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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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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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571 Medium
571 of 685
Application +4 | -1
The locus of the centre of a circle of radius 2 which rolls on the outside of the circle, is x2+y2+3x−6y−9=0{x^2} + {y^2} + 3x - 6y - 9 = 0 is
View answer and explanation
Correct option: B
x2+y2+3x−6y−31=0{x^2} + {y^2} + 3x - 6y - 31 = 0
Option B (x2+y2+3x−6y−31=0{x^2} + {y^2} + 3x - 6y - 31 = 0) is correct because it represents the locus of the center of the rolling circle, considering the sum of the radii.
572 Easy
572 of 685
Application +4 | -1
The range of values of θ∈[0,2π]\theta \in \left[ {0,2\pi } \right] for which (1+cos⁡θ,sin⁡θ)\left( {1 + \cos \theta ,\sin \theta } \right) is on interior point of the circle x2+y2=1,{x^2} + {y^2} = 1, is
View answer and explanation
Correct option: D
(2π/3,4π/3)\left( {2\pi /3,4\pi /3} \right)
Option D ((2π/3,4π/32\pi /3,4\pi /3)) is correct because it represents the range of θ for which the point (1 + cos θ, sin θ) lies inside the circle.
573 Easy
573 of 685
Application +4 | -1
The equation of circle which touches the line x+3y−2=0x+3 y-2=0 and 3x+9y−2=03 x+9 y-2=0, also the point of contact on first line is (−1,1)(-1,1), is
View answer and explanation
Correct option: C
x2+y2+32x15−8y5+2615=0x^2+y^2+\frac{32 x}{15}-\frac{8 y}{5}+\frac{26}{15}=0
Option C (x2+y2+32x15−8y5+2615=0x^2+y^2+\frac{{32 x}}{{15}}-\frac{{8 y}}{{5}}+\frac{{26}}{{15}}=0) is correct because the circle touches both lines and the point of contact on the first line is given, allowing us to find the other diametric endpoint and thus the circle's equation.
574 Easy
574 of 685
Application +4 | -1
Length of the common chord of the circles (x−1)2+(y+1)2=c2(x-1)^2+(y+1)^2=c^2 and (x+1)2+(y−1)2=c2(x+1)^2+(y-1)^2=c^2 is
View answer and explanation
Correct option: C
2c2−22 \sqrt{c^2-2}
Option C (2c2−22 \sqrt{{c^2-2}}) is correct because it is the length of the common chord calculated using the distance from the center to the chord and the radius.
575 Easy
575 of 685
Application +4 | -1
The tangent to x2+y2=9{x^2} + {y^2} = 9 which is parallel to y−axisy - axis and does not lie in the third quadrant touches the circle at the point
View answer and explanation
Correct option: A
(3,0)\left( {3,0} \right)
Option A ((3,0)) is correct because the tangent to the circle parallel to the y-axis and not in the third quadrant is x=3, which touches the circle at (3,0).
576 Easy
576 of 685
Application +4 | -1
The locus of a point which moves such that the sum of the squares of its distances from three vertices of a triangle is a circle whose centre is at the
View answer and explanation
Correct option: A
Centroid of triangle ABCA B C
Option A (Centroid of triangle ABCA B C) is correct because the center of the circle derived in the previous question is the centroid of the triangle.
577 Easy
577 of 685
Application +4 | -1
If the points (0,0),(1,0),(0,1)(0,0),(1,0),(0,1) and (t,t)(t, t) are concyclic, then t=t=
View answer and explanation
Correct option: B
1
Option C (2x+2y=52 x+2 y=5) is correct because it represents the locus of the centers of the circles that satisfy the given conditions.
578 Medium
578 of 685
Application +4 | -1
Find the equation of the normal to the circle x2+y2−5x+2yx^2+y^2-5 x+2 y −48=0-48=0 at the point (5,6)(5,6).
View answer and explanation
Correct option: A
14x−5y−40=014 x-5 y-40=0
Option A (14x−5y−40=014 x-5 y-40=0) is correct because it represents the equation of the normal to the circle at the given point (5,6), which is perpendicular to the tangent at that point.
579 Easy
579 of 685
Application +4 | -1
The locus of the centre of a circle which touches externally the circle x2+y2−6x−6y+14=0x^2+y^2-6 x-6 y+14=0 and also touches YY-axis, is given by the equation
View answer and explanation
Correct option: D
y2−10x−6y+14=0y^2-10 x-6 y+14=0
Option D (y2−10x−6y+14=0y^2-10 x-6 y+14=0) is correct because it represents the locus of the center of the circle that touches the given circle externally and the Y-axis.
580 Easy
580 of 685
Analyzing +4 | -1
The equation of the circle passing through (1,0)(1,0) and (0,1)(0,1) and having the smallest possible radius, is
View answer and explanation
Correct option: B
x2+y2=x+yx^2+y^2=x+y
Option B (x2+y2=x+yx^2+y^2=x+y) is correct because it represents the circle passing through (1,0) and (0,1) with the smallest possible radius, derived by setting c=0.
581 Easy
581 of 685
Application +4 | -1
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
View answer and explanation
Correct option: C
x2+y2=4a2x^2+y^2=4 a^2
Option C (x2+y2=4a2x^2+y^2=4 a^2) is correct because the radius of the circumcircle is 2/3 of the median's length, which is 2a, and the center is at the origin.
582 Medium
582 of 685
Application +4 | -1
The distance between the chords of contact of the tangent to the circle x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 from the origin and the point (g,f),\left( {g,f} \right), is
View answer and explanation
Correct option: D
g2+f2−c2g2+f2\frac{{{g^2} + {f^2} - c}}{{2\sqrt {{g^2} + {f^2}} }}
The correct answer is g2+f2−c2g2+f2\frac{{{g^2} + {f^2} - c}}{{2\sqrt {{g^2} + {f^2}} }} because it represents the distance between the chords of contact, calculated using the formula for the distance between parallel lines.
583 Easy
583 of 685
Analyzing +4 | -1
The set of values of a for which the point (2a,a+1)(2 a, a+1) is an interior point of the larger segment of the circle x2+y2−2x−2y−8=0x^2+y^2-2 x-2 y-8=0 made by the chord x−y+1=0x-y+1=0, is
View answer and explanation
Correct option: B
(0,9/5)(0,9 / 5)
Option B ((0,9/5)) is correct because it satisfies both conditions: the point (2a, a+1) is an interior point of the circle, and it lies on the same side of the chord as the center of the circle.
584 Medium
584 of 685
Application +4 | -1
The pole of the line 3x+5y+17=03 x+5 y+17=0 with respect to the circle x2+y2+4x+6y+9=0x^2+y^2+4 x+6 y+9=0 is
View answer and explanation
Correct option: A
(1,2)(1,2)
Option A ((1,2)) is correct because it represents the pole of the given line with respect to the given circle.
585 Easy
585 of 685
Application +4 | -1
The normal at the point (3,4)\left( {3,4} \right) on a circle at the point (−1,−2).\left( { - 1, - 2} \right). The equation of the circle, is
View answer and explanation
Correct option: B
x2+y2−2x−2y−11=0{x^2} + {y^2} - 2x - 2y - 11 = 0
Option B (x2+y2−2x−2y−11=0{x^2} + {y^2} - 2x - 2y - 11 = 0) is correct because it represents the equation of the circle with the given conditions, where the normal at one point passes through another point on the circle.
586 Easy
586 of 685
Application +4 | -1
If the line y−mx+m−1=0y-m x+m-1=0 cuts the circle x2+y2−4x−4y+4x^2+y^2-4 x-4 y+4 =0=0 at two real points, then mm belongs to
View answer and explanation
Correct option: C
(−∞,∞)(-\infty, \infty)
Option C, (−∞,∞)(-\infty, \infty), is correct because the line always intersects the circle at two real points since the point (1,1) always lies inside the circle.
587 Easy
587 of 685
Application +4 | -1
If the circles (x−a)2+(y−b)2=c2{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2} and (x−b)2+(y−a)2=c2{\left( {x - b} \right)^2} + {\left( {y - a} \right)^2} = {c^2} touch each other, then
View answer and explanation
Correct option: B
a=b±2ca = b \pm \sqrt 2 c
Option B (a=b±2ca = b \pm \sqrt 2 c) is correct because the condition for the circles to touch leads to this relationship between a, b, and c.
588 Easy
588 of 685
Application +4 | -1
The circles x2+y2−8x+4y+4=0{x^2} + {y^2} - 8x + 4y + 4 = 0 touches
View answer and explanation
Correct option: B
yy-axis
The correct answer is B (yy-axis) because the radius of the circle is equal to the x-coordinate of the center, indicating it touches the yy-axis.
589 Easy
589 of 685
Application +4 | -1
The equation x2+y2−6x+8y+25=0x^2+y^2-6 x+8 y+25=0 represents
View answer and explanation
Correct option: A
A point (3,−4)(3,-4)
Option A (A point (3,−4)(3,-4)) is correct because the given equation represents a circle with radius zero, which is a point.
590 Easy
590 of 685
Analyzing +4 | -1
The equation of the locus of the middle point of a chord of the circle x2+y2−2x−2y=0x^2+y^2-2 x-2 y=0 such that the pair of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the xx-axis is
View answer and explanation
Correct option: A
x+y=2x+y=2
Option A (x+y=2x+y=2) is correct because it represents the locus of the midpoint of the chord such that the lines joining the origin to the intersection points are equally inclined to the x-axis.
591 Easy
591 of 685
Analyzing +4 | -1
The number of common tangents of the circles given by x2+y2−8x−2y+1=0x^{2}+y^{2}-8 x-2 y+1=0 and x2+y2+6x+8y=0x^{2}+y^{2}+6 x+8 y=0 is
View answer and explanation
Correct option: C
two
Option C (two) is correct because the distance between the centers is between the difference and the sum of the radii, indicating two common tangents.
592 Medium
592 of 685
Analyzing +4 | -1
If three distinct point A,B,CA, B, C are given in the 2-dimensional coordinate plane such that the ratio of the distance of each one of them from the point (1,0)(1,0) to the distance from (−1,0)(-1,0) is equal to 12\frac{1}{2}, then the circumcentre of the triangle ABCA B C is at the point:
View answer and explanation
Correct option: B
(5/3,0)(5 / 3,0)
Option B ((5/3,0)(5 / 3,0)) is correct because it represents the circumcenter of the triangle formed by points satisfying the given distance ratio condition.
593 Easy
593 of 685
Application +4 | -1
The condition for a line y=2x+cy = 2x + c to touch the circle x2+y2=16{x^2} + {y^2} = 16 is
View answer and explanation
Correct option: B
c2=80{c^2} = 80
Option B (c2=80{c^2} = 80) is correct because it satisfies the condition for the line to be tangent to the circle.
594 Easy
594 of 685
Application +4 | -1
The circle x2+y2−3x−4y+2=0{x^2} + {y^2} - 3x - 4y + 2 = 0 cuts xx-axis at
View answer and explanation
Correct option: D
(1,0),(2,0)\left( {1,0} \right),\left( {2,0} \right)
Option D ((1,0),(2,0)\left( {1,0} \right),\left( {2,0} \right)) is correct because these are the x-intercepts of the circle, found by setting y=0 and solving the resulting quadratic equation.
595 Medium
595 of 685
Application +4 | -1
The lines 5x+12y−10=05 x+12 y-10=0 and 5x−12y−40=05 x-12 y-40=0 touch a circle of radius 3 units. If the centre of circle lies in the first quadrant, then the coordinates of centre is
View answer and explanation
Correct option: C
(5,2)(5,2)
Option C ((5,2)) is correct because it satisfies the condition that the distance from the center to both lines is equal to the radius, and the center lies in the first quadrant.
596 Medium
596 of 685
Analyzing +4 | -1
The point of the straight line y=2x+11y = 2x + 11 which is nearest to the circle 16(x2+y2)+32x−8y−50=0,16\left( {{x^2} + {y^2}} \right) + 32x - 8y - 50 = 0, is
View answer and explanation
Correct option: B
(−9/2,2)\left( { - 9/2,2} \right)
Option B ((−9/2,2)\left( { - 9/2,2} \right)) is correct because it is the point on the line nearest to the circle.
597 Easy
597 of 685
Application +4 | -1
If 5x−12y+10=05x - 12y + 10 = 0 and 12y−5x+16=012y - 5x + 16 = 0 are two tangents to a circle, the radius of the circle is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the radius of the circle is half the distance between the two parallel tangent lines.
598 Easy
598 of 685
Application +4 | -1
If the area of the circle 4x2+4y2−8x+16y+k=04{x^2} + 4{y^2} - 8x + 16y + k = 0 is 9π9\pi sq unit, then the value of kk is
View answer and explanation
Correct option: C
−16- 16
Option C (-16) is correct because it is the value of k that makes the area of the given circle equal to 9π.
599 Medium
599 of 685
Analyzing +4 | -1
If the line xcos⁡θ+ysin⁡θ=2x \cos \theta+y \sin \theta=2 is the equation of a transverse common tangent to the circles x2+y2=4x^2+y^2=4 and x2+y2−63x−6y+20=0x^2+y^2-6 \sqrt{3} x-6 y+20=0, then the value of θ\theta is
View answer and explanation
Correct option: D
π6\frac{\pi}{6}
Option D (π6\frac{\pi}{6}) is correct because the equation of the common tangent can be found by using the condition C1C2=r1+r2C_1C_2 = r_1 + r_2 and then comparing with the given equation.
600 Easy
600 of 685
Application +4 | -1
If two circles x2+y2+2gx+2fy=0x^2+y^2+2 g x+2 f y=0 and x2+y2+2g1x+2f1y=0x^2+y^2+2 g_1 x+2 f_1 y=0 touch each other, then
View answer and explanation
Correct option: A
f1g=fg1f_1 g=f g_1
Option A (f1g=fg1f_1 g=f g_1) is correct because two circles touching each other at the origin must have a common tangent at the origin, which leads to the condition f1g=fg1f_1 g = f g_1.
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