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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 301-330 of 480 questions

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301 Easy
301 of 480
Application +4 | -1
If a chord of the circle x2+y2=32x^2+y^2=32 makes equal intercepts of length ll on the coordinate axes, then
View answer and explanation
Correct option: A
l∈(−8,8)l \in(-8,8)
The correct answer is l∈(−8,8)l \in(-8,8) because the length of the perpendicular from the center to the chord must be less than the radius.
302 Easy
302 of 480
Analyzing +4 | -1
Centre of circle whose normal’s are x2−2xy−3x+6y=0{x^2} - 2xy - 3x + 6y = 0, is
View answer and explanation
Correct option: A
(3,32)\left( {3,\frac{3}{2}} \right)
Option A (3,32)\left( {3,\frac{3}{2}} \right) is correct because the center of the circle is the intersection point of its normals.
303 Easy
303 of 480
Analyzing +4 | -1
Centre of circle whose normals are, x2−2xy+3x+6y=0{x^2} - 2xy + 3x + 6y = 0, is
View answer and explanation
Correct option: A
(3,32)\left( {3,\frac{3}{2}} \right)
Option A (3,32)\left( {3,\frac{3}{2}} \right) is correct because the normals of a circle pass through its center, and the intersection of the given normals gives the center.
304 Easy
304 of 480
Application +4 | -1
CC the centre of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. The tangents at any point PP on this hyperbola meets the straight lines bx−ay=0b x-a y=0 and bx+ay=0b x+a y=0 in the points QQ and RR respectively. Then CQ⋅CR=C Q \cdot C R=
View answer and explanation
Correct option: A
a2+b2a^2+b^2
Option A (a2+b2a^2+b^2) is correct because the product of the distances CQ and CR simplifies to a2+b2a^2 + b^2 using the equation of the tangent to the hyperbola.
305 Easy
305 of 480
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.
306 Easy
306 of 480
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.
307 Easy
307 of 480
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.
308 Easy
308 of 480
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.
309 Easy
309 of 480
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.
310 Easy
310 of 480
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.
311 Easy
311 of 480
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.
312 Easy
312 of 480
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.
313 Easy
313 of 480
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.
314 Easy
314 of 480
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.
315 Easy
315 of 480
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.
316 Easy
316 of 480
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.
317 Easy
317 of 480
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.
318 Easy
318 of 480
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.
319 Easy
319 of 480
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.
320 Easy
320 of 480
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.
321 Easy
321 of 480
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.
322 Easy
322 of 480
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.
323 Easy
323 of 480
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.
324 Easy
324 of 480
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.
325 Easy
325 of 480
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.
326 Easy
326 of 480
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.
327 Easy
327 of 480
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.
328 Easy
328 of 480
Application +4 | -1
If g2+f2=c{\mathrm{g}^2} + {f^2} = c, then the equation x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 will represent
View answer and explanation
Correct option: D
A circle of radius zero
Option D (A circle of radius zero) is correct because the radius of the circle is zero when g2+f2=c{\mathrm{g}^2} + {f^2} = c.
329 Easy
329 of 480
Application +4 | -1
The area (in square unit) of the circle which touches the lines 4x+3y=154x + 3y = 15 and 4x+3y=54x + 3y = 5 is
View answer and explanation
Correct option: D
π\pi
Option D (π\pi) is correct because the area of the circle is calculated using the radius, which is half the distance between the two parallel lines.
330 Easy
330 of 480
Analyzing +4 | -1
If the circles x2+y2=4x^2+y^2=4 and x2+y2−8x+12=0x^2+y^2-8 x+12=0 touch each other, then the equation of their common tangent at the point, where they touch, is given by
View answer and explanation
Correct option: C
x=2x=2
Option C (x=2x=2) is correct because the equation of the common tangent is found by subtracting the equations of the two circles, which touch each other externally.
Revision notes and key takeaways

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