {(y - 1)^2} = 1$$<br><br>so $$P(1 + \cos \theta ,1 + \sin \theta )$$<br><br>A(1, 4), B(1, $$-$$5)<br><br>$${(PA)^2} + {(PB)^2}$$<br><br>$$ = {(\cos \theta )^2} + {(\sin \theta - 3)^2} + {(\cos \theta )^2} + {(\sin \theta + 6)^2}$$<br><br>$$ = 47 + 6\sin \theta $$<br><br>It is maximum if $$\sin \theta = 1$$<br><br>When $$ \sin \theta = 1,\cos \theta = 0$$<br><br>So P(1, 2), A(1, 4), B(1, $$-$$5)<br><br>P, A, B are collinear points. | mcq | jee-main-2021-online-26th-february-evening-slot |
yqOGL0vZWY5owc1JTs1kmliqxwh | maths | circle | position-of-a-point-with-respect-to-circle | For the four circles M, N, O and P, following four equations are given :<br/><br/>Circle M : x<sup>2</sup> + y<sup>2</sup> = 1<br/><br/>Circle N : x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x = 0<br/><br/>Circle O : x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x $$-$$ 2y + 1 = 0<br/><br/>Circle P : x<sup>2</sup> + y<sup>2</sup> $$-$$ 2y = 0<br/><br/>If the centre of circle M is joined with centre of the circle N, further center of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a : | [{"identifier": "A", "content": "Rhombus"}, {"identifier": "B", "content": "Square"}, {"identifier": "C", "content": "Rectangle"}, {"identifier": "D", "content": "Parallelogram"}] | ["B"] | null | $${C_M} = (0,0)$$<br><br>$${C_N} = (1,0)$$<br><br>$${C_O} = (1,1)$$<br><br>$${C_P} = (0,1)$$<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266015/exam_images/tynzdnx9ipskg5hfjo3r.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Morning Shift Mathematics - Circle Question 79 English Explanation"> | mcq | jee-main-2021-online-18th-march-morning-shift |
1kru4cxvo | maths | circle | position-of-a-point-with-respect-to-circle | Let the circle S : 36x<sup>2</sup> + 36y<sup>2</sup> $$-$$ 108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x $$-$$ 2y = 4 and 2x $$-$$ y = 5 lies inside the circle S, then : | [{"identifier": "A", "content": "$${{25} \\over 9} < C < {{13} \\over 3}$$"}, {"identifier": "B", "content": "100 < C < 165"}, {"identifier": "C", "content": "81 < C < 156"}, {"identifier": "D", "content": "100 < C < 156"}] | ["D"] | null | S : 36x<sup>2</sup> + 36y<sup>2</sup> $$-$$ 108x + 120y + C = 0<br><br>$$\Rightarrow$$ x<sup>2</sup> + y<sup>2</sup> $$-$$ 3x + $${{10} \over 3}$$y + $${C \over {36}}$$ = 0<br><br>Centre $$ \equiv ( - g, - f) \equiv \left( {{3 \over 2},{{ - 10} \over 6}} \right)$$<br><br>radius = $$r = \sqrt {{9 \over 4} + {{100} \over {36}} - {C \over {36}}} $$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266423/exam_images/ulkyx4q33vkkr3loyvyz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - Circle Question 76 English Explanation"><br>Now, <br><br>$$ \Rightarrow r < {3 \over 2}$$<br><br>$$ \Rightarrow {9 \over 4} + {{100} \over {36}} - {C \over {36}} < {9 \over 4}$$<br><br>$$\Rightarrow$$ C > 100 ...... (1)<br><br>Now, point of intersection of x $$-$$ 2y = 4 and 2x $$-$$ y = 5 is (2, $$-$$1), which lies inside the circle S.<br><br>$$\therefore$$ S(2, $$-$$1) < 0<br><br>$$\Rightarrow$$ (2)<sup>2</sup> + ($$-$$1)<sup>2</sup> $$-$$ 3(2) + $${{10} \over 3}$$($$-$$1) + $${C \over {36}}$$ < 0<br><br>$$\Rightarrow$$ 4 + 1 $$-$$ 6 $$-$$ $${{10} \over 3}$$ + $${C \over {36}}$$ < 0<br><br>C < 156 ..... (2)<br><br>From (1) & (2)<br><br>100 < C < 156 Ans. | mcq | jee-main-2021-online-22th-july-evening-shift |
1ktislkyb | maths | circle | position-of-a-point-with-respect-to-circle | If the variable line 3x + 4y = $$\alpha$$ lies between the two <br/>circles (x $$-$$ 1)<sup>2</sup> + (y $$-$$ 1)<sup>2</sup> = 1 <br/>and (x $$-$$ 9)<sup>2</sup> + (y $$-$$ 1)<sup>2</sup> = 4, without intercepting a chord on either circle, then the sum of all the integral values of $$\alpha$$ is ___________. | [] | null | 165 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263671/exam_images/tmz0n9h2htagrkfozwru.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Circle Question 65 English Explanation"><br><br>Both centers should lie on either side of the line as well as line can be tangent to circle.<br><br>(3 + 4 $$-$$ $$\alpha$$) . (27 + 4 $$-$$ $$\alpha$$) < 0<br><br>(7 $$-$$ $$\alpha$$) . (31 $$-$$ $$\alpha$$) < 0 $$\Rightarrow$$ $$\alpha$$ $$\in$$ (7, 31) ....... (1)<br><br>d<sub>1</sub> = distance of (1, 1) from line<br><br>d<sub>2</sub> = distance of (9, 1) from line<br><br>$${d_1} \ge {r_1} \Rightarrow {{|7 - \alpha |} \over 5} \ge 1 \Rightarrow \alpha \in ( - \infty ,2] \cup [12,\infty )$$ .... (2)<br><br>$${d_2} \ge {r_2} \Rightarrow {{|31 - \alpha |} \over 5} \ge 2 \Rightarrow \alpha \in ( - \infty ,21] \cup [41,\infty )$$ ....(3)<br><br>(1) $$\cap$$ (2) $$\cap$$ (3) $$\Rightarrow$$ $$\alpha$$ $$\in$$ [12, 21]<br><br>Sum of integers = 165 | integer | jee-main-2021-online-31st-august-morning-shift |
1l56r83vp | maths | circle | position-of-a-point-with-respect-to-circle | <p>The set of values of k, for which the circle $$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$ lies inside the fourth quadrant and the point $$\left( {1, - {1 \over 3}} \right)$$ lies on or inside the circle C, is :</p> | [{"identifier": "A", "content": "an empty set"}, {"identifier": "B", "content": "$$\\left( {6,{{65} \\over 9}} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {{{80} \\over 9},10} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {9,{{92} \\over 9}} \\right]$$"}] | ["D"] | null | <p>$$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$</p>
<p>$$\because$$ $$\left( {1, - {1 \over 3}} \right)$$ lies on or inside the C</p>
<p>then $$4 + {4 \over 9} - 12 - {8 \over 3} + k \le 0$$</p>
<p>$$ \Rightarrow k \le {{92} \over 9}$$</p>
<p>Now, circle lies in 4<sup>th</sup> quadrant centre $$ \equiv \left( {{3 \over 2}, - 1} \right)$$</p>
<p>$$\therefore$$ $$r < 1 \Rightarrow \sqrt {{9 \over 4} + 1 - {k \over 4}} < 1$$</p>
<p>$$ \Rightarrow {{13} \over 4} - {k \over 4} < 1$$</p>
<p>$$ \Rightarrow {k \over 4} > {9 \over 4}$$</p>
<p>$$ \Rightarrow k > 9$$</p>
<p>$$\therefore$$ $$k \in \left( {9,{{92} \over 9}} \right)$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
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