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JEE • MATHEMATICS • CIRCLES

Circles practice · Easy

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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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121 Easy
121 of 480
Application +4 | -1
The normal to the circle x2+y2−2x−2y=0x^2+y^2-2 x-2 y=0 passing through (2,2)(2,2) is
View answer and explanation
Correct option: A
x=yx=y
Option A (x=yx=y) is correct because the normal to a circle passes through its center, and the line x=y passes through both the center and the given point.
122 Easy
122 of 480
Application +4 | -1
If the tangent from a point PP to the circle x2+y2=1x^2+y^2=1 is perpendicular to the tangent from PP to the circle x2+y2=3x^2+y^2=3, then the locus of PP, is
View answer and explanation
Correct option: A
a circle of radius 2
The correct answer is a circle of radius 2 because the locus of P is the director circle of both circles.
123 Easy
123 of 480
Application +4 | -1
If the lines 3x−4y−7=03x - 4y - 7 = 0 and 2x−3y−5=02x - 3y - 5 = 0 are two diameters of a circle of area 49π49\pi sq unit, the equation of the circle is
View answer and explanation
Correct option: C
x2+y2−2x+2y−47=0{x^2} + {y^2} - 2x + 2y - 47 = 0
Option C (x2+y2−2x+2y−47=0{x^2} + {y^2} - 2x + 2y - 47 = 0) is correct because the center of the circle is the intersection of the diameters, and the radius can be found from the area.
124 Easy
124 of 480
Application +4 | -1
The image of the circle x2+y2+16x−24y+183=0x^2+y^2+16 x-24 y+183=0 in the line mirror 4x+7y+13=04 x+7 y+13=0, is
View answer and explanation
Correct option: A
(x+16)2+(y+2)2=52(x+16)^2+(y+2)^2=5^2
Option A ((x+16)2+(y+2)2=52(x+16)^2+(y+2)^2=5^2) is correct because it represents the correct transformation of the original circle across the given line mirror.
125 Easy
125 of 480
Application +4 | -1
The locus of a point, which moves such that the lengths of the tangents from it two concentric circles of radii aa and bb are inversely proportional to their radii is a circle with radius
View answer and explanation
Correct option: C
a2+b2\sqrt{a^2+b^2}
Option C (a2+b2\sqrt{a^2+b^2}) is correct because it represents the radius of the circle formed by the locus of points where the tangent lengths are inversely proportional to the radii of the two concentric circles.
126 Easy
126 of 480
Application +4 | -1
If the circle (x−2)2+(y−3)2=a2(x-2)^2+(y-3)^2=a^2 lies entirely with in the circle x2+y2=b2x^2+y^2=b^2, then
View answer and explanation
Correct option: C
b−a>13b-a > \sqrt{13}
Option C (b−a>13b-a > \sqrt{{13}}) is correct because the distance between the centers of the circles must be less than the difference of their radii for one circle to lie entirely within the other.
127 Easy
127 of 480
Application +4 | -1
One of the diameter of the circle x2+y2−12x+4y+6=0{x^2} + {y^2} - 12x + 4y + 6 = 0 is given by
View answer and explanation
Correct option: B
x+3y=0x + 3y = 0
Option B (x+3y=0x + 3y = 0) is correct because it passes through the center of the circle, which is a necessary condition for a diameter.
128 Easy
128 of 480
Application +4 | -1
The tangent at PP, any point on the circle x2+y2=4,{x^2} + {y^2} = 4, meets the coordinate axes in AA and BB, then
View answer and explanation
Correct option: C
The locus of the mid-point of ABAB is x2+y2=x2y2{x^2} + {y^2} = {x^2}{y^2}
Option C (The locus of the mid-point of ABAB is x2+y2=x2y2{x^2} + {y^2} = {x^2}{y^2}) is correct because it's derived from finding the equation of the tangent to the circle and then determining the locus of the midpoint of the intercepts on the axes.
129 Easy
129 of 480
Application +4 | -1
The equation of the image of the circle (x−3)2+(y−2)2=1(x-3)^2+(y-2)^2=1 by the mirror x+y=19x+y=19 is
View answer and explanation
Correct option: D
(x−17)2+(y−16)2=1(x-17)^2+(y-16)^2=1
Option D ((x−17)2+(y−16)2=1(x-17)^2+(y-16)^2=1) is correct because it represents the circle with the center being the reflection of the original center across the given line.
130 Easy
130 of 480
Application +4 | -1
The locus of a point which moves in a plane so that the sum of the squares of its distances from the line ax+by+c=0a x+b y+c=0 and bx−ay+d=0b x-a y+d=0 is r2r^2, is a circle of radius
View answer and explanation
Correct option: A
rr
Option A (rr) is correct because the locus of the point is a circle, and its radius is found by simplifying the given equation and comparing it to the standard circle equation.
131 Easy
131 of 480
Application +4 | -1
If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2+y2=1{x^2} + {y^2} = 1 is a circle of radius rr, then rr is equal to
View answer and explanation
Correct option: B
01-Feb
Option B (0.5) is correct because the radius of the locus of the midpoint is half the radius of the original circle.
132 Easy
132 of 480
Application +4 | -1
The lines 3x−4y+λ=03 x-4 y+\lambda=0 and 6x−8y+μ=06 x-8 y+\mu=0 are tangents to the same circle. The radius of the circle is
View answer and explanation
Correct option: A
∣2λ−μ20∣\left|\frac{2 \lambda-\mu}{20}\right|
Option A, ∣2λ−μ20∣\left|\frac{{2 \lambda-\mu}}{{20}}\right|, is correct because it represents half the distance between the two parallel tangent lines, which is the radius of the circle.
133 Easy
133 of 480
Application +4 | -1
The circular wire of diameter 10  cm\mathrm{10}\,\,\mathrm{cm} is cut and placed along the circumference of a circle of diameter 1m\mathrm{1 m}. The angle subtended by the wire at the center of the circle is equal to
View answer and explanation
Correct option: C
π5  rad\frac{\pi }{5}\,\,\mathrm{rad}
Option C (π5 rad\frac{\pi }{5}\,\mathrm{rad}) is correct because the angle subtended is calculated by dividing the arc length (circumference of the small circle) by the radius of the larger circle.
134 Easy
134 of 480
Application +4 | -1
If the lines 3x−4y−7=03 x-4 y-7=0 and 2x−3y−5=02 x-3 y-5=0 are two diameters of a circle of area 49π49 \pi square units, the equation of the circle is
View answer and explanation
Correct option: D
x2+y2−2x+2y−47=0x^2+y^2-2 x+2 y-47=0
Option D (x2+y2−2x+2y−47=0x^2+y^2-2 x+2 y-47=0) is correct because it represents the equation of a circle with the given area and center at the intersection of the two diameters.
135 Easy
135 of 480
Understanding +4 | -1
The centre of the circle x=2+3cos⁡θx=2+3 \cos \theta, y=3sin⁡θ−1y=3 \sin \theta-1, is
View answer and explanation
Correct option: B
(2,−1)(2,-1)
Option B (2,-1) is correct because the center can be directly identified from the parametric equations by rewriting them in the standard circle equation form.

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136 Easy
136 of 480
Application +4 | -1
The equation of circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a3 {a} is x2+y2=ka2{x}^2+{y}^2={ka}^2, where k={k}=
View answer and explanation
Correct option: A
4
Option A (4) is correct because the radius of the circle is 2a, and the equation of the circle is x2 + y2 = 4a2, thus k=4.
137 Easy
137 of 480
Application +4 | -1
A line drawn through the point P(4,7)\mathrm{P}(4,7) cuts the circle x2+y2=9x^{2}+y^{2}=9 at the points A and B\mathrm{B}. Then PA.PB is equal to :
View answer and explanation
Correct option: B
56
Option B (56) is correct because PA.PB is equal to PT², where PT is the length of the tangent from P to the circle.
138 Easy
138 of 480
Understanding +4 | -1
The curve parametrically described by the equations x=2+3cos⁡θ,y=3+3sin⁡θx=2+3 \cos \theta, y=3+3 \sin \theta represents a circle which touches
View answer and explanation
Correct option: A
xx-axis at (2,0)(2,0)
Option A (xx-axis at (2,0)(2,0)) is correct because the circle has center (2,3) and radius 3, touching the x-axis at (2,0).
139 Easy
139 of 480
Application +4 | -1
The radius of the circle x2+y2+4x+6y+13=0{x^2} + {y^2} + 4x + 6y + 13 = 0 is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the equation represents a point circle, where the radius is zero.
140 Easy
140 of 480
Understanding +4 | -1
The centre of the circ le x=2+3cos⁡θ,y=3sin⁡θ−1x = 2 + 3\cos \mathrm{\theta },y = 3\sin \theta - 1 is
View answer and explanation
Correct option: B
(2,−1)\left( {2, - 1} \right)
The correct answer is B ((2,−1)\left( {2, - 1} \right)) because the parametric equations directly provide the center of the circle in the form x=h+rcos⁡θx = h + r\cos{\theta}, y=k+rsin⁡θy = k + r\sin{\theta}, where (h,k)(h, k) is the center.
141 Easy
141 of 480
Understanding +4 | -1
Assertion: Centre of the circle x2+y2−6x+4y−12=0x^2+y^2-6 x+4 y-12=0 is (3,−2)(3,-2).

Reason: The coordinates of the centre of the circle x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 are (−12 coefficient of x,−12 coefficient of y)\left(-\frac{1}{2} \text{ coefficient of } x,-\frac{1}{2} \text{ coefficient of } y\right)
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A is correct because the assertion is correct, the reason is correct, and the reason correctly explains the assertion. The center of the circle can be found using the coefficients of x and y.
142 Easy
142 of 480
Application +4 | -1
If the circle x2+y2−6x−8y+(25−a2)=0x^{2}+y^{2}-6 x-8 y+\left(25-a^{2}\right)=0 touches the axis of yy, then aa equals.
View answer and explanation
Correct option: D
±3\pm 3
The correct answer is D (±3\pm 3) because if the circle touches the y-axis, the absolute value of the x-coordinate of the center equals the radius.
143 Easy
143 of 480
Application +4 | -1
The equation of the circle of radius 5 and touching the coordinate axes in third quadrant, is
View answer and explanation
Correct option: D
(x+5)2+(y+5)2=25(x+5)^2+(y+5)^2=25
Option D ((x+5)2+(y+5)2=25(x+5)^2+(y+5)^2=25) is correct because the center is at (-5, -5) and the radius is 5, touching the axes in the third quadrant.
144 Easy
144 of 480
Analyzing +4 | -1
The maximum number of points with rational coordinates on a circle whose centre is (3,0)\left( {\sqrt 3 ,0} \right) is
View answer and explanation
Correct option: B
Two
Option B (Two) is correct because a circle centered at (3,0)(\sqrt{3}, 0) can have at most two points with rational coordinates.
145 Easy
145 of 480
Application +4 | -1
C1C_1 is a circle with centre at the origin and radius equal to rr and C2C_2 is a circle with centre at (3r,0)(3 r, 0) and radius equal to 2r2 r. The number of common tangents that can be drawn to the two circles are
View answer and explanation
Correct option: C
3
Option C (3) is correct because the circles touch each other externally, resulting in three common tangents.
146 Easy
146 of 480
Application +4 | -1
If (3,−2)\left( {3, - 2} \right) is the centre of a circle and 4x+3y+19=04x + 3y + 19 = 0 is a tangent to the circle, then the equation of the circle is
View answer and explanation
Correct option: C
x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0
Option C (x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0) is correct because it represents the circle with the given center and tangent.
147 Easy
147 of 480
Understanding +4 | -1
The equation (x−x1)(x−x2)+(y−y1)(y−y2)=0\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0 represents a circle whose centre is
View answer and explanation
Correct option: B
(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)
Option B is correct because the equation represents a circle with the given line segment as a diameter, and the center is the midpoint of that diameter.
148 Easy
148 of 480
Application +4 | -1
The image of circle x2+y2−2x+4y−4=0x^2+y^2-2 x+4 y-4=0 with respect to line x−y+5=0x-y+5=0 is
View answer and explanation
Correct option: A
x2+y2+14x−12y+76=0x^2+y^2+14 x-12 y+76=0
Option A (x2+y2+14x−12y+76=0x^2+y^2+14 x-12 y+76=0) is correct. The image of the circle is found by finding the image of the center with respect to the line and keeping the same radius.
149 Easy
149 of 480
Application +4 | -1
The tangent at (1,7)\left( {1,7} \right) to the curve x2=y−6{x^2} = y - 6 touches the circle x2+y2+16x+12y+c=0{x^2} + {y^2} + 16x + 12y + c = 0 at
View answer and explanation
Correct option: D
(−6,−7)\left( { - 6, - 7} \right)
Option D ((-6, -7)) is correct because the tangent to the curve at (1,7) touches the circle at (-6,-7) after solving the equations of the tangent and the circle.
150 Easy
150 of 480
Application +4 | -1
The length of the chord of the circle x2+y2+4x−3y+2=0{x^2} + {y^2} + 4x - 3y + 2 = 0 along the yy-axis, is
View answer and explanation
Correct option: A
1
The correct answer is A (1) because it represents the length of the chord of the circle along the y-axis, calculated using the y-intercept formula.
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