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JEE • MATHEMATICS • CIRCLES

Circles practice · Medium

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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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Showing 1-30 of 189 questions
1 Medium
1 of 189
Application +4 | -1
The locus of the middle point of the chords of the circle x2+y2=a2{x^2} + {y^2} = {a^2} such that the chords pass through a given point (x1,y1)\left( {{x_1},{y_1}} \right), is
View answer and explanation
Correct option: A
x2+y2−xx1−yy1=0{x^2} + {y^2} - x{x_1} - y{y_1} = 0
Option A (x2+y2−xx1−yy1=0{x^2} + {y^2} - x{x_1} - y{y_1} = 0) is correct because it represents the locus of the midpoint of the chords passing through a given point.
2 Medium
2 of 189
Application +4 | -1
If a chord of contact of tangents drawn from a point PP with respect to the circle x2+y2=9x^2+y^2=9 is x=2x=2, then area, in square units, of triangle formed by tangents drawn from PP to the circle and their chord of contact is equal to
View answer and explanation
Correct option: C
552\frac{5 \sqrt{5}}{2}
Option C (552\frac{{5 \sqrt 5 }}{{2}}) is correct because it is the area of the triangle formed by the tangents and chord of contact, calculated using the formula with the given circle and point P.
3 Medium
3 of 189
Application +4 | -1
If the angle between tangents drawn to x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 from (0,0)(0,0) is π2\frac{\pi}{2}, then
View answer and explanation
Correct option: B
g2+f2=2cg^2+f^2=2 c
Option B (g2+f2=2cg^2+f^2=2c) is correct because it represents the condition for the angle between tangents from (0,0) to be π2\frac{{\pi}}{{2}}.
4 Medium
4 of 189
Analyzing +4 | -1
If each of the lines 5x+8y=135 x+8 y=13 and 4x−y=34 x-y=3 contains a diameter of the circle x2+y2−2(a2−7a+11)x−2(a2−6a+6)y+b3+1=0x^{2}+y^{2}-2\left(a^{2}-7 a+11\right) x-2\left(a^{2}-6 a+6\right) y+b^{3}+1=0, then
View answer and explanation
Correct option: D
a=5a=5 and b∈(−∞,1)b \in(-\infty, 1)
Option D (a=5a=5 and b∈(−∞,1)b \in(-\infty, 1)) is correct because the center of the circle must be the intersection of the diameters, and the radius must be real.
5 Medium
5 of 189
Application +4 | -1
If a circles x2+y2+2g1x+2f1y=0x^2+y^2+2 g_1 x+2 f_1 y=0 and x2+y2+2g2x+2f2y=0x^2+y^2+2 g_2 x+2 f_2 y=0 touch each other, then
View answer and explanation
Correct option: B
f1g1=f2g2\frac{f_1}{g_1}=\frac{f_2}{g_2}
Option B (f1g1=f2g2\frac{f_1}{g_1}=\frac{f_2}{g_2}) is correct because when two circles touch each other, the line joining their centers passes through the point of contact, implying the ratio of their y-coordinates to x-coordinates are equal.
6 Medium
6 of 189
Analyzing +4 | -1
Let ϕ(x,y)=0\phi(x, y)=0 be the equation of a circle. If ϕ(0,λ)=0\phi(0, \lambda)=0 has equal roots λ=2,2\lambda=2,2 and ϕ(λ,0)=0\phi(\lambda, 0)=0 has roots λ=45,5\lambda=\frac{4}{5}, 5, then the centre of the circle is
View answer and explanation
Correct option: B
(2910,2)\left(\frac{29}{10}, 2\right)
Option B ((2910\frac{{29}}{{10}}, 2)) is correct because it represents the center of the circle that satisfies the given conditions for ϕ(0,λ)=0\phi(0, \lambda)=0 and ϕ(λ,0)=0\phi(\lambda, 0)=0.
7 Medium
7 of 189
Analyzing +4 | -1
Let L1L_1 be a straight line passing through the origin and L2L_2 be the straight line x+y=1x+y=1. If the intercepts made by the circle x2+y2−x+3y=0x^2+y^2-x+3 y=0 on L1L_1 and L2L_2 are equal, then which of the following equations can represent L1L_1 ?
View answer and explanation
Correct option: B
x−y=0,x+7y=0x-y=0, x+7 y=0
Option B (x−y=0,x+7y=0x-y=0, x+7 y=0) is correct because these lines satisfy the condition that the intercepts made by the circle on them are equal to the intercept made on the line x+y=1x+y=1.
8 Medium
8 of 189
Application +4 | -1
The equation of the tangent from the point (0,1)\left( {0,1} \right) to the circle x2+y2−2x−6y+6=0{x^2} + {y^2} - 2x - 6y + 6 = 0, is
View answer and explanation
Correct option: A
y−1=0y - 1 = 0
Option A (y−1=0y - 1 = 0) is correct because it represents a tangent line from the point (0, 1) to the given circle.
9 Medium
9 of 189
Application +4 | -1
The circles x2+y2=4x^2+y^2=4 and x2+y2−4λx+9=0x^2+y^2-4 \lambda x+9=0 have two common tangents if
View answer and explanation
Correct option: B
λ<−138\lambda < \frac{-13}{8} or λ>138\lambda > \frac{13}{8}
Option B (λ<−138\lambda < \frac{{-13}}{{8}} or λ>138\lambda > \frac{{13}}{{8}}) is correct because it represents the condition where the two circles intersect, allowing for two common tangents.
10 Medium
10 of 189
Analyzing +4 | -1
Let S≡0S \equiv 0 and S′≡0S^{\prime} \equiv 0 be two circles belonging to the family of circles x2+y2+λx−4y+3=0:(λx^2+y^2+\lambda x-4 y+3=0:(\lambda ∈R)\in \mathrm{R}) such that S≡0S \equiv 0 bisects the circumference of circle S′≡0S^{\prime} \equiv 0 and has radius equal to twice of radius of S′≡0{S}^{\prime} \equiv 0. Then distance between their centres is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the distance between the centers of the two circles is calculated to be 1 unit based on the given conditions.
11 Medium
11 of 189
Application +4 | -1
The circle x2+y2−2x−3ky−2=0x^2+y^2-2 x-3 k y-2=0 passes through some fixed point. One of them may be:
View answer and explanation
Correct option: B
(1+3,0)(1+\sqrt{3}, 0)
Option B ((1+3,0)(1+\sqrt{3}, 0)) is correct because substituting y=0 into the circle's equation allows us to find the fixed points that the circle passes through regardless of the value of k.
12 Medium
12 of 189
Application +4 | -1
The equation of sphere concentric with the sphere x2+y2+z2−4x−6y−8z−5=0x^2+y^2+z^2-4 x-6 y-8 z-5=0 and which passes through the point (0,1,0)(0,1,0) is
View answer and explanation
Correct option: C
x2+y2+z2−4x−6y−8z+5=0x^2+y^2+z^2-4 x-6 y-8 z+5=0
The correct answer is C (x2+y2+z2−4x−6y−8z+5=0x^2+y^2+z^2-4 x-6 y-8 z+5=0) because it represents a sphere with the same center as the given sphere and passing through the point (0,1,0).
13 Medium
13 of 189
Application +4 | -1
One hundred concentric circles with radii 1,2,3−1001,2,3-100 are drawn in a plane. The interior of circle with radius 1 is colored red & each region bounded by consecutive circles is coloured either red or green with no two adjacent regions the same color, then sum of total area of green regions will be
View answer and explanation
Correct option: C
5050π5050 \pi
Option D (175) is correct because it represents the number of different throws where the largest number appearing is 4.
14 Medium
14 of 189
Application +4 | -1
Line (x−3)cos⁡θ+(y−3)sin⁡θ=1(x-3) \cos \theta+(y-3) \sin \theta=1 touches a circle (x−3)2+(y−3)2=1(x-3)^2+(y-3)^2=1, then find the number of values of θ\theta.
View answer and explanation
Correct option: A
1
Option D (Infinite) is incorrect. The question states that the line *touches* the circle. The line is already given as tangent to the circle, so any value of θ\theta will satisfy the equation.
15 Medium
15 of 189
Application +4 | -1
The locus of the point (3h+2,3k)(\sqrt{3 h+2}, \sqrt{3 k}). If (h,k)(h, k) lies on x+y=1x+y=1 is
View answer and explanation
Correct option: B
A circle
Option A circle is correct because substituting the expressions for h and k into x+y=1 results in the equation of a circle.

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16 Medium
16 of 189
Application +4 | -1
The equation of tangent drawn from the origin to the circle x2+y2−2rx+2hy+h2=0{x^2} + {y^2} - 2rx + 2hy + {h^2} = 0 are
View answer and explanation
Correct option: D
(h2−r2)x−2rhy=0,x=0\left( {{h^2} - {r^2}} \right)x - 2rhy = 0,x = 0
Option D ((h2−r2)x−2rhy=0,x=0\left( {{h^2} - {r^2}} \right)x - 2rhy = 0,x = 0) is correct because these are the equations of the tangents drawn from the origin to the given circle.
17 Medium
17 of 189
Analyzing +4 | -1
For the two circles x2+y2=16{x^2} + {y^2} = 16 and x2+y2−2y=0{x^2} + {y^2} - 2y = 0 there is/are
View answer and explanation
Correct option: D
No common tangent
The correct answer is D. The second circle lies completely inside the first circle, so there are no common tangents.
18 Medium
18 of 189
Application +4 | -1
A circle is described on the line joining the points (2,−3)(2,-3) and (−4,7)(-4,7) as a diameter. The circle
View answer and explanation
Correct option: D
origin lies inside the circle
The correct answer is that the origin lies inside the circle because the distance from the center of the circle to the origin is less than the radius of the circle.
19 Medium
19 of 189
Application +4 | -1
A tangent PTP T is drawn to the circle x2+y2=4x^2+y^2=4 at the point P(3,1)P(\sqrt{3}, 1). A straight line LL, perpendicular to PTP T is a tangent to the circle (x−3)2+y2=1(x-3)^2+y^2=1. A possible equation of LL is
View answer and explanation
Correct option: A
x−3y=1x-\sqrt{3} y=1
Option A (x+2y=0x+2 y=0) is correct because it is the locus of the centers of the circles passing through the intersection of the given circles.
20 Medium
20 of 189
Application +4 | -1
The radius of inscribed circle in the quadrilateral formed by tangents drawn from (3,−1)(3,-1) to circle x2+y2−2x−2y−2=0x^2+y^2-2 x-2 y-2=0 and radius formed by point of contact of these tangents is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the point (3,-1) lies on the director circle, making the radius of the inscribed circle half the radius of the given circle.
21 Medium
21 of 189
Application +4 | -1
The tangents to the circle x2+y2=169x^2+y^2=169 at the points (5,12)(5,12) and (−5,−12)(-5,-12) are
View answer and explanation
Correct option: A
parallel
Option A (parallel) is correct. The slopes of the tangents at the given points are equal, indicating that the tangents are parallel.
22 Medium
22 of 189
Analyzing +4 | -1
The centre of the circle passing through (0,2)(0,2) and touching the parabola y=2x2y=2 x^2 at (2,2)(2,2) is
View answer and explanation
Correct option: B
(−2,52)\left(-2, \frac{5}{2}\right)
Option B ((-2, 5/2)) is correct because it represents the center of the circle that satisfies the given tangency and point conditions.
23 Medium
23 of 189
Application +4 | -1
In what ratio the line y−x+2=0y-x+2=0 divides the line joining the points (3,−1)(3,-1) and (8,9)(8,9) ?
View answer and explanation
Correct option: C
2:32: 3
Option C (2:3) is correct because the ratio is found by substituting the intersection point of the line and the line segment into the line's equation.
24 Medium
24 of 189
Application +4 | -1
A circle of radius 4 units touches the coordinate axes in the first quadrant. The equation of its image with respect to the line mirror y=0y=0 is
View answer and explanation
Correct option: A
x2+y2−8x+8y+16=0x^2+y^2-8 x+8 y+16=0
Option A (x2+y2−8x+8y+16=0x^2+y^2-8 x+8 y+16=0) is correct because it represents the equation of the circle obtained by reflecting the original circle about the line y=0.
25 Medium
25 of 189
Application +4 | -1
The locus of the middle point of chords of the circle x2+y2=a2{x^2} + {y^2} = {a^2} which pass through the fixed point (h,k)\left( {h,k} \right) is
View answer and explanation
Correct option: A
x2+y2−hx−ky=0{x^2} + {y^2} - hx - ky = 0
Option A (x2+y2−hx−ky=0{x^2} + {y^2} - hx - ky = 0) is correct because it is the locus of the middle point of the chords, derived by using the equation of the chord and the given fixed point.
26 Medium
26 of 189
Application +4 | -1
Equation of chord of minimum length passing through (1,2)(1,2) of circle x2+y2−4x−2y−4=0x^2+y^2-4 x-2 y-4=0 is ax+by+c=0a x+b y+c=0 then a+b+c=a+b+c=
View answer and explanation
Correct option: C
±8\pm 8
Option C (±8\pm 8) is correct because the equation of the chord of minimum length passing through (1,2) is found using T=S1, leading to x-y+8=0, and thus a+b+c = ±\pm8.
27 Medium
27 of 189
Analyzing +4 | -1
(a,b)(a, b) is the mid-point of the chord AB‾\overline{A B} of the circle x2+y2=r2x^2+y^2=r^2. The tangent at A, B meet at C{C}. then area of ΔABC=\Delta A B C=
View answer and explanation
Correct option: B
(r2−a2−b2)32a2+b2\frac{\left(r^2-a^2-b^2\right)^{\frac{3}{2}}}{\sqrt{a^2+b^2}}
Option B is correct because the area of the triangle ABC is derived using the midpoint of the chord and the radius of the circle.
28 Medium
28 of 189
Application +4 | -1
If the term independent of xx in the expansion of (x3+32x2)10\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10} is mn\frac{m}{n} (m, nm, \, {n} are relatively prime) then the length of the tangent from (0,0)(0,0) to a circle with centre (m,0)(m, 0) and radius 'nn' units is
View answer and explanation
Correct option: D
3
The correct answer is 3 because it represents the length of the tangent from the origin to the circle, calculated after finding the term independent of x and determining the circle's equation.
29 Medium
29 of 189
Application +4 | -1
If one of the diameters of the circle x2+y2−2x−6y+6=0x^2+y^2-2 x-6 y+6=0 is a chord to the circle with centre (2,1)(2,1), then the radius of circle, is
View answer and explanation
Correct option: C
3
The correct answer is 3 because the radius of the larger circle is found using the distance formula and the radius of the smaller circle.
30 Medium
30 of 189
Application +4 | -1
If one end of the diameter is (1,1)\left( {1,1} \right) and the other end lies on the line x+y=3x + y = 3, then locus of centre of circle is
View answer and explanation
Correct option: C
2x+2y=52x + 2y = 5
Option C (2x+2y=52x + 2y = 5) is correct because the locus of the center of the circle is found by expressing the center's coordinates in terms of a parameter and eliminating the parameter.
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