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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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451 Easy
451 of 685
Analyzing +4 | -1
The number of common tangents to the circles one of which passes through the origin and cuts off intercepts 2 from each of the axes and the other circle has the segment joining the origin and the point (1,1)(1,1) as a diameter, 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 equal to the difference of their radii, indicating they touch internally, hence one common tangent.
452 Easy
452 of 685
Application +4 | -1
If a tangent to the circle x2+y2=1{x^2} + {y^2} = 1 intersects the coordinate axes at distinct points PP and QQ, then the locus of the mid-point of PQPQ is
View answer and explanation
Correct option: C
x2+y2−4x2y2=0{x^2} + {y^2} - 4{x^2}{y^2} = 0
Option C (x2+y2−4x2y2=0{x^2} + {y^2} - 4{x^2}{y^2} = 0) is correct because it represents the locus of the midpoint of the line segment PQ, where P and Q are the points where the tangent to the circle intersects the coordinate axes.
453 Medium
453 of 685
Analyzing +4 | -1
f(x,y)≡x2+y2+2ax+2by+c=0f(x, y) \equiv x^2+y^2+2 a x+2 b y+c=0 represent a circle. If f(x,0)=0f(x, 0)=0 has equal roots, each being 2 and f(0,y)=0f(0, y)=0 has 2 and 3 as its roots, then the centre of the circle is
View answer and explanation
Correct option: B
Data are not consistent
Option D (Data are inconsistent) is correct because the given conditions lead to conflicting values for the constant term 'c' in the circle's equation.
454 Medium
454 of 685
Analyzing +4 | -1
If in a △ABC\triangle A B C (whose circumcentre is origin), a≤sin⁡Aa \leq \sin A, then for any point (x,y)(x, y) inside the circumcircle of △ABC\triangle A B C
View answer and explanation
Correct option: A
∣xy∣<18|x y|<\frac{1}{8}
Option A (∣xy∣<18|xy| < \frac{{1}}{{8}}) is correct because it is derived from the given condition a≤sin⁡Aa \leq \sin A and applying AM-GM inequality inside the circumcircle.
455 Easy
455 of 685
Application +4 | -1
Two diameters of the circle 3x2+3y2−6x−18y−7=03{x^2} + 3{y^2} - 6x - 18y - 7 = 0 are along the lines 3x+y=c13x + y = {c_1} and x−3y=c2x - 3y = {c_2}. Then, the value of c1c2{c_1}{c_2} is
View answer and explanation
Correct option: A
−48- 48
Option A (−48- 48) is correct because it is the product of the constants c1c_1 and c2c_2 obtained by substituting the center of the circle into the diameter equations.
456 Easy
456 of 685
Analyzing +4 | -1
The equation of the circumcircle of an equilateral triangle is x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 and one vertex of the triangle is (1,1)(1,1). The equation of the incircle of the triangle, is
View answer and explanation
Correct option: C
4(x2+y2)+8gx+8fy=(1−g)(1+3g)+(1−f)(1+3f)4\left(x^2+y^2\right)+8 g x+8 f y=(1-g)(1+3 g)+(1-f)(1+3 f)
Option C (4(x2+y2)+8gx+8fy=(1−g)(1+3g)+(1−f)(1+3f)4(x^2+y^2)+8 g x+8 f y=(1-g)(1+3 g)+(1-f)(1+3 f)) is correct because it represents the equation of the incircle of the equilateral triangle, derived using the relationship between circumcenter, incenter, circumradius, and inradius.
457 Easy
457 of 685
Application +4 | -1
The line 2x -y + 1 = 0 is a tangent to the circle at the point (2, 5) and the center of the circle lies on x –2y = 4. Then the radius of the circle is
View answer and explanation
Correct option: A
353\sqrt 5
Option A (353\sqrt 5) is correct because it is the distance between the center of the circle and the point of tangency, calculated using the given conditions.
458 Easy
458 of 685
Application +4 | -1
The equation to the chord of the circle x2+y2=9{x^2} + {y^2} = 9 whose middle point is (1,−2)\left( {1, - 2} \right) is
View answer and explanation
Correct option: C
x−2y−5=0x - 2y - 5 = 0
Option D (x−2y+5=0x - 2y + 5 = 0) is correct because the equation of the chord with a given midpoint (1, -2) is found using the formula T = S', where T is the equation of the tangent-like expression and S' is the value of the circle equation at the midpoint.
459 Easy
459 of 685
Application +4 | -1
The centre of a circle passing through the points (0,0),(1,0)(0,0),(1,0) and touching the circle x2+y2=9x^2+y^2=9 is
View answer and explanation
Correct option: D
(12,−2)\left(\frac{1}{2},-\sqrt{2}\right)
Option D, (12,−2)\left(\frac{{1}}{{2}},-\sqrt{{2}}\right), is correct because the center of the circle passing through (0,0) and (1,0) and touching x2+y2=9x^2+y^2=9 satisfies the given conditions.
460 Easy
460 of 685
Application +4 | -1
The equation of the circle with centre (0,2)(0,2) and radius 2 is x2+y2−my=0x^2+y^2-m y=0. The value of mm is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the value of m in the circle's equation is 4, derived from the center and radius.
461 Easy
461 of 685
Application +4 | -1
The equation of a circle which touches the line y=xy=x at (1,1)(1,1) and having y=x−3y=x-3 as a normal, is
View answer and explanation
Correct option: A
4x2+4y2−20x+4y+8=04 x^2+4 y^2-20 x+4 y+8=0
Option A (4x2+4y2−20x+4y+8=04 x^2+4 y^2-20 x+4 y+8=0) is correct because it represents the equation of a circle that touches the line y=x at (1,1) and has y=x-3 as a normal.
462 Easy
462 of 685
Application +4 | -1
The equation of the circle passing through the point (−1,2)(-1,2) and having two diameters along the pair of lines x2−y2−4x+2y+3=0x^2-y^2-4 x+2 y+3=0, is
View answer and explanation
Correct option: C
x2+y2−4x−2y−5=0x^2+y^2-4 x-2 y-5=0
Option C (x2+y2−4x−2y−5=0x^2+y^2-4 x-2 y-5=0) is correct because it represents the circle passing through the given point and having diameters along the given pair of lines.
463 Easy
463 of 685
Application +4 | -1
The lines 2x−3y=52 x-3 y=5 and 3x−4y=73 x-4 y=7 are diameters of a circle having area as 154 sq.units. Then the equation of the circle is
View answer and explanation
Correct option: D
x2+y2−2x+2y=47x^{2}+y^{2}-2 x+2 y=47.
Option D (x2+y2−2x+2y=47x^{2}+y^{2}-2 x+2 y=47) is correct because it represents the equation of the circle with the given area and diameters, found by determining the center and radius.
464 Medium
464 of 685
Analyzing +4 | -1
The number of common tangents to two circles x2+y2=4{x^2} + {y^2} = 4 and x2+y2−8x+12=0{x^2} + {y^2} - 8x + 12 = 0 is
View answer and explanation
Correct option: D
3
Option D (3) is correct because the distance between the centers equals the sum of the radii, indicating the circles touch externally, allowing for three common tangents.
465 Medium
465 of 685
Application +4 | -1
A line is drawn through a fixed point P(α,β)P\left( {\alpha ,\beta } \right) to cut the circle x2+y2=r2{x^2} + {y^2} = {r^2} at AA and BB. Then, PA.PBPA.PB is equal to
View answer and explanation
Correct option: B
α2+β2−r2{\alpha ^2} + {\beta ^2} - {r^2}
Option B (α2+β2−r2{\alpha ^2} + {\beta ^2} - {r^2}) is correct because it represents the power of the point P with respect to the circle.
466 Easy
466 of 685
Application +4 | -1
The point diametrically opposite to the point P(1,0)P(1,0) on the circle x2+y2+2x+4y−3=0x^2+y^2+2 x+4 y-3=0, is
View answer and explanation
Correct option: A
(−3,−4)(-3,-4)
Option A ((-3,-4)) is correct because it is the point diametrically opposite to P(1,0) on the given circle.
467 Easy
467 of 685
Application +4 | -1
Tangents drawn from the point (4,3)(4,3) to the circle x2+y2−2x−4y=0x^2+y^2-2 x-4 y=0 are inclined at an angle
View answer and explanation
Correct option: D
π/2\pi / 2
Option D (π/2\pi / 2) is correct because the point (4,3) lies on the director circle of the given circle.
468 Easy
468 of 685
Application +4 | -1
If θ\theta is the angle between the tangents from (−1,0)\left( { - 1,0} \right) to the circle x2+y2−5x+4y−2=0{x^2} + {y^2} - 5x + 4y - 2 = 0, then θ\theta is equal to
View answer and explanation
Correct option: A
2tan⁡−1(74)2{\tan ^{ - 1}}\left( {\frac{7}{4}} \right)
The correct answer is 2tan⁡−1(74)2{\tan ^{ - 1}}\left( {\frac{7}{4}} \right) because it represents the angle between the tangents from the given point to the circle.
469 Easy
469 of 685
Application +4 | -1
The equation of the circle with centre (2,1)\left( {2,1} \right) and touching the line 3x+4y=53x + 4y = 5 is
View answer and explanation
Correct option: C
x2+y2−4x−2y+4=0{x^2} + {y^2} - 4x - 2y + 4 = 0
Option C (x2+y2−4x−2y+4=0{x^2} + {y^2} - 4x - 2y + 4 = 0) is correct because the equation of the circle with center (2,1) and touching the line 3x + 4y = 5 is derived by finding the radius and substituting into the circle equation.
470 Easy
470 of 685
Analyzing +4 | -1
If the circles x2+y2−16x−20y+164=r2{x^2} + {y^2} - 16x - 20y + 164 = {r^2} and (x−4)2+(y−7)2=36{(x - 4)^2} + {(y - 7)^2} = 36 intersect at two distinct points, then
View answer and explanation
Correct option: C
1<r<111 < r < 11
Option C (1<r<111 < r < 11) is correct because it satisfies the condition for two circles to intersect at two distinct points, which is the distance between their centers being between the absolute difference and the sum of their radii.
471 Easy
471 of 685
Application +4 | -1
The equation of the circle whose centre is (2,−3)(2,-3) and radius 5, is
View answer and explanation
Correct option: B
x2+y2−4x+6y−12=0x^2+y^2-4 x+6 y-12=0
The correct answer is B (x2+y2−4x+6y−12=0x^2+y^2-4 x+6 y-12=0) because it represents the equation of a circle with center (2, -3) and radius 5.
472 Easy
472 of 685
Analyzing +4 | -1
The diameter of the circle having the pair of lines x2+2xy+3x+6y=0x^2+2 x y+3 x+6 y=0 as its normals and having the size just sufficient to contain the circle x(x−4)+y(y−3)=0x(x-4)+y(y-3)=0 is
View answer and explanation
Correct option: B
15
Option B (15) is correct because it represents the diameter of the circle, calculated as the distance between the normals plus the diameter of the inner circle.
473 Easy
473 of 685
Application +4 | -1
The circle passing through the point (−1,0)(-1,0) and touching the yy-axis at (0,2)(0,2) also passes through the point:
View answer and explanation
Correct option: D
(−4,0)(-4,0)
Option D ((-4,0)) is correct because the circle equation is found using the given conditions, and substituting (-4,0) into the equation satisfies it.
474 Easy
474 of 685
Analyzing +4 | -1
Consider L1:2x+3y+p−3=0;L2:2x+3y+p+3=0L_1: 2 x+3 y+p-3=0 ; L_2: 2 x+3 y+p+3=0 where, pp is a real number, and C:x2+y2+6x−10y+30=0C: x^2+y^2+6 x-10 y+30=0

Statement I If line L1L_1 is a chord of circle CC, then line L2L_2 is not always a diameter of circle CC and
Statement II If line L1L_1 is a diameter of circle CC, then line L2L_2 is not a chord of circle CC.
View answer and explanation
Correct option: C
Statement I is True, statement II is False
Option C (Statement I is True, statement II is False) is correct because if L1 is a chord, L2 may or may not be a diameter, and if L1 is a diameter, L2 cannot be a chord.
475 Medium
475 of 685
Application +4 | -1
The locus of the point from which mutually perpendicular tangents can be drawn to the circle x2+y2=36x^2+y^2=36 is
View answer and explanation
Correct option: D
x2+y2=72x^2+y^2=72.
Option D (x2+y2=72x^2+y^2=72) is correct because the locus of the point from which mutually perpendicular tangents can be drawn to a circle is the director circle, and its radius squared is twice the radius squared of the original circle.
476 Easy
476 of 685
Application +4 | -1
The inverse of the point (1,2)\left( {1,2} \right) with respect to the circle x2+y2−4x−6y+9=0{x^2} + {y^2} - 4x - 6y + 9 = 0, is
View answer and explanation
Correct option: C
(0,1)\left( {0,1} \right)
The correct answer is D ((0,1)) because finding the foot of the perpendicular from the point (1,2) to the tangent line of the circle at that point gives the inverse point.
477 Medium
477 of 685
Application +4 | -1
The equations of the tangents to the circle x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0 which are parallel to the line 4x+3y+5=04x + 3y + 5 = 0, are
View answer and explanation
Correct option: C
4x+3y+19=04x + 3y + 19 = 0 and 4x+3y−31=04x + 3y - 31 = 0
Option C (4x + 3y + 19 = 0 and 4x + 3y - 31 = 0) is correct because these are the tangent lines found by using the perpendicular distance from the center of the circle to the tangent equals the radius.
478 Easy
478 of 685
Analyzing +4 | -1
The equation 2x2+2y2+4x+8y+15=02 x^2+2 y^2+4 x+8 y+15=0 represents
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because completing the square results in a negative radius squared, indicating the equation does not represent a real circle or a pair of straight lines.
479 Medium
479 of 685
Application +4 | -1
If (α,β)\left( {\alpha ,\beta } \right) is a point on the chord PQPQ of the circle x2+y2=25,{x^2} + {y^2} = 25, where the coordinates of PP and QQ are (3,−4)\left( {3, - 4} \right) and (4,3)\left( {4,3} \right) respectively, then
View answer and explanation
Correct option: A
3≤α≤4 and−4≤β≤33 \le \alpha \le 4\,\mathrm{and} - 4 \le \beta \le 3
Option A is correct because if (α,β) is a point on the chord PQ, then α must lie between the x-coordinates of P and Q, and β must lie between the y-coordinates of P and Q.
480 Easy
480 of 685
Application +4 | -1
Two tangents are drawn from the origin to a circle with centre at (2,−1)(2,-1). If the equation of one of the tangents is 3x+y=03 x+y=0, the equation of the other tangent is
View answer and explanation
Correct option: C
x−3y=0x-3 y=0
Option C (x−3y=0x-3y=0) is correct because the other tangent from the origin to the circle can be found by using the fact that the perpendicular distance from the center to both tangents is the same.
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