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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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Showing 31-60 of 480 questions
31 Easy
31 of 480
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.
32 Easy
32 of 480
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.
33 Easy
33 of 480
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.
34 Easy
34 of 480
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.
35 Easy
35 of 480
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.
36 Easy
36 of 480
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.
37 Easy
37 of 480
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.
38 Easy
38 of 480
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.
39 Easy
39 of 480
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.
40 Easy
40 of 480
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.
41 Easy
41 of 480
Understanding +4 | -1
The angle between the tangents drawn from a point on the director circle x2+y2=50x^2+y^2=50 to the circle x2+y2=25x^2+y^2=25, is
View answer and explanation
Correct option: C
90∘90^{\circ}
Option C (90∘90^{\circ}) is correct because the director circle is the locus of points from which tangents to the circle are perpendicular.
42 Easy
42 of 480
Application +4 | -1
If 3x+y+k=03x + y + k = 0 is a tangent to the circle x2+y2=10{x^2} + {y^2} = 10, the values of kk are
View answer and explanation
Correct option: C
±10\pm 10
Option C (±10\pm 10) is correct because it represents the values of kk for which the line 3x+y+k=03x + y + k = 0 is tangent to the circle x2+y2=10{x^2} + {y^2} = 10.
43 Easy
43 of 480
Application +4 | -1
Equation of a circle which touches yy-axis at (0,2)(0,2) and cuts off an intercept of 3 units from the xx-axis is x2+y2−2αx−4y+4=0x^2+y^2-2 \alpha x-4 y+4=0 where α2=\alpha^2=
View answer and explanation
Correct option: B
25/425 / 4
Option B (25/425 / 4) is correct because it is the value of α2\alpha^2 that satisfies the given conditions for the circle.
44 Easy
44 of 480
Application +4 | -1
A point PP moves such that PA/PB=pP A / P B=p where AA and BB are two fixed points with AB=aA B=a, the locus of PP is a circle with radius
View answer and explanation
Correct option: C
ap/(1−p2)a p /\left(1-p^2\right)
Option C (ap/(1−p2)a p / (1-p^2)) is correct because it represents the radius of the circle that is the locus of P.
45 Easy
45 of 480
Application +4 | -1
If (1,  a),  (b,  2)\left( {1,\;a} \right),\;\left( {b,\;2} \right) are conjugate points with respect to the circle x2+y2=25{x^2} + {y^2} = 25, then 4a+2b4a + 2b is equal to
View answer and explanation
Correct option: B
50
Option B (50) is correct because it is the value of 4a + 2b when (1, a) and (b, 2) are conjugate points with respect to the given circle.
46 Easy
46 of 480
Application +4 | -1
The radius of the circle passing through the point P(6,2)P\left( {6,2} \right) and two of whose diameters are x+y=6x + y = 6 and x+2y=4x + 2y = 4, is
View answer and explanation
Correct option: D
20\sqrt {20}
Option D (20\sqrt {20}) is correct because the radius is the distance between the center of the circle and the point P(6,2), where the center is the intersection of the two diameters.
47 Easy
47 of 480
Application +4 | -1
ABCDABCD is a square whose side is a. The equation of the circle circumscribing the square, taking ABAB and ADAD as axes of reference, is
View answer and explanation
Correct option: C
x2+y2−ax−ay=0{x^2} + {y^2} - ax - ay = 0
Option C (x2+y2−ax−ay=0{x^2} + {y^2} - ax - ay = 0) is correct because the circle passes through the vertices of the square when AB and AD are the axes of reference.
48 Easy
48 of 480
Application +4 | -1
Equation of circle passing through (1,5)(1,5) and (4,1)(4,1) and touching yy-axis is x2+y2−5x−6y+9+λ(4x+3y−19)=0x^2+y^2-5 x-6 y+9+\lambda(4 x+3 y-19)=0 where λ\lambda is equal to
View answer and explanation
Correct option: A
0,−40/90,-40 / 9
Option A (0,−40/90, -40/9) is correct because these are the values of λ\lambda for which the circle passes through the given points and touches the y-axis.
49 Easy
49 of 480
Application +4 | -1
The area of square inscribed in a circle x2+y2−6x−8y=0{x^2} + {y^2} - 6x - 8y = 0 is
View answer and explanation
Correct option: B
50 sq units
Option B (50 sq units) is correct because the area of the square inscribed in the circle is half the square of the circle's diameter.
50 Easy
50 of 480
Application +4 | -1
The line 4x+3y−4=04 x+3 y-4=0 divides the circumference of the circle centred at (5,3)(5,3) in the ratio 1:21: 2. Then the equation of the circle is
View answer and explanation
Correct option: A
x2+y2−10x−6y−66=0x^2+y^2-10 x-6 y-66=0
Option A (x2+y2−10x−6y−66=0x^2+y^2-10 x-6 y-66=0) is correct because the radius of the circle is calculated to be 10, and the equation is derived accordingly.
51 Easy
51 of 480
Application +4 | -1
The centres of those circles which touch the circle, x2+y2−8x−8y−4=0{x^2} + {y^2} - 8x - 8y - 4 = 0, externally and also touch the XX-axis, lie on
View answer and explanation
Correct option: D
A parabola
Option D is correct because the locus of the centers of the circles that touch the given circle externally and the x-axis is a parabola.
52 Easy
52 of 480
Application +4 | -1
The coordinates of the middle point of the chord 2x−5y+18=02 x-5 y+18=0 cut off by the circle x2+y2−6x+2y−54=0x^2+y^2-6 x+2 y-54=0 is
View answer and explanation
Correct option: A
(1,4)(1,4)
Option C, (4,1), is correct because it represents the midpoint of the chord, which lies on the line perpendicular to the chord passing through the center of the circle.
53 Easy
53 of 480
Application +4 | -1
The number of common tangents to the circles
x2+y2−2x−4y+1=0{x^2} + {y^2} - 2x - 4y + 1 = 0 and x2+y2−12x−16y+91=0{x^2} + {y^2} - 12x - 16y + 91 = 0, is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the distance between the centers of the circles is greater than the sum of their radii, indicating four common tangents.
54 Easy
54 of 480
Application +4 | -1
If circumference of circle x2+y2+8x+8y−b=0x^2+y^2+8 x+8 y-b=0 is bisected by circle x2+y2−2x+4y+a=0x^2+y^2-2 x+4 y+a=0 then (a+b)(a+b) is equal to
View answer and explanation
Correct option: C
-56
The correct answer is C (-56) because the condition for a circle to bisect the circumference of another circle is that the radical axis passes through the center of the first circle.
55 Easy
55 of 480
Application +4 | -1
The equations of the circles which touch both the axes and the line x=ax=a are
View answer and explanation
Correct option: C
x2+y2−ax±ay+a24=0x^2+y^2-a x \pm a y+\frac{a^2}{4}=0
Option C (x2+y2−ax±ay+a24=0x^2+y^2-a x \pm a y+\frac{a^2}{4}=0) is correct because it represents a circle touching both axes and the line x=a.
56 Easy
56 of 480
Application +4 | -1
A variable circle passes through the fixed point A(p,q)A\left( {p,q} \right) and touches xx-axis. The locus of the other end of the diameter through AA is
View answer and explanation
Correct option: A
(x−p)2=4qy{\left( {x - p} \right)^2} = 4qy
Option A ((x−p)2=4qy{\left( {x - p} \right)^2} = 4qy) is correct because it represents the locus of the other end of the diameter through A.
57 Easy
57 of 480
Analyzing +4 | -1
If 2x+3y−6=02x + 3y - 6 = 0 and 9x+6y−18=09x + 6y - 18 = 0 cuts the axes in concyclic points, then the centre of the circle, is
View answer and explanation
Correct option: D
(5/2,5/2)\left( {5/2,5/2} \right)
Option D ((5/2,5/2)\left( {5/2,5/2} \right)) is correct because the center of the circle passing through the concyclic points is found by equating the equation of the circle to zero and finding the center.
58 Easy
58 of 480
Analyzing +4 | -1
The lengths of the tangents from any point on the circle 15x2+15y2−48x+64y=015 x^2+15 y^2-48 x+64 y=0 to the two circles 5x2+5y2−24x+32y+75=0,5 x^2+5 y^2-24 x+32 y+75=0,

5x2+5y2−48x+64y+300=05 x^2+5 y^2-48 x+64 y+300=0 are in the ratio
View answer and explanation
Correct option: A
1:21: 2
Option A (1:21:2) is correct because the ratio of the lengths of the tangents from a point on the circle to the other two circles is calculated to be 1:21:2.
59 Easy
59 of 480
Application +4 | -1
The line segment joining the points (4,7)\left( {4,7} \right) and (−2,  −1)\left( { - 2,\; - 1} \right) is a diameter of a circle. If the circle intersects the xx-axis at AA and BB, then ABAB is equal to
View answer and explanation
Correct option: D
8
Option D (8) is correct because the length of the chord AB, where the circle intersects the x-axis, is calculated using the formula 2g2−c2\sqrt{{g^2 - c}} after finding the equation of the circle.
60 Easy
60 of 480
Application +4 | -1
Coordinates of the mid-point of the segment cut by the circle x2+y2−6x+2y−54=0x^2+y^2-6 x+2 y-54=0 on the line 2x−5y+18=02 x-5 y+18=0, is
View answer and explanation
Correct option: C
(1,4)(1,4)
Option C (1,4) is correct because the midpoint of the chord is found by solving the equation of the line perpendicular to the chord and passing through the circle's center.
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