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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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391 Medium
391 of 685
Application +4 | -1
If S1:x2+y2=4S_1: x^2+y^2=4 and S2:x2+y2−2ax−2by+2=0S_2: x^2+y^2-2 a x-2 b y+2=0 touches each other, then 4 times of radius of the circle S2S_2 is:
View answer and explanation
Correct option: A
2
Option A (2) is correct because the radius of the second circle can be found using the condition for touching circles and then multiplied by 4.
392 Medium
392 of 685
Application +4 | -1
The two circles x2+y2−2x−2y+1=0{x^2} + {y^2} - 2x - 2y + 1 = 0 and x2+y2−4x−6y−8=0{x^2} + {y^2} - 4x - 6y - 8 = 0 are such that
View answer and explanation
Correct option: B
They intersect each other
Option B (They intersect each other) is correct because the distance between the centers is less than the sum of the radii and greater than the absolute difference of the radii.
393 Medium
393 of 685
Application +4 | -1
Tangents at two points AA and BB on the circle x2+x^2+ y2−6x−4y−11=0y^2-6 x-4 y-11=0 intersect at the point (1,8)(1,8). Equation of the chord ABA B is
View answer and explanation
Correct option: C
x−3y+15=0x-3 y+15=0
Option C (x−3y+15=0x-3 y+15=0) is correct because the equation of the chord of contact is found by using the formula T=0, where T is obtained by substituting the point (1,8) into the circle equation.
394 Easy
394 of 685
Application +4 | -1
Find the length of the tangent drawn from the point (5,1)(5,1) to the circle x2+y2+6x−4y−3=0x^2+y^2+6 x-4 y-3=0.
View answer and explanation
Correct option: C
7
Option C (7) is correct because the length of the tangent from a point to a circle is calculated by substituting the point's coordinates into the circle's equation and taking the square root.
395 Easy
395 of 685
Application +4 | -1
Combined equation of tangents drawn to x2+y2=1x^2+y^2=1 from (−1,−1)(-1,-1), is
View answer and explanation
Correct option: D
x+y+xy+1=0x+y+x y+1=0
Option D (x+y+xy+1=0x+y+x y+1=0) is correct because the combined equation of the tangents is found by multiplying the equations of the individual tangents.
396 Easy
396 of 685
Application +4 | -1
The locus of the centre of the circle for which one end of a diameter is (1,1)\left( {1,1} \right) while the other end is on the line x+y=3x + y = 3, is
View answer and explanation
Correct option: C
2x+2y=52x + 2y = 5
Option C (2x+2y=52x + 2y = 5) is correct because it represents the locus of the center of the circle derived from the given conditions.
397 Medium
397 of 685
Application +4 | -1
A circle having centre (0,k)(0, k) with k>6k>6 is tangent to the line pair x2−y2=0x^2-y^2=0 and y=6y=6. the radius of the circle is
View answer and explanation
Correct option: D
62+66 \sqrt{2}+6
Option D (62+66 \sqrt{2}+6) is correct because it correctly calculates the radius of the circle tangent to the line pair and the line y=6.
398 Medium
398 of 685
Application +4 | -1
If the incentre of an equilateral triangle is (1,1)(1,1) and the equation of its one side is 3x+4y+3=03 x+4 y+3=0, then the equation of the circumcircle of this triangle is :
View answer and explanation
Correct option: A
x2+y2−2x−2y−14=0x^{2}+y^{2}-2 x-2 y-14=0
Option A (x2+y2−2x−2y−14=0x^{2}+y^{2}-2 x-2 y-14=0) is correct because it represents the circumcircle equation derived from the incenter and side equation of the equilateral triangle.
399 Easy
399 of 685
Application +4 | -1
If the lines 2x−3y−5=02 x-3 y-5=0 and 3x−4y=73 x-4 y=7 are diameters of a circle of area 154 square units, then the equation of the circle is
View answer and explanation
Correct option: C
x2+y2−2x+2y−47=0x^2+y^2-2 x+2 y-47=0
Option C (x2+y2−2x+2y−47=0x^2+y^2-2 x+2 y-47=0) is correct because the center and radius are correctly calculated from the given information.
400 Easy
400 of 685
Application +4 | -1
The equation of the tangents to the circle x2+y2=4{x^2} + {y^2} = 4, which are parallel to x+2y+3=0x + 2y + 3 = 0, are
View answer and explanation
Correct option: C
x+2y=±25x + 2y = \pm 2\sqrt 5
Option C (x+2y=±25x + 2y = \pm 2\sqrt 5) is correct because these are the equations of the tangents parallel to the given line, found by equating the perpendicular distance from the center to the radius.
401 Easy
401 of 685
Application +4 | -1
The length of the longest ray drawn from the point (4,3)(4,3) to the circle x2+y2+16x+18y+1=0x^2+y^2+16 x+18 y+1=0 is equal to
View answer and explanation
Correct option: A
the radius of the circle
Option A (the radius of the circle) is correct because the longest ray from a point to a circle is along the line connecting the point to the center of the circle, extending to the circle's circumference. The length of the tangent is equal to the radius in this case.
402 Medium
402 of 685
Application +4 | -1
The coordinates of the centre of the smallest circle passing through the origin and having y=x+1y = x + 1 as a diameter are
View answer and explanation
Correct option: D
(−12,12)\left( { - \frac{1}{2},\frac{1}{2}} \right)
Option D ((−12,12)\left( { - \frac{1}{2},\frac{1}{2}} \right)) is correct because the center of the smallest circle lies on the perpendicular bisector of the line segment joining the origin and the intersection of the line y = x + 1 with the perpendicular bisector.
403 Easy
403 of 685
Application +4 | -1
If the square of the length of the tangent drawn from any point on the circle x2+y2+70x−45y+6789=0x^2+y^2+70 x-45 y+6789=0 to the circle x2+y2+70x−45y+9876=0x^2+y^2+70 x-45 y+9876=0 is KK then find K−3080.K-3080.
View answer and explanation
Correct option: A
7
Option A (7) is correct because the square of the tangent length is found to be 3087, and 3087 - 3080 = 7.
404 Easy
404 of 685
Application +4 | -1
The radius of the circle r2−22r(cos⁡θ+sin⁡θ)−5=0,{r^2} - 2\sqrt 2 r\left( {\cos \theta + \sin \theta } \right) - 5 = 0, is
View answer and explanation
Correct option: C
3
Option C (3) is correct because converting the polar equation to Cartesian form allows for easy identification of the circle's radius using the standard equation.
405 Easy
405 of 685
Application +4 | -1
Find the equation to the circle touching the yy-axis at a distance -3 from the origin and intercepting a length 8 on the xx-axis.
View answer and explanation
Correct option: C
x2+y2±10x+6y+9=0x^2+y^2 \pm 10 x+6 y+9=0
Option C (x2+y2±10x+6y+9=0x^2+y^2 \pm 10 x+6 y+9=0) is correct because it satisfies the conditions of touching the y-axis at (0, -3) and intercepting a length of 8 on the x-axis.
406 Medium
406 of 685
Analyzing +4 | -1
If (2,4)(2,4) is a point interior to the circle x2+y2−6x−10y+λ=0x^2+y^2-6 x-10 y+\lambda=0 and the circle does not cut the axes at any point, then
View answer and explanation
Correct option: A
λ∈(25,32)\lambda \in(25,32)
Option A (λ∈(25,32)\lambda \in(25,32)) is correct because it satisfies both conditions: the point (2,4) lies inside the circle, and the circle does not cut the axes.
407 Medium
407 of 685
Application +4 | -1
In the two circles (x+7)2+(y−3)2=36{\left( {x + 7} \right)^2} + {\left( {y - 3} \right)^2} = 36 and (x−5)2+(y+2)2=49{\left( {x - 5} \right)^2} + {\left( {y + 2} \right)^2} = 49 touch each other externally, then the point of contact is
View answer and explanation
Correct option: B
(−1913,913)\left( {\frac{{ - 19}}{{13}},\frac{9}{{13}}} \right)
Option B (−1913,913)\left( {\frac{{ - 19}}{{13}},\frac{9}{{13}}} \right) is correct because it represents the point of contact when two circles touch each other externally, calculated using the section formula.
408 Easy
408 of 685
Analyzing +4 | -1
α,β\alpha, \beta and γ\gamma are parametric angles of three points P,QP, Q and RR respectively, on the circle x2+y2=1x^2+y^2=1 and AA is the point (−1,0)(-1,0). If the lengths of the chords AP,AQA P, A Q and ARA R are in GPG P, then cos⁡(α2),cos⁡(β2)\cos \left(\frac{\alpha}{2}\right), \cos \left(\frac{\beta}{2}\right) and cos⁡(γ2)\cos \left(\frac{\gamma}{2}\right) are in
View answer and explanation
Correct option: B
GPG P
Option B (GP) is correct because if the lengths of the chords AP, AQ, and AR are in GP, then the cosines of half the angles are also in GP.
409 Hard
409 of 685
Application +4 | -1
If P(−1,−3)P\left( { - 1, - 3} \right) is a centre of similitude for the circles x2+y2=1{x^2} + {y^2} = 1 and x2+y2−2x−6y+6=0,{x^2} + {y^2} - 2x - 6y + 6 = 0, then the length of the common tangent through PP to the circles is
View answer and explanation
Correct option: B
3
The correct answer is 3 because it represents the length of the common tangent through P to the circles, calculated using the lengths of tangents from P to each circle.
410 Easy
410 of 685
Application +4 | -1
If the points (2,0),(0,1),(4,5)\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right) and (0,c)\left( {0,c} \right) are concyclic, then the value of cc is
View answer and explanation
Correct option: B
143\frac{{14}}{3}
Option B (14/3) is correct because it's the value of c that satisfies the concyclic condition after finding the circle's equation.
411 Medium
411 of 685
Application +4 | -1
The equation of the tangent to circle 5x2+5y2=15{x^2} + 5{y^2} = 1, parallel to line 3x+4y=13x + 4y = 1 are
View answer and explanation
Correct option: C
3x+4y=±53x + 4y = \pm \sqrt 5
Option C (3x+4y=±53x + 4y = \pm \sqrt 5) is correct because these are the equations of the tangents to the circle that are parallel to the given line.
412 Easy
412 of 685
Analyzing +4 | -1
Let PQP Q and RSR S be tangents at the extremities of diameter PRP R of a circle of radius rr. If PSP S and RQR Q intersect at a point XX on the circumference of the circle, then 2r2 r equals
View answer and explanation
Correct option: A
PQ⋅RS\sqrt{P Q \cdot R S}
Option A (PQ⋅RS\sqrt{P Q \cdot R S}) is correct because using similar triangles formed by the intersecting tangents and the diameter, we can show that 2r=PQ⋅RS2r = \sqrt{PQ \cdot RS}.
413 Easy
413 of 685
Application +4 | -1
The radius of any circle touching the lines 3x−4y+5=03x - 4y + 5 = 0 and 6x−8y−9=06x - 8y - 9 = 0 is
View answer and explanation
Correct option: B
0.950.95
Option B (0.95) is correct because the radius of the circle touching two parallel lines is half the distance between the lines, which is calculated to be 0.95.
414 Medium
414 of 685
Application +4 | -1
If the centroid of an equilateral triangle is (2,−2)(2,-2) and its one vertex is (−1,1)(-1,1), then the equation of its circumcircle is
View answer and explanation
Correct option: A
x2+y2−4x+4y−10=0x^2+y^2-4 x+4 y-10=0
Option A (x2+y2−4x+4y−10=0x^2+y^2-4 x+4 y-10=0) is correct because the circumcircle passes through the vertex and has its center at the centroid.
415 Easy
415 of 685
Application +4 | -1
Let SS be the circle in the xyx y-plane defined by the equation x2+y2=4x^2+y^2=4. Let PP be a point on the circle with both coordinates being positive. Let the tangent to SS at PP intersect the coordinate axes at MM and NN. Then the mid-point of the line segment MN must lie on the curve
View answer and explanation
Correct option: D
x2+y2=x2y2x^2+y^2=x^2 y^2
Option D (x+22y=6x+2 \sqrt{2} y=6) is correct because it is a common tangent to both circles.
416 Easy
416 of 685
Application +4 | -1
Radius of bigger circle touching the circle x2+y2−4x−4y+4=0x^2+y^2-4 x-4 y+4=0 and both the coordinate axes, is
View answer and explanation
Correct option: B
2(3+22)2(3+2 \sqrt{2}) units
Option B (2(3+22)2(3+2 \sqrt{2}) units) is correct because it represents the radius of the larger circle touching the given circle and both coordinate axes, derived from geometric considerations and solving for the unknown radius.
417 Medium
417 of 685
Analyzing +4 | -1
The number of points with integral coordinates (2a,a−1)\left( {2a,a - 1} \right) that fall in the interior of the larger segment of the circle x2+y2=25{x^2} + {y^2} = 25 cut off by the parabola x2+4y=0,{x^2} + 4y = 0, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because there are two integral values of 'a' that satisfy the given conditions for the point to lie inside the larger segment.
418 Easy
418 of 685
Application +4 | -1
The angle between the tangents drawn at the points (5,12)\left( {5,12} \right) and (12,−5)\left( {12, - 5} \right) to the circle x2+y2=169{x^2} + {y^2} = 169 is
View answer and explanation
Correct option: D
900{90^0}
Option D (900{90^0}) is correct because the tangents at the given points are perpendicular to each other, which can be verified by checking the product of their slopes.
419 Easy
419 of 685
Application +4 | -1
The values of λ\lambda so that the line 3x−4y=λ3x - 4y = \lambda touches x2+y2−4x−8y−5=0{x^2} + {y^2} - 4x - 8y - 5 = 0 are
View answer and explanation
Correct option: A
−35,  15- 35,\;15
Option A (−35,15-35, 15) is correct because these are the values of λ\lambda for which the line is tangent to the circle, found by equating the perpendicular distance from the center to the radius.
420 Easy
420 of 685
Analyzing +4 | -1
Let the tangent to the circle C1:x2+y2=2C_1: x^2+y^2=2 at the point M(−1,1)M(-1,1) intersect the circle C2:(x−3)2+(y−C_2:(x-3)^2+(y- 2)2=52)^2=5, at two distinct points AA and BB. If the tangents to C2C_2 at the points AA and BB intersect at NN, then the area of the triangle ANBA N B is equal to
View answer and explanation
Correct option: C
16\frac{1}{6}
Option C (16\frac{1}{6}) is correct because the area of the triangle ANB is found by determining the intersection points A and B, finding the intersection point N of the tangents, and then calculating the area using determinant formula.
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