Circle
187 Questions
Start JEE Mains Test
2021
Q101
JEE Mains
MCQ
14 Mar 2026
Let the tangent to the circle x2 + y2 = 25 at the point R(3, 4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to :
A.
${{585} \over {66}}$
B.
${{625} \over {72}}$
C.
${{529} \over {64}}$
D.
${{125} \over {72}}$
2021
Q102
JEE Mains
MCQ
14 Mar 2026
Two tangents are drawn from a point P to the circle x2 + y2 $-$ 2x $-$ 4y + 4 = 0, such that the angle between these tangents is ${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$, where ${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$ $\in$(0, $\pi$). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of $\Delta$PAB and $\Delta$CAB is :
A.
3 : 1
B.
9 : 4
C.
2 : 1
D.
11 : 4
2021
Q103
JEE Mains
MCQ
14 Mar 2026
The line 2x $-$ y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x $-$ 2y = 4. Then, the radius of the circle is :
A.
5$\sqrt 3 $
B.
4$\sqrt 5 $
C.
3$\sqrt 5 $
D.
5$\sqrt 4 $
2021
Q104
JEE Mains
MCQ
14 Mar 2026
Choose the incorrect statement about the two circles whose equations are given below :
x2 + y2 $-$ 10x $-$ 10y + 41 = 0 and
x2 + y2 $-$ 16x $-$ 10y + 80 = 0
x2 + y2 $-$ 10x $-$ 10y + 41 = 0 and
x2 + y2 $-$ 16x $-$ 10y + 80 = 0
A.
Distance between two centres is the average of radii of both the circles.
B.
Both circles pass through the centre of each other.
C.
Circles have two intersection points.
D.
Both circle's centers lie inside region of one another.
2021
Q105
JEE Mains
MCQ
14 Mar 2026
Let the lengths of intercepts on x-axis and y-axis made by the circle
x2 + y2 + 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 :
x2 + y2 + 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 :
A.
${\sqrt {10} }$
B.
${\sqrt {6} }$
C.
${\sqrt {11} }$
D.
${\sqrt {7} }$
2021
Q106
JEE Mains
MCQ
14 Mar 2026
Let A(1, 4) and B(1, $-$5) be two points. Let P be a point on the circle
(x $-$ 1)2 + (y $-$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :
(x $-$ 1)2 + (y $-$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :
A.
a straight line
B.
an ellipse
C.
a parabola
D.
a hyperbola
2021
Q107
JEE Mains
MCQ
14 Mar 2026
If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of radius r, then r is equal to :
A.
${1 \over 4}$
B.
${1 \over 2}$
C.
1
D.
${1 \over 3}$
2021
Q108
JEE Mains
MCQ
14 Mar 2026
In the circle given below, let OA = 1 unit, OB = 13 unit and PQ $ \bot $ OB. Then, the area of the triangle PQB (in square units) is :
A.
24$\sqrt 2 $
B.
24$\sqrt 3 $
C.
26$\sqrt 2 $
D.
26$\sqrt 3 $
2021
Q109
JEE Mains
Numerical
14 Mar 2026
Let B be the centre of the circle x2 + y2 $-$ 2x + 4y + 1 = 0. Let the tangents at two points P and Q on the circle intersect at the point A(3, 1). Then 8.$\left( {{{area\,\Delta APQ} \over {area\,\Delta BPQ}}} \right)$ is equal to _____________.
Correct Answer: 18
Explanation:
Radius = $\sqrt {1 + 4 - 1} = 2$
$AB = \sqrt {{3^2} + {2^2}} = \sqrt {13} $
In $\Delta$ABP
$A{P^2} = A{B^2} - B{P^2} = 13 - 4 = 9$
AP = 3
AQ = AP = 3
Let $\angle$ABP = $\theta$, $\angle$BAP = 90$-$ $\theta$
In $\Delta$ABP, tan$\theta$ = 3/2
$\sin \theta = {3 \over {\sqrt {13} }}$, $\cos \theta = {2 \over {\sqrt {13} }}$
In $\Delta$ARP,
$\cos (90 - \theta ) = {{AR} \over {AP}} \Rightarrow AR = 3\sin \theta $
In $\Delta$BRP,
$\cos \theta = {{BR} \over {BP}}$
$ \Rightarrow BR = 2\cos \theta = {{Area\,(\Delta APQ)} \over {Area\,(\Delta BPQ)}} = {{{1 \over 2} \times PQ \times AR} \over {{1 \over 2} \times PQ \times BR}}$
$ = {{AR} \over {RB}} = {{3\sin \theta } \over {2\cos \theta }} = {9 \over 4}$
$ \Rightarrow 8\left( {{{Area\,(\Delta APQ)} \over {Area\,(\Delta BPQ)}}} \right) = 18$
2021
Q110
JEE Mains
Numerical
14 Mar 2026
If the variable line 3x + 4y = $\alpha$ lies between the two
circles (x $-$ 1)2 + (y $-$ 1)2 = 1
and (x $-$ 9)2 + (y $-$ 1)2 = 4, without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is ___________.
circles (x $-$ 1)2 + (y $-$ 1)2 = 1
and (x $-$ 9)2 + (y $-$ 1)2 = 4, without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is ___________.
Correct Answer: 165
Explanation:
Both centers should lie on either side of the line as well as line can be tangent to circle.
(3 + 4 $-$ $\alpha$) . (27 + 4 $-$ $\alpha$) < 0
(7 $-$ $\alpha$) . (31 $-$ $\alpha$) < 0 $\Rightarrow$ $\alpha$ $\in$ (7, 31) ....... (1)
d1 = distance of (1, 1) from line
d2 = distance of (9, 1) from line
${d_1} \ge {r_1} \Rightarrow {{|7 - \alpha |} \over 5} \ge 1 \Rightarrow \alpha \in ( - \infty ,2] \cup [12,\infty )$ .... (2)
${d_2} \ge {r_2} \Rightarrow {{|31 - \alpha |} \over 5} \ge 2 \Rightarrow \alpha \in ( - \infty ,21] \cup [41,\infty )$ ....(3)
(1) $\cap$ (2) $\cap$ (3) $\Rightarrow$ $\alpha$ $\in$ [12, 21]
Sum of integers = 165
2021
Q111
JEE Mains
Numerical
14 Mar 2026
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 C1($\alpha$, $\beta$) and C2($\gamma$, $\delta$), C1 $\ne$ C2 are their centres, then |($\alpha$ + $\beta$) ($\gamma$ + $\delta$)| is equal to ___________.
Correct Answer: 40
Explanation:
Slope of line joining centres of circles = ${4 \over 3} = \tan \theta $
$ \Rightarrow \cos \theta = {3 \over 5},\sin \theta = {4 \over 5}$
Now using parametric form
${{x - 1} \over {\cos \theta }} = {{y - 2} \over {\sin \theta }} = \pm \,5$
(x, y) = (1 + 5cos$\theta$, 2 + 5sin$\theta$)
($\alpha$, $\beta$) = (4, 6)
(x, y) = ($\gamma$, $\delta$) = (1 $-$ 5cos$\theta$, 2 $-$ 5sin$\theta$)
($\gamma$, s) = ($-$2, $-$2)
$\Rightarrow$ |($\alpha$ + $\beta$) ($\gamma$ + $\delta$)| = | 10x $-$ 4 | = 40
$ \Rightarrow \cos \theta = {3 \over 5},\sin \theta = {4 \over 5}$
Now using parametric form
${{x - 1} \over {\cos \theta }} = {{y - 2} \over {\sin \theta }} = \pm \,5$
(x, y) = (1 + 5cos$\theta$, 2 + 5sin$\theta$)
($\alpha$, $\beta$) = (4, 6)
(x, y) = ($\gamma$, $\delta$) = (1 $-$ 5cos$\theta$, 2 $-$ 5sin$\theta$)
($\gamma$, s) = ($-$2, $-$2)
$\Rightarrow$ |($\alpha$ + $\beta$) ($\gamma$ + $\delta$)| = | 10x $-$ 4 | = 40
2021
Q112
JEE Mains
Numerical
14 Mar 2026
Let the equation x2 + y2 + 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 = p2 and q is an integer} is __________.
Correct Answer: 61
Explanation:
$r = \sqrt {{{{p^2}} \over 4} + {{{{(1 - p)}^2}} \over 4} - 5} = {{\sqrt {2{p^2} - 2p - 19} } \over 2}$
Since, $r \in (0,5]$
So, $0 < 2{p^2} - 2p - 19 \le 100$
$ \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]$
so, number of integral values of p2 is 61.
Since, $r \in (0,5]$
So, $0 < 2{p^2} - 2p - 19 \le 100$
$ \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]$
so, number of integral values of p2 is 61.
2021
Q113
JEE Mains
Numerical
14 Mar 2026
The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d2 is equal to _____________.
Correct Answer: 16
Explanation:
Let point P(x, y)
A(0, 0), B(1, 0), C(0, 1), D(1, 1)
(PA)2 + (PB)2 + (PC)2 + (PD)2 = 18
${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$ = 18
$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$
$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$
$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $
$ \Rightarrow {d^2} = 16$
A(0, 0), B(1, 0), C(0, 1), D(1, 1)
(PA)2 + (PB)2 + (PC)2 + (PD)2 = 18
${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$ = 18
$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$
$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$
$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $
$ \Rightarrow {d^2} = 16$
2021
Q114
JEE Mains
Numerical
14 Mar 2026
The minimum distance between any two points P1 and P2 while considering point P1 on one circle and point P2 on the other circle for the given circles' equations
x2 + y2 $-$ 10x $-$ 10y + 41 = 0
x2 + y2 $-$ 24x $-$ 10y + 160 = 0 is ___________.
x2 + y2 $-$ 10x $-$ 10y + 41 = 0
x2 + y2 $-$ 24x $-$ 10y + 160 = 0 is ___________.
Correct Answer: 1
Explanation:
${S_1}:{(x - 5)^2} + {(y - 5)^2} = 9$
Centre (5, 5), r1 = 3
${S_2}:{(x - 12)^2} + {(y - 5)^2} = 9$
Centre (12, 5), r2 = 3
So (P1P2)min = 1
Centre (5, 5), r1 = 3
${S_2}:{(x - 12)^2} + {(y - 5)^2} = 9$
Centre (12, 5), r2 = 3
So (P1P2)min = 1
2021
Q115
JEE Mains
Numerical
14 Mar 2026
Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point ($-$5, 0). If the locus of the point P is a circle of radius r, then 4r2 is equal to ________
Correct Answer: 56
Explanation:
Let P(h, k)
Given
PA = 3PB
PA2 = 9PB2
$ \Rightarrow $ (h $-$ 5)2 + k2 = 9[(h + 5)2 + k2]
$ \Rightarrow $ 8h2 + 8k2 + 100h + 200 = 0
$ \therefore $ Locus
${x^2} + {y^2} + \left( {{{25} \over 2}} \right)x + 25 = 0$
$ \therefore $ $c \equiv \left( {{{ - 25} \over 4},0} \right)$
$ \therefore $ ${r^2} = {\left( {{{ - 25} \over 4}} \right)^2} - 25$
$ = {{625} \over {16}} - 25$
$ = {{225} \over {16}}$
$ \therefore $ $4{r^2} = 4 \times {{225} \over {16}} = {{225} \over 4} = 56.25$
After Round of 4r2 = 56
Given
PA = 3PB
PA2 = 9PB2
$ \Rightarrow $ (h $-$ 5)2 + k2 = 9[(h + 5)2 + k2]
$ \Rightarrow $ 8h2 + 8k2 + 100h + 200 = 0
$ \therefore $ Locus
${x^2} + {y^2} + \left( {{{25} \over 2}} \right)x + 25 = 0$
$ \therefore $ $c \equiv \left( {{{ - 25} \over 4},0} \right)$
$ \therefore $ ${r^2} = {\left( {{{ - 25} \over 4}} \right)^2} - 25$
$ = {{625} \over {16}} - 25$
$ = {{225} \over {16}}$
$ \therefore $ $4{r^2} = 4 \times {{225} \over {16}} = {{225} \over 4} = 56.25$
After Round of 4r2 = 56
2021
Q116
JEE Mains
Numerical
14 Mar 2026
If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x $-$ 2)2 + (y $-$ 3)2 = 25 at the point (5, 7) is A, then 24A is equal to _________.
Correct Answer: 1225
Explanation:
This question is bonus if we consider poistive x axis.If we consider only x axis for this question then it is right question.
2021
Q117
JEE Mains
Numerical
14 Mar 2026
If one of the diameters of the circle x2 + y2 - 2x - 6y + 6 = 0 is a chord of another circle 'C',
whose center is at (2, 1), then its radius is ________.
Correct Answer: 3
Explanation:
Circle x2 + y2 - 2x - 6y + 6 = 0 has centre
O1(1, 3) and radius r
= 2.
Let centre O2 (2, 1) of required circle and its radius being r.
Distance between (1, 3) and (2, 1) is $\sqrt 5 $
$ \therefore $ ${\left( {\sqrt 5 } \right)^2} + {(2)^2} = {r^2}$
$ \Rightarrow r = 3$
Let centre O2 (2, 1) of required circle and its radius being r.
Distance between (1, 3) and (2, 1) is $\sqrt 5 $
$ \therefore $ ${\left( {\sqrt 5 } \right)^2} + {(2)^2} = {r^2}$
$ \Rightarrow r = 3$
2020
Q118
JEE Mains
MCQ
14 Mar 2026
If the length of the chord of the circle,
x2 + y2 = r2 (r > 0) along the line, y – 2x = 3 is r,
then r2 is equal to :
x2 + y2 = r2 (r > 0) along the line, y – 2x = 3 is r,
then r2 is equal to :
A.
${9 \over 5}$
B.
${{24} \over 5}$
C.
${{12} \over 5}$
D.
12
2020
Q119
JEE Mains
MCQ
14 Mar 2026
The circle passing through the intersection of the circles,
x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its centre on
the line, 2x – 3y + 12 = 0, also passes through the point :
x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its centre on
the line, 2x – 3y + 12 = 0, also passes through the point :
A.
(–3, 1)
B.
(1, –3)
C.
(–1, 3)
D.
(–3, 6)
2020
Q120
JEE Mains
MCQ
14 Mar 2026
A circle touches the y-axis at the point (0, 4)
and passes through the point (2, 0). Which of
the following lines is not a tangent to this circle?
A.
3x – 4y – 24 = 0
B.
4x + 3y – 8 = 0
C.
3x + 4y – 6 = 0
D.
4x – 3y + 17 = 0
Equation of family of circle touching y-axis at
2020
Q121
JEE Mains
MCQ
14 Mar 2026
If a line, y = mx + c is a tangent to the circle,
(x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2 + y2 = 1 at the point $\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$, then :
A.
c2 + 6c + 7 = 0
B.
c2 - 7c + 6 = 0
C.
c2 – 6c + 7 = 0
D.
c2 + 7c + 6 = 0
2020
Q122
JEE Mains
MCQ
14 Mar 2026
Let the tangents drawn from the origin to the circle,
x2 + y2 - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)2 is equal to :
x2 + y2 - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)2 is equal to :
A.
${{56} \over 5}$
B.
${{32} \over 5}$
C.
${{52} \over 5}$
D.
${{64} \over 5}$
2020
Q123
JEE Mains
Numerical
14 Mar 2026
Let PQ be a diameter of the circle x2 + y2 = 9. If $\alpha $ and $\beta $ are the lengths of the perpendiculars from P and Q on the straight line,
x + y = 2 respectively, then the maximum value of $\alpha\beta $ is _____.
x + y = 2 respectively, then the maximum value of $\alpha\beta $ is _____.
Correct Answer: 7
Explanation:
Let $P(3\cos \theta ,\,3\sin \theta )$
$Q( - 3\cos \theta ,\, - 3\sin \theta )$
$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$
$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$
$\alpha \beta = \left| {{{{{\left( {3\cos \theta + 3\sin \theta } \right)}^2} - 4} \over 2}} \right|$
$ = \left| {{{5 + 9\sin 2\theta } \over 2}} \right|$
$\alpha {\beta _{\max }}$$ = {{5 + 9} \over 2} = 7$ (when sin2$\theta $ = 1)
$Q( - 3\cos \theta ,\, - 3\sin \theta )$
$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$
$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$
$\alpha \beta = \left| {{{{{\left( {3\cos \theta + 3\sin \theta } \right)}^2} - 4} \over 2}} \right|$
$ = \left| {{{5 + 9\sin 2\theta } \over 2}} \right|$
$\alpha {\beta _{\max }}$$ = {{5 + 9} \over 2} = 7$ (when sin2$\theta $ = 1)
2020
Q124
JEE Mains
Numerical
14 Mar 2026
The diameter of the circle, whose centre lies on
the line x + y = 2 in the first quadrant and which
touches both the lines x = 3 and y = 2, is
_______ .
Correct Answer: 3
Explanation:
$ \because $ center lies on x + y = 2 and in 1st quadrant center = ($\alpha $, 2 $-$ $\alpha $)
where $\alpha $ > 0 and 2 $-$ $\alpha $ > 0 $ \Rightarrow $ 0 < $\alpha $ < 2
$ \because $ circle touches x = 3 and y = 2
$ \therefore $ ${{\left| {\alpha - 3} \right|} \over 1} = r$
and ${{\left| {2 - (2 - \alpha )} \right|} \over 1} = r$
$ \Rightarrow \,|\alpha |\, = r$
$ \therefore $ $|\alpha - 3|\, = \,|\alpha |$
$ \Rightarrow $ ${\alpha ^2} - 6\alpha + 9 = {\alpha ^2}$
$ \Rightarrow \alpha = {3 \over 2}$
$ \therefore $ $r = {3 \over 2}$
$ \Rightarrow $ 2r = 3 = diameter.
2020
Q125
JEE Mains
Numerical
14 Mar 2026
The number of integral values of k for which
the line, 3x + 4y = k intersects the circle,
x2 + y2 – 2x – 4y + 4 = 0 at two distinct points is ______.
x2 + y2 – 2x – 4y + 4 = 0 at two distinct points is ______.
Correct Answer: 9
Explanation:
Circle x2
+ y2
– 2x – 4y + 4 = 0
$ \Rightarrow $ (x – 1)2 + (y – 2)2 = 1
Centre: (1, 2), radius = 1
Line 3x + 4y – k = 0 intersects the circle at two distinct points.
$ \Rightarrow $ distance of centre from the line < radius
$ \Rightarrow $ $\left| {{{3 \times 1 + 4 \times 2 - k} \over {\sqrt {{3^2} + {4^2}} }}} \right| < 1$
$ \Rightarrow $ |11 - k| < 5
$ \Rightarrow $ 6 < k < 5
$ \Rightarrow $ k $ \in $ {7, 8, 9, ……15} since k $ \in $ I
$ \therefore $ Total 9 integral value of k.
$ \Rightarrow $ (x – 1)2 + (y – 2)2 = 1
Centre: (1, 2), radius = 1
Line 3x + 4y – k = 0 intersects the circle at two distinct points.
$ \Rightarrow $ distance of centre from the line < radius
$ \Rightarrow $ $\left| {{{3 \times 1 + 4 \times 2 - k} \over {\sqrt {{3^2} + {4^2}} }}} \right| < 1$
$ \Rightarrow $ |11 - k| < 5
$ \Rightarrow $ 6 < k < 5
$ \Rightarrow $ k $ \in $ {7, 8, 9, ……15} since k $ \in $ I
$ \therefore $ Total 9 integral value of k.
2020
Q126
JEE Mains
Numerical
14 Mar 2026
If the curves, x2 – 6x + y2 + 8 = 0 and
x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the largest value of k is ______.
x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the largest value of k is ______.
Correct Answer: 36
Explanation:
C1 : x2 + y2 – 6x + + 8 = 0
C1(3, 0) and r1 = 1
C2 : x2 + y2 – 8y + 16 – k = 0
C2(0, 4) and r2 = $\sqrt k $
Two circles touch each other
$ \therefore $ C1C2 = | r1 $ \pm $ r2 |
$ \Rightarrow $ 5 = | 1 $ \pm $ $\sqrt k $ |
$ \therefore $ 1 + $\sqrt k $ = 5 or $\sqrt k $ - 1 = 5
$ \Rightarrow $ k = 16 or k = 36
So largest value of k = 36.
C1(3, 0) and r1 = 1
C2 : x2 + y2 – 8y + 16 – k = 0
C2(0, 4) and r2 = $\sqrt k $
Two circles touch each other
$ \therefore $ C1C2 = | r1 $ \pm $ r2 |
$ \Rightarrow $ 5 = | 1 $ \pm $ $\sqrt k $ |
$ \therefore $ 1 + $\sqrt k $ = 5 or $\sqrt k $ - 1 = 5
$ \Rightarrow $ k = 16 or k = 36
So largest value of k = 36.
2019
Q127
JEE Mains
MCQ
14 Mar 2026
A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the
point :
A.
(1, 5)
B.
( 2, 3)
C.
(3, 5)
D.
(3, 10)
2019
Q128
JEE Mains
MCQ
14 Mar 2026
If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90o, then
the length (in cm) of their common chord is :
A.
${{13} \over 5}$
B.
${{60} \over {13}}$
C.
${{120} \over {13}}$
D.
${{13} \over 2}$
C1C2 = $\sqrt {{{12}^2} + {5^2}} $ = 13
2019
Q129
JEE Mains
MCQ
14 Mar 2026
The locus of the centres of the circles, which touch the circle, x2
+ y2
= 1 externally, also touch the y-axis and
lie in the first quadrant, is :
A.
$x = \sqrt {1 + 2y} ,y \ge 0$
B.
$y = \sqrt {1 + 2x} ,x \ge 0$
C.
$y = \sqrt {1 + 4x} ,x \ge 0$
D.
$x = \sqrt {1 + 4y} ,y \ge 0$
2019
Q130
JEE Mains
MCQ
14 Mar 2026
If the circles x2
+ y2
+ 5Kx + 2y + K = 0 and 2(x2
+ y2) + 2Kx + 3y –1 = 0, (K$ \in $R), intersect at the points
P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for :
A.
exactly two values of K
B.
no value of K
C.
exactly one value of K
D.
infinitely many values of K
2019
Q131
JEE Mains
MCQ
14 Mar 2026
The line x = y touches a circle at the point (1,1). If the circle also passes through the point (1, – 3), then its
radius is :
A.
3
B.
2
C.
2$\sqrt 2 $
D.
3$\sqrt 2 $
2019
Q132
JEE Mains
MCQ
14 Mar 2026
A rectangle is inscribed in a circle with a diameter
lying along the line 3y = x + 7. If the two adjacent
vertices of the rectangle are (–8, 5) and (6, 5), then
the area of the rectangle (in sq. units) is :
A.
72
B.
84
C.
56
D.
98
2019
Q133
JEE Mains
MCQ
14 Mar 2026
The common tangent to the circles x 2 + y2 = 4 and
x2 + y2 + 6x + 8y – 24 = 0 also passes through the
point :
A.
(6, –2)
B.
(4, –2)
C.
(–4, 6)
D.
(–6, 4)
2019
Q134
JEE Mains
MCQ
14 Mar 2026
If a tangent to the circle x2 + y2 = 1 intersects
the coordinate axes at distinct points P and Q,
then the locus of the mid-point of PQ is :
A.
x2 + y2 – 4x2y2 = 0
B.
x2 + y2 - 2xy = 0
C.
x2 + y2 – 2x2y2 = 0
D.
x2 + y2 - 16x2y2 = 0
2019
Q135
JEE Mains
MCQ
14 Mar 2026
The tangent and the normal lines at the point
( $\sqrt 3 $, 1) to the circle x2
+ y2 = 4 and the x-axis form a triangle. The area of this triangle (in
square units) is :
A.
${4 \over {\sqrt 3 }}$
B.
${1 \over {\sqrt 3 }}$
C.
${2 \over {\sqrt 3 }}$
D.
${1 \over {3 }}$
2019
Q136
JEE Mains
MCQ
14 Mar 2026
The sum of the squares of the lengths of the chords
intercepted on the circle, x2 + y2 = 16, by the lines,
x + y = n, n $ \in $ N, where N is the set of all natural
numbers, is :
A.
210
B.
160
C.
320
D.
105
2019
Q137
JEE Mains
MCQ
14 Mar 2026
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 :
A.
(x2 + y2)2 = 4R2x2y2
B.
(x2 + y2) (x + y) = R2xy
C.
(x2 + y2)2 = 4Rx2y2
D.
(x2 + y2)3 = 4R2x2y2
2019
Q138
JEE Mains
MCQ
14 Mar 2026
If a variable line, 3x + 4y – $\lambda $ = 0 is such that the two circles x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of $\lambda $ is the interval :
A.
(23, 31)
B.
(2, 17)
C.
[13, 23]
D.
[12, 21]
2019
Q139
JEE Mains
MCQ
14 Mar 2026
Let C1 and C2 be the centres of the circles x2 + y2 – 2x – 2y – 2 = 0 and x2 + y2 – 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 PC1QC2 is :
A.
4
B.
6
C.
9
D.
8
2019
Q140
JEE Mains
MCQ
14 Mar 2026
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 :
A.
$2\sqrt 2 $
B.
$\sqrt 2 $
C.
2
D.
1
2019
Q141
JEE Mains
MCQ
14 Mar 2026
A square is inscribed in the circle x2 + y2
– 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 :
A.
$\sqrt {137} $
B.
6
C.
$\sqrt {41} $
D.
13
2019
Q142
JEE Mains
MCQ
14 Mar 2026
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 :
A.
$4\sqrt 5 $
B.
${{\sqrt 5 } \over 2}$
C.
$2\sqrt 5 $
D.
${{\sqrt 5 } \over 4}$
2019
Q143
JEE Mains
MCQ
14 Mar 2026
If the area of an equilateral triangle inscribed in the circle x2 + y2
+ 10x + 12y + c = 0 is $27\sqrt 3 $ sq units then c is equal to :
A.
20
B.
25
C.
$-$ 25
D.
13
2019
Q144
JEE Mains
MCQ
14 Mar 2026
If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, – 1), then the radius of C is :
A.
5
B.
2$\sqrt {5} $
C.
4
D.
$\sqrt {37} $
2019
Q145
JEE Mains
MCQ
14 Mar 2026
If the circles
x2 + y2 $-$ 16x $-$ 20y + 164 = r2
and (x $-$ 4)2 + (y $-$ 7)2 = 36
intersect at two distinct points, then :
x2 + y2 $-$ 16x $-$ 20y + 164 = r2
and (x $-$ 4)2 + (y $-$ 7)2 = 36
intersect at two distinct points, then :
A.
r > 11
B.
0 < r < 1
C.
r = 11
D.
1 < r < 11
2019
Q146
JEE Mains
MCQ
14 Mar 2026
Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent, then :
A.
a, b, c are in A.P.
B.
$\sqrt a ,\sqrt b ,\sqrt c $ are in A.P
C.
${1 \over {\sqrt b }} + {1 \over {\sqrt c }}$ = ${1 \over {\sqrt a }}$
D.
${1 \over {\sqrt b }} = {1 \over {\sqrt a }} + {1 \over {\sqrt c }}$
2018
Q147
JEE Mains
MCQ
14 Mar 2026
If a circle C, whose radius is 3, touches externally the circle,
${x^2} + {y^2} + 2x - 4y - 4 = 0$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :
${x^2} + {y^2} + 2x - 4y - 4 = 0$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :
A.
$2\sqrt 5 $
B.
$3\sqrt 2 $
C.
$\sqrt 5 $
D.
$2\sqrt 3 $
2018
Q148
JEE Mains
MCQ
14 Mar 2026
If the tangent at (1, 7) to the curve x2 = y - 6
touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is :
touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is :
A.
95
B.
195
C.
185
D.
85
2018
Q149
JEE Mains
MCQ
14 Mar 2026
The tangent to the circle C1 : x2 + y2 $-$ 2x $-$ 1 = 0 at the point (2, 1) cuts off a chord of length 4 from a circle C2 whose center is (3, $-$2). The radius of C2 is :
A.
2
B.
$\sqrt 2 $
C.
3
D.
$\sqrt 6 $
2018
Q150
JEE Mains
MCQ
14 Mar 2026
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 :
A.
2
B.
$\sqrt 5 $
C.
$\sqrt 2 $
D.
1