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JEE • MATHEMATICS • CIRCLES

Circles practice · Easy

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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 Easy
391 of 480
Application +4 | -1
The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6{x^2} + 3{y^2} = 6 on any tangent to it is
View answer and explanation
Correct option: C
(x2+y2)2=6x2+2y2{({x^2} + {y^2})^2} = 6{x^2} + 2{y^2}
Option C ((x2+y2)2=6x2+2y2{\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}) is correct because deriving the equation of the tangent and using the perpendicularity condition leads to this locus.
392 Easy
392 of 480
Application +4 | -1
The points (5,−7)\left( {5, - 7} \right) lies outside the circle
View answer and explanation
Correct option: A
x2+y2−8x=0{x^2} + {y^2} - 8x = 0
Option A (x2+y2−8x=0{x^2} + {y^2} - 8x = 0) is correct because substituting the point (5, -7) into the equation results in a positive value, indicating the point lies outside the circle.
393 Easy
393 of 480
Application +4 | -1
The locus of a point which moves so that the ratio of the length of the tangents to the circles x2+y2+4x+3=0{x^2} + {y^2} + 4x + 3 = 0 and x2+y2−6x+5=0{x^2} + {y^2} - 6x + 5 = 0 is 2:32:3, is
View answer and explanation
Correct option: D
5x2+5y2+60x+7=05{x^2} + 5{y^2} + 60x + 7 = 0
Option D (5x2+5y2+60x+7=05{x^2} + 5{y^2} + 60x + 7 = 0) is correct because it represents the locus of a point where the ratio of tangent lengths to the two circles is 2:3.
394 Easy
394 of 480
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.
395 Easy
395 of 480
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.
396 Easy
396 of 480
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.
397 Easy
397 of 480
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).
398 Easy
398 of 480
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.
399 Easy
399 of 480
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.
400 Easy
400 of 480
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.
401 Easy
401 of 480
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.
402 Easy
402 of 480
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.
403 Easy
403 of 480
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.
404 Easy
404 of 480
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.
405 Easy
405 of 480
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.
406 Easy
406 of 480
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.
407 Easy
407 of 480
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.
408 Easy
408 of 480
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.
409 Easy
409 of 480
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.
410 Easy
410 of 480
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.
411 Easy
411 of 480
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.
412 Easy
412 of 480
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.
413 Easy
413 of 480
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.
414 Easy
414 of 480
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.
415 Easy
415 of 480
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.
416 Easy
416 of 480
Analyzing +4 | -1
Suppose f(x,y)=0f(x, y)=0 is the equation of a circle such that f(x,1)=0f(x, 1)=0 has equal roots (each equal to 2) and f(1,x)=0f(1, x)=0 also has equal roots (each equal to zero). The equation of circle is
View answer and explanation
Correct option: D
x2+y2−4x+3=0x^2+y^2-4 x+3=0
Option D (x2+y2−4x+3=0x^2+y^2-4 x+3=0) is correct because it satisfies the given conditions of equal roots for both f(x,1)=0f(x, 1)=0 and f(1,x)=0f(1, x)=0.
417 Easy
417 of 480
Application +4 | -1
An equation of the circle described on the chord xcos⁡α+ysin⁡α−p=0x \cos \alpha+y \sin \alpha-p=0 of the circle x2+y2=a2x^2+y^2=a^2 as diameter is
View answer and explanation
Correct option: B
x2+y2−2pxcos⁡α−2pysin⁡α+2p2−a2=0x^2+y^2-2 p x \cos \alpha-2 p y \sin \alpha+2 p^2-a^2=0
Option B (x2+y2−2pxcos⁡α−2pysin⁡α+2p2−a2=0x^2+y^2-2 p x \cos \alpha-2 p y \sin \alpha+2 p^2-a^2=0) is correct because the equation of the circle is found by considering the given chord as the diameter and using the condition that the center lies on the chord.
418 Easy
418 of 480
Application +4 | -1
The smallest circle with centre on y-axis and passing through the point(7,3)has radius
View answer and explanation
Correct option: B
7
Option B (7) is correct because the smallest circle centered on the y-axis and passing through (7,3) will have its center at (0,3), making the radius 7.
419 Easy
419 of 480
Application +4 | -1
Consider four circles (x±1)2+(y±1)2=1(x \pm 1)^2+(y \pm 1)^2=1. Equation of smaller circle touching these four circles is
View answer and explanation
Correct option: D
x2+y2=3−22x^2+y^2=3-2 \sqrt{2}
Option D (x2+y2=3−22x^2+y^2=3-2 \sqrt{2}) is correct because it represents the equation of a circle centered at the origin with a radius equal to the distance from the origin to the point where the smaller circle touches the four given circles.
420 Easy
420 of 480
Analyzing +4 | -1
The circle x2+y2=4x+8y+5x^2+y^2=4 x+8 y+5 intersects the line 3x−4y=m3 x-4 y=m at two distinct points, if
View answer and explanation
Correct option: D
−35<m<15-35 < m < 15
Option D (-35 < m < 15) is correct because the condition for the line to intersect the circle at two distinct points is satisfied within this range of m values.
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