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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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31 Medium
31 of 685
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.
32 Easy
32 of 685
Application +4 | -1
The angular between the tangent drawn from the origin to the circle (x−7)2+(y+1)2=25{\left( {x - 7} \right)^2} + {\left( {y + 1} \right)^2} = 25 is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because the product of the slopes of the tangents from the origin to the circle is -1, indicating perpendicularity.
33 Easy
33 of 685
Application +4 | -1
Circles are drawn through the point (3,0)(3,0) to cut an intercept of length 6 units on the negative direction of the xx-axis. The equation of the locus of their centres, is
View answer and explanation
Correct option: C
x=0x=0
Option C (x=0) is correct because the locus of the centers of the circles is found to be x=0.
34 Easy
34 of 685
Application +4 | -1
If the straight line x−2y+1=0x - 2y + 1 = 0 intersects the circle x2+y2=25{x^2} + {y^2} = 25 in points PP and QQ, then the coordinates of the point of intersection of tangents drawn at PP and QQ to the circle x2+y2=25{x^2} + {y^2} = 25 are
View answer and explanation
Correct option: C
(−25,50)\left( { - 25,50} \right)
Option C ((-25,50)) is correct because it represents the point of intersection of the tangents to the circle at points P and Q, where the line x - 2y + 1 = 0 intersects the circle.
35 Easy
35 of 685
Application +4 | -1
If a square with length of each side equal to aa is inscribed in the circle x2+y2+4x+10y+21=0x^2+y^2+4 x+10 y+21=0, then aa is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because the side length of the inscribed square is calculated using the circle's radius and the relationship between the square's diagonal and side.
36 Easy
36 of 685
Analyzing +4 | -1
The number of common tangents to the circles x2+y2−x=0,x2+y2+x=0{x^2} + {y^2} - x = 0,{x^2} + {y^2} + x = 0 is
View answer and explanation
Correct option: D
3
The correct answer is 3 because the circles touch each other externally, resulting in three common tangents: one direct common tangent and two transverse common tangents.
37 Medium
37 of 685
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.
38 Easy
38 of 685
Application +4 | -1
If the lengths of the tangents from two points AA and BB to a circle are ll and l′l^{\prime} respectively and points are conjugate with respect to the circle (i.e. each passes through the polar of the other with respect to the circle), then (AB)2=(A B)^2=
View answer and explanation
Correct option: C
l2+l2l^2+l^2
Option C (l2+l′2l^2+l^{\prime 2}) is correct because it represents the square of the distance between two conjugate points with respect to a circle.
39 Easy
39 of 685
Application +4 | -1
The value of kk for which two tangents can be drawn from (k,k)(k, k) to the circle x2+y2+2x+2y−16=0x^2+y^2+2 x+2 y-16=0 is
View answer and explanation
Correct option: C
k∈(−∞,−4)∪(2,∞)k \in(-\infty,-4) \cup(2, \infty)
Option C, k∈(−∞,−4)∪(2,∞)k \in(-\infty,-4) \cup(2, \infty), is correct because it represents the values of k for which the point (k, k) lies outside the circle, allowing two tangents to be drawn.
40 Hard
40 of 685
Analyzing +4 | -1
The set of values of 'cc' so that the equation y=∣x∣+cy = \left| x \right| + c and x2+y2−8∣x∣−9=0{x^2} + {y^2} - 8\left| x \right| - 9 = 0 have no solution, is
View answer and explanation
Correct option: D
(52−4,  ∞)\left( {5\sqrt 2 - 4,\;\infty } \right)
Option C is correct because eliminating the parameter 't' results in the equation of a parabola.
41 Easy
41 of 685
Application +4 | -1
The locus of the point of intersection of perpendicular tangents to the circles x2+y2=a2x^2+y^2=a^2 and x2+y2=b2x^2+y^2=b^2, is
View answer and explanation
Correct option: B
x2+y2=a2+b2x^2+y^2=a^2+b^2
The correct answer is x2+y2=a2+b2x^2+y^2=a^2+b^2 because the locus of the intersection of perpendicular tangents is the sum of the squares of the radii.
42 Easy
42 of 685
Application +4 | -1
The locus of a point such that the tangents drawn from it to the circle x2+y2−6x−8y=0x^2+y^2-6 x-8 y=0 are perpendicular to each other is
View answer and explanation
Correct option: A
x2+y2−6x−8y−25=0x^2+y^2-6 x-8 y-25=0
Option A (x2+y2−6x−8y−25=0x^2+y^2-6 x-8 y-25=0) is correct because the locus of the point of intersection of perpendicular tangents to a circle is the director circle, which has the same center as the original circle and a radius of r2r\sqrt{{2}}.
43 Easy
43 of 685
Application +4 | -1
If the chord of contact of tangents from a point P(x1,y1)P\left( {{x_1},{y_1}} \right) to the circle x2+y2=a2{x^2} + {y^2} = {a^2} touches the circle (x−a)2+y2=a2,{\left( {x - a} \right)^2} + {y^2} = {a^2}, then the locus of (x1,y1),\left( {{x_1},{y_1}} \right), is
View answer and explanation
Correct option: B
A parabola
The correct answer is A parabola because the locus of the point P results in an equation of the form y2=kx{y^2} = kx, which represents a parabola.
44 Easy
44 of 685
Application +4 | -1
The equation of normal of x2+y2−2x+4y−5=0{x^2} + {y^2} - 2x + 4y - 5 = 0 at (2,1)\left( {2,1} \right) is
View answer and explanation
Correct option: A
y=3x−5y = 3x - 5
Option A (y=3x−5y = 3x - 5) is correct because it represents the equation of the line passing through the center of the circle and the given point.
45 Medium
45 of 685
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.
46 Medium
46 of 685
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.
47 Medium
47 of 685
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).
48 Medium
48 of 685
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.
49 Easy
49 of 685
Application +4 | -1
The tangents drawn from the origin to the circle x2+y2−2rx−2hy+h2=0x^2+y^2-2 r x-2 h y+h^2=0 are perpendicular, if
View answer and explanation
Correct option: A
h=±rh=\pm r
Option A (h=±rh=\pm r) is correct because the condition for perpendicular tangents from the origin leads to the relationship h2=r2h^2 = r^2.
50 Easy
50 of 685
Application +4 | -1
If the lines 2x+3y+1=02x + 3y + 1 = 0 and 3x−y−4=03x - y - 4 = 0 lie along diameters of a circle of circumference 10π10\pi, then the equation of the circle is
View answer and explanation
Correct option: A
x2+y2−2x+2y−23=0{x^2} + {y^2} - 2x + 2y - 23 = 0
Option A (x2+y2−2x+2y−23=0{x^2} + {y^2} - 2x + 2y - 23 = 0) is correct because it represents the equation of a circle with center (1, -1) and radius 5, derived from the intersection of diameters and the given circumference.
51 Medium
51 of 685
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.
52 Easy
52 of 685
Application +4 | -1
The centre and radius of the circle 3x2+3y2−8x−10y+3=03 x^2+3 y^2-8 x-10 y+3=0 are
View answer and explanation
Correct option: D
(43,53)432\left(\frac{4}{3}, \frac{5}{3}\right) \frac{4}{3} \sqrt{2}
Option D ((43,53)432\left(\frac{{4}}{{3}}, \frac{{5}}{{3}}\right) \frac{{4}}{{3}} \sqrt{{2}}) is correct because it accurately represents the center and radius calculated from the given circle equation after converting it to standard form.
53 Medium
53 of 685
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.
54 Easy
54 of 685
Application +4 | -1
A square is inscribed in the circle x2+y2−2x+4y−93=0x^2+y^2-2 x+4 y-93=0 with its sides parallel to the axes of coordinates. The coordinates of the vertex farthest from the origin are
View answer and explanation
Correct option: B
(8,−9)(8,-9)
Option B ((8,−9)(8,-9)) is correct because it represents the vertex of the square farthest from the origin, calculated using the circle's center and radius.
55 Easy
55 of 685
Analyzing +4 | -1
A variable circle CC has the equation x2+y2−2(t2−3t+1)x−2(t2+2t)y+t=0x^2+y^2-2\left(t^2-3 t+1\right) x-2\left(t^2+2 t\right) y+t=0, where tt is aa parameter. If the power of point (a,b)(a, b) w.r.t. the circle CC is constant, then the ordered pair (a,b)(a, b) is
View answer and explanation
Correct option: C
(−110,110)\left(-\frac{1}{10}, \frac{1}{10}\right)
Option C (−110,110)\left(-\frac{{1}}{{10}}, \frac{{1}}{{10}}\right) is correct because the power of the point (a,b)(a, b) with respect to the circle is constant only when a+b=0a + b = 0 and 6a−4b+1=06a - 4b + 1 = 0.
56 Easy
56 of 685
Application +4 | -1
The abscissae of two points AA and BB are the roots of the equation x2−2ax−b2=0,{x^2} - 2ax - {b^2} = 0, and their ordinates are the roots of the equation x2+2px−q2=0.{x^2} + 2px - {q^2} = 0. The radius of the circle with ABAB as diameter is
View answer and explanation
Correct option: D
a2+b2+p2+q2\sqrt {{a^2} + {b^2} + {p^2} + {q^2}}
Option D (a2+b2+p2+q2\sqrt {{a^2} + {b^2} + {p^2} + {q^2}}) is correct because it's derived from the distance formula using the roots of the given quadratic equations.
57 Medium
57 of 685
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.
58 Easy
58 of 685
Analyzing +4 | -1
If a1a2=b1b2a_1 a_2=b_1 b_2, then the lines a1x+b1y+c1=0a_1 x+b_1 y+c_1=0 and a2x+b2y+c2=0a_2 x+b_2 y+c_2=0 cut the coordinate axes in concyclic points. The equation of the circle, is
View answer and explanation
Correct option: C
(a1x+b1y+c1)(a2x+b2y+c2)−(a1b2+a2b1)xy=0\left(a_1 x+b_1 y+c_1\right)\left(a_2 x+b_2 y+c_2\right)-\left(a_1 b_2+a_2 b_1\right) x y=0
Option C (a1x+b1y+c1)(a2x+b2y+c2)−(a1b2+a2b1)xy=0\left(a_1 x+b_1 y+c_1\right)\left(a_2 x+b_2 y+c_2\right)-\left(a_1 b_2+a_2 b_1\right) x y=0 is correct because it represents the equation of the circle passing through the points of intersection of the given lines with the coordinate axes, given the condition a1a2=b1b2a_1 a_2 = b_1 b_2.
59 Medium
59 of 685
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.
60 Easy
60 of 685
Analyzing +4 | -1
Tangents are drawn from the point (17,7)(17,7) to the circle x2+y2=169x^2+y^2=169
Statement I The tangents are mutually perpendicular. because

Statement II The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2+y2=338x^2+y^2=338
View answer and explanation
Correct option: A
Statement I is True, statement II is True; statement II is a
correct explanation for statement I
Option A (Statement I is True, statement II is True; statement II is a correct explanation for statement I) is correct because the locus of points from which perpendicular tangents can be drawn to the circle is the director circle, and (17,7) lies on it.
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