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JEE • MATHEMATICS • SYSTEM OF CIRCLES

System of 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 1-30 of 75 questions
1 Easy
1 of 75
Application +4 | -1
x=1x = 1 is the radical axis of the two orthogonally intersecting circles. If x2+y2=4{x^2} + {y^2} = 4 is one of the circles, then the other circle is
View answer and explanation
Correct option: B
x2+y2−8x+4=0{x^2} + {y^2} - 8x + 4 = 0
Option B (x2+y2−8x+4=0{x^2} + {y^2} - 8x + 4 = 0) is correct because it is orthogonal to the given circle and has x = 1 as the radical axis.
2 Easy
2 of 75
Application +4 | -1
The circle x2+y2+8y−4=0{x^2} + {y^2} + 8y - 4 = 0, cuts the real circle x2+y2+gx+4=0{x^2} + {y^2} + gx + 4 = 0 orthogonally, if
View answer and explanation
Correct option: A
For any real value of gg
Option A (For any real value of gg) is correct because the orthogonality condition is satisfied regardless of the value of g.
3 Easy
3 of 75
Application +4 | -1
The equation of the circle whose diameter is the common chord of the circles x2+y2+3x+2y+1=0x^2+y^2+3 x+2 y+1=0 and x2+y2+3x+4y+2=0x^2+y^2+3 x+4 y+2=0, is
View answer and explanation
Correct option: C
2x2+2y2+6x+2y+1=02 x^2+2 y^2+6 x+2 y+1=0
Option C (2x2+2y2+6x+2y+1=02 x^2+2 y^2+6 x+2 y+1=0) is correct because it represents the equation of the circle whose diameter is the common chord of the given circles.
4 Easy
4 of 75
Application +4 | -1
If the circles x2+y2+2ky+2x+6=0{x^2} + {y^2} + 2ky + 2x + 6 = 0 and x2+y2+2ky+k=0{x^2} + {y^2} + 2ky + k = 0 intersects orthogonally. Then, the value of kk is
View answer and explanation
Correct option: C
−32- \frac{3}{2}
Option C (−32- \frac{3}{2}) is correct because it is one of the values of k obtained by solving the quadratic equation derived from the orthogonality condition of the circles.
5 Easy
5 of 75
Analyzing +4 | -1
If the circles x2+y2+2ax+cy+a=0x^2+y^2+2 a x+c y+a=0 and x2+y2−3ax+dy−1=0x^2+y^2-3 a x+d y-1=0 intersect in two distinct points PP and QQ, then the line 5x+by−a=05 x+b y-a=0 passes through PP and QQ for
View answer and explanation
Correct option: B
no value of aa
Option B (no value of aa) is correct because the condition for the line 5x+by−a=05x + by - a = 0 to pass through the intersection points P and Q leads to a quadratic equation with no real solutions for aa.
6 Easy
6 of 75
Analyzing +4 | -1
The equation of the circle having its centre on the line x+2y−3=0x + 2y - 3 = 0 and passing through the point of intersection of the circles x2+y2−4y+1=0{x^2} + {y^2} - 4y + 1 = 0 and x2+y2−4x−2y+4=0{x^2} + {y^2} - 4x - 2y + 4 = 0 is
View answer and explanation
Correct option: A
x2+y2−6x+7=0{x^2} + {y^2} - 6x + 7 = 0
The correct answer is x2+y2−6x+7=0{x^2} + {y^2} - 6x + 7 = 0 (Option A) because it satisfies the condition of passing through the intersection of the given circles and having its center on the line x+2y−3=0x + 2y - 3 = 0.
7 Easy
7 of 75
Application +4 | -1
The length of the common chord of the circles x2+y2+4x+1=0{x^2} + {y^2} + 4x + 1 = 0 and x2+y2+4y−1=0{x^2} + {y^2} + 4y - 1 = 0
View answer and explanation
Correct option: A
152\sqrt {\frac{{15}}{2}}
Option A (152\sqrt {\frac{{15}}{2}}) is correct because it represents the length of the common chord of the two circles.
8 Easy
8 of 75
Application +4 | -1
The equation of circle concentric with circle x2+y2−6x−4y+1=0x^2+y^2-6 x-4 y+1=0 and passing through origin is
View answer and explanation
Correct option: A
x2+y2−6x−4y=0x^2+y^2-6 x-4 y=0
Option A (x2+y2−6x−4y=0x^2+y^2-6 x-4 y=0) is correct because it has the same center as the given circle and passes through the origin.
9 Easy
9 of 75
Analyzing +4 | -1
The radical axis of the circles x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 and 2(x2+y2)+3x+2y+2c=02\left(x^2+y^2\right)+3 x+2 y+2 c=0 touches the circle x2+y2+2x+2y+1=0x^2+y^2+2 x+2 y+1=0, if
View answer and explanation
Correct option: B
g=43g=\frac{4}{3} and f=12f=\frac{1}{2}
Option B (g=43g=\frac{4}{3} and f=12f=\frac{1}{2}) is correct because these values satisfy the condition for the radical axis to touch the third circle.
10 Easy
10 of 75
Analyzing +4 | -1
The circles x2+y2+x+y=0{x^2} + {y^2} + x + y = 0 and x2+y2+x−y{x^2} + {y^2} + x - y intersect at an angle of
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D π2\frac{\pi }{2} is correct because the angle of intersection between the two circles is 90 degrees, indicating they are orthogonal.
11 Easy
11 of 75
Application +4 | -1
If a circle passes through the point (a,b)\left( {a,b} \right) and cuts the circle x2+y2=p2{x^2} + {y^2} = {p^2} orthogonally, then the equation of the locus of its centre is
View answer and explanation
Correct option: A
2ax+2by−(a2+b2+p2)=02ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0
The correct answer is A (2ax+2by−(a2+b2+p2)=02ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0) because it represents the locus of the center of the circle passing through (a, b) and cutting the given circle orthogonally.
12 Easy
12 of 75
Application +4 | -1
If a circle passes through the point (a,b)(a, b) and cuts the circle x2+y2=p2x^2+y^2=p^2 orthogonally, then the equation of the locus of its centre is
View answer and explanation
Correct option: D
2ax+2by−(a2+b2+p2)=02 a x+2 b y-\left(a^2+b^2+p^2\right)=0
Option D (2ax+2by−(a2+b2+p2)=02 a x+2 b y-\left(a^2+b^2+p^2\right)=0) is correct because it represents the locus of the center of the circle that passes through (a, b) and cuts the circle x2+y2=p2x^2+y^2=p^2 orthogonally.
13 Easy
13 of 75
Analyzing +4 | -1
The angle of intersection of the circles x2+y2=4{x^2} + {y^2} = 4 and x2+y2=2x+2y,{x^2} + {y^2} = 2x + 2y, is
View answer and explanation
Correct option: D
π4\frac{\pi }{4}
Option D (π4\frac{\pi }{4}) is correct because it is the angle between the circles, calculated using the radii and the distance between their centers.
14 Easy
14 of 75
Application +4 | -1
The equation of the circle whose diameter is the common chord of the circles x2+y2+2x+3y+2=0{x^2} + {y^2} + 2x + 3y + 2 = 0 and x2+y2+2x−3y−4=0{x^2} + {y^2} + 2x - 3y - 4 = 0 is
View answer and explanation
Correct option: C
x2+y2+2x+2y+1=0{x^2} + {y^2} + 2x + 2y + 1 = 0
Option C (x2+y2+2x+2y+1=0{x^2} + {y^2} + 2x + 2y + 1 = 0) is correct because it represents the circle whose diameter is the common chord of the given circles.
15 Easy
15 of 75
Analyzing +4 | -1
x2+y2−6x−6y+4=0,x2+y2−2x−4y+3=0,{x^2} + {y^2} - 6x - 6y + 4 = 0,{x^2} + {y^2} - 2x - 4y + 3 = 0, x2+y2+2kx+2y+1=0{x^2} + {y^2} + 2kx + 2y + 1 = 0. If the radical centre of the above three circles exists, then which of the following cannot be the value of kk
View answer and explanation
Correct option: C
5
Option C (5) is correct because the radical center exists only when the determinant of the coefficients of the radical axes is non-zero, leading to k≠5k \ne 5.

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16 Easy
16 of 75
Application +4 | -1
If the circle x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 bisects the circumference of the circle x2+y2+2g′x+2f′y+c′=0x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0, then the length of the common chord of these two circles, is
View answer and explanation
Correct option: B
2g′2+f′2−c′2 \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}
Option B (2g′2+f′2−c′2 \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}) is correct because when one circle bisects the circumference of another, the common chord is the diameter of the bisected circle.
17 Easy
17 of 75
Application +4 | -1
The centre of the circle, which cuts orthogonally each of the three circles x2+y2+2x+17y+4=0{x^2} + {y^2} + 2x + 17y + 4 = 0 and x2+y2+7x+6y+11=0,x2+y2−x+22y+3=0{x^2} + {y^2} + 7x + 6y + 11 = 0,{x^2} + {y^2} - x + 22y + 3 = 0, is
View answer and explanation
Correct option: A
(3,2)\left( {3,2} \right)
Option A (3,2) is correct because it represents the center of the circle that intersects the three given circles orthogonally, found by solving the equations derived from the orthogonality condition.
18 Easy
18 of 75
Analyzing +4 | -1
A circle CC touches the line x=2yx = 2y at the point (2, 1) and intersects the circle C1:x2+y2+2y−5=0{C_1}:{x^2} + {y^2} + 2y - 5 = 0 at two points PP and QQ such that PQPQ is a diameter of C1{C_1}. Then the diameter of CC is
View answer and explanation
Correct option: A
757\sqrt 5
Option A (757\sqrt 5) is correct because it is the diameter of the circle C, which is found by using the condition that the common chord PQ is a diameter of circle C1.
19 Easy
19 of 75
Analyzing +4 | -1
A circle SS passes through the point (0, 1) and is orthogonal to the circles (x−1)2+y2=16(x-1)^2+y^2=16 and x2+y2=1x^2+y^2=1. Then,
View answer and explanation
Correct option: B
Radius of SS is 7
The correct answer is B (Radius of SS is 7) because after finding the center and using the formula for radius, the radius is calculated to be 7.
20 Easy
20 of 75
Application +4 | -1
The radical axis of the coaxial system of circles with limiting points (1,2)\left( {1,2} \right) and (−2,1)\left( { - 2,1} \right) is
View answer and explanation
Correct option: B
3x+y=03x + y = 0
Option B (3x+y=03x + y = 0) is correct because it represents the equation of the radical axis derived from the limiting points of the coaxial system.
21 Easy
21 of 75
Analyzing +4 | -1
If PP and QQ are the points of intersection of the circles x2+y2+3x+7y+2p−5=0x^2+y^2+3 x+7 y+2 p-5=0 and x2+y2+2x+2y−p2=0x^2+y^2+2 x+2 y-p^2=0 then there is a circle passing through P,QP, Q and (1,1)(1,1) for :
View answer and explanation
Correct option: A
all except one value of pp
Option A (all except one value of pp) is correct because the equation for λ\lambda is undefined when p=−1p = -1, indicating that there is no circle passing through P, Q, and (1,1) for that specific value of pp.
22 Easy
22 of 75
Application +4 | -1
The equation of the common tangent of the two touching circles, y2+x2−6x−12y+37=0{y^2} + {x^2} - 6x - 12y + 37 = 0 and x2+y2−6y+7=0{x^2} + {y^2} - 6y + 7 = 0 is
View answer and explanation
Correct option: C
x−y−5=0x - y - 5 = 0
Option C (x−y−5=0x - y - 5 = 0) is correct because it is the equation obtained by subtracting the equations of the two circles.
23 Easy
23 of 75
Analyzing +4 | -1
If PP and QQ are the points of intersection of the circles
x2+y2+3x+7y+2p−5=0{x^2} + {y^2} + 3x + 7y + 2p - 5 = 0
x2+y2+2x+2y−p2=0{x^2} + {y^2} + 2x + 2y - {p^2} = 0, then there is a circle passing through P,QP,Q and (1,1)\left( {1,1} \right) and
View answer and explanation
Correct option: B
All except one value of pp
Option B (All except one value of pp) is correct because there is a circle passing through P, Q, and (1,1) for all values of p except when the equation becomes undefined.
24 Easy
24 of 75
Application +4 | -1
If S≡x2+y2+2x+3y+1=0S \equiv x^2+y^2+2 x+3 y+1=0 and S′≡x2+y2+4x+3y+2=0S^{\prime} \equiv x^2+y^2+4 x+3 y+2=0 are two circles, then the point (−3,−2)(-3,-2) lies
View answer and explanation
Correct option: A
inside S′S^{\prime} only
Option A (inside S′S^{\prime} only) is correct because substituting the point into the circle equations shows it's outside S and inside S'.
25 Easy
25 of 75
Application +4 | -1
The point from which the tangents to the circles

x2+y2−8x+40=0x^2+y^2-8 x+40=0

5x2+5y2−25x+80=05 x^2+5 y^2-25 x+80=0

x2+y2−8x+16y+160=0x^2+y^2-8 x+16 y+160=0

are equal in length, is
View answer and explanation
Correct option: C
(8,−15/2)(8,-15 / 2)
Option C ((8,−15/2)(8,-15 / 2)) is correct because it represents the radical center of the three circles, where the lengths of tangents from that point to each circle are equal.
26 Easy
26 of 75
Application +4 | -1
The equation of the circle passing through (1, -3) and the points common to the two circles x2+y2−6x+8y−16=0,x2+y2+4x−2y−8=0x^2+y^2-6 x+8 y-16=0, x^2+y^2+4 x-2 y-8=0, is
View answer and explanation
Correct option: B
2x2+2y2+3x+y−20=02 x^2+2 y^2+3 x+y-20=0
Option B (2x2+2y2+3x+y−20=02 x^2+2 y^2+3 x+y-20=0) is correct because it represents the equation of a circle passing through the given point and the intersection of the two given circles.
27 Easy
27 of 75
Analyzing +4 | -1
The locus of the centres of circles passing through the origin and intersecting the fixed circle x2+y2−5x+3y−1=0x^2+y^2-5 x+3 y-1=0 orthogonally is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the locus of the centers is a straight line with the equation 5x - 3y = -1, which does not match any of the other options.
28 Easy
28 of 75
Application +4 | -1
If the circle x2+y2+2x+3y+1=0{x^2} + {y^2} + 2x + 3y + 1 = 0 cuts x2+y2+4x+3y+2=0{x^2} + {y^2} + 4x + 3y + 2 = 0 in AA and BB, then the equation of the circle on ABAB as diameter is
View answer and explanation
Correct option: B
2x2+2y2+2x+6y+1=02{x^2} + 2{y^2} + 2x + 6y + 1 = 0
Option D (2x2+2y2+2x+6y+1=02{x^2} + 2{y^2} + 2x + 6y + 1 = 0) is correct because it represents the equation of the circle with the common chord as a diameter, found by using the equation of the common chord and the equation of the first circle.
29 Easy
29 of 75
Application +4 | -1
The equation of a circle which passes through (2a,0)\left( {2a,0} \right) and whose radical axis in relation to the circle x2+y2=a2{x^2} + {y^2} = {a^2} is x=a/2,x = a/2, is
View answer and explanation
Correct option: C
x2+y2−2ax=0{x^2} + {y^2} - 2ax = 0
Option C (x2+y2−2ax=0{x^2} + {y^2} - 2ax = 0) is correct because it represents the equation of a circle passing through (2a,0) and having the given radical axis.
30 Easy
30 of 75
Application +4 | -1
The radical axis of the circles, belonging to the coaxal system of circles whose limiting points are (1,3)(1,3) and (2,6)(2,6), is
View answer and explanation
Correct option: B
x+3y−15=0x+3 y-15=0
Option B (x+3y−15=0x+3 y-15=0) is correct because it represents the radical axis of the coaxial system of circles with the given limiting points.
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  • ✓ This set contains 75 single-correct-answer MCQs on SYSTEM OF CIRCLES.
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