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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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Showing 241-270 of 685 questions

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241 Easy
241 of 685
Application +4 | -1
Equation of a tangent to the circle with centre (2,−1)(2,-1) is 3x+y=03{x}+y=0. If the square of the length of the tangent to the circle from the point (23,17)(23,17) is kk then find k−1520k-1520.
View answer and explanation
Correct option: A
5
Option C (7) is correct because after calculating the radius and using the formula for the length of the tangent, k is found to be 1525, so k-1520 = 5.
242 Easy
242 of 685
Application +4 | -1
The other end of the diameter through the point (−1,1)\left( { - 1,1} \right) on the circle x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0 is
View answer and explanation
Correct option: C
(7,−5)\left( {7, - 5} \right)
Option A (x2+y2−x−y=0{x^2} + {y^2} - x - y = 0) is correct because it represents the equation of the circle with AB as its diameter, where A and B are the intersection points of the line y=x and the given circle.
243 Easy
243 of 685
Application +4 | -1
A circle of radius 5 touches another circle x2+y2−2x−4y−20=0{x^2} + {y^2} - 2x - 4y - 20 = 0 at (5,5)\left( {5,5} \right), then its equation is
View answer and explanation
Correct option: B
x2+y2−18x−16y+120=0{x^2} + {y^2} - 18x - 16y + 120 = 0
Option B (x2+y2−18x−16y+120=0{x^2} + {y^2} - 18x - 16y + 120 = 0) is correct because the point (5,5) lies on this circle and it has a radius of 5, touching the given circle at (5,5).
244 Easy
244 of 685
Application +4 | -1
The locus of a point from which the length of tangents to the circles x2+y2=4x^2+y^2=4 and 2(x2+y2)−10x+3y−2=02\left(x^2+y^2\right)-10 x+3 y-2=0 are equal is
View answer and explanation
Correct option: D
A straight line perpendicular to the line joining the centres of the circles
Option D (A straight line perpendicular to the line joining the centres of the circles) is correct because the locus of points from which the lengths of tangents to two circles are equal is the radical axis, which is a straight line perpendicular to the line joining the centers of the circles.
245 Easy
245 of 685
Application +4 | -1
If the locus of a point which moves so that the line joining the points of contact of the tangents drawn from it to the circle x2+y2=b2x^2+y^2=b^2 touches the circle x2+y2x^2+y^2 =a2=a^2, is the circle x2+y2=c2x^2+y^2=c^2, then a,b,ca, b, c are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because the condition for the chord of contact to touch another circle leads to a geometric progression relationship between a, b, and c.
246 Easy
246 of 685
Analyzing +4 | -1
The locus of the point, the chord of contact of tangents from which to the circle x2+y2=a2x^2+y^2=a^2 subtends a right angle at the centre is a circle of radius
View answer and explanation
Correct option: C
2a\sqrt{2} a
Option C (2a\sqrt{2} a) is correct because the locus of the point is a circle with radius 2a\sqrt{2}a, derived by using the chord of contact and the condition that it subtends a right angle at the center.
247 Easy
247 of 685
Analyzing +4 | -1
The chord of contact of tangents from 3 points A,B,CA, B, C to the circle x2+y2=a2x^2+y^2=a^2 are concurrent, then A,B,CA, B, C will
View answer and explanation
Correct option: B
be collinear
Option B (be collinear) is correct because if the chords of contact from three points to a circle are concurrent, the points must be collinear, as shown by the determinant condition.
248 Medium
248 of 685
Application +4 | -1
The equation of the line which is tangent to both the circle x2+y2=5{x^2} + {y^2} = 5 and the parabola y2=40x{y^2} = 40x is
View answer and explanation
Correct option: A
2x−y±5=02x - y \pm 5 = 0
Option A (2x−y±5=02x - y \pm 5 = 0) is correct because it represents the tangent line that satisfies the tangency conditions for both the circle and the parabola.
249 Easy
249 of 685
Application +4 | -1
If the equation px2+(2−q)xy+3y2p x^2+(2-q) x y+3 y^2 −6qx+30y+6q=0-6 q x+30 y+6 q=0 represents a circle, then the values of pp and qq are
View answer and explanation
Correct option: B
p=3,q=2p=3, q=2
Option B (p=3,q=2p=3, q=2) is correct because for the given equation to represent a circle, the coefficient of x2x^2 must equal the coefficient of y2y^2, and the coefficient of xyxy must be zero.
250 Medium
250 of 685
Application +4 | -1
If the line y=7x−25y = 7x - 25 meets the circle x2+y2=25{x^2} + {y^2} = 25 in the points A,  BA,\;B, then the distance between AA and BB is
View answer and explanation
Correct option: C
525\sqrt 2
Option C (525\sqrt 2) is correct because the distance between the intersection points A and B is calculated using the distance formula.
251 Medium
251 of 685
Application +4 | -1
If y=f(x)=ax+by=f(x)=a x+b is a tangent to circle x2+y2+2x+2y−2=0x^2+y^2+2 x+2 y-2=0 then the value of (a+b)2+2(a−b)(a+b+1)(a+b)^2+2(a-b)(a+b+1) is equal to
View answer and explanation
Correct option: D
-3
Option D (-3) is correct because by substituting the equation of the line into the circle's equation and setting the discriminant to zero (tangency condition), we can simplify the expression to -3.
252 Easy
252 of 685
Application +4 | -1
A circle touches the line y=xy=x at (4,4)(4,4) and makes an intercept of length 6 units on the line x+y=0x+y=0. One possible coordinates of its centre are
View answer and explanation
Correct option: C
(4−52,4+52)\left(4-\frac{5}{\sqrt{2}}, 4+\frac{5}{\sqrt{2}}\right)
Option C ((4−52,4+52)\left(4-\frac{5}{\sqrt{2}}, 4+\frac{5}{\sqrt{2}}\right)) is correct because it represents one of the possible centers of the circle based on the given conditions.
253 Medium
253 of 685
Analyzing +4 | -1
The intercept on the line y=xy = x by the circle x2+y2−2x=0{x^2} + {y^2} - 2x = 0 is AB.AB. The equation of the circle with ABAB as diameter is
View answer and explanation
Correct option: B
x2+y2=x+y{x^2} + {y^2} = x + y
Option B (x2+y2=x+y{x^2} + {y^2} = x + y) is correct because it represents the circle with the intercept AB as its diameter, found by substituting the intersection points into the general circle equation.
254 Easy
254 of 685
Analyzing +4 | -1
The two circles x2+y2−2x−4y=0x^2+y^2-2 x-4 y=0 and x2+y2−8y−4=0x^2+y^2-8 y-4=0
View answer and explanation
Correct option: B
touch each other internally
Option B (touch each other internally) is correct because the distance between the centers is equal to the difference of the radii, indicating internal tangency.
255 Easy
255 of 685
Application +4 | -1
If the line 3x−4y−k=0,  (k>0)3x - 4y - k = 0,\;(k > 0) touches the circle x2+y2−4x−8y−5=0{x^2} + {y^2} - 4x - 8y - 5 = 0 at (a,  b)\left( {a,\;b} \right), then k+a+bk + a + b is equal to
View answer and explanation
Correct option: A
20
Option A (20) is correct because it is found by determining the value of k using the tangency condition, then finding the point of tangency (a, b) and summing k + a + b.
256 Easy
256 of 685
Application +4 | -1
If a circle C\mathrm{C}, whose radius is 3 , touches externally the circle, x2+y2+2x−4y−4=0x^{2}+y^{2}+2 x-4 y-4=0 at the point (2,2)(2,2), then the length of the intercept cut by this circle cc, on the xx-axis is equal to
View answer and explanation
Correct option: D
252 \sqrt{5}
Option D (252 \sqrt{5}) is correct because the length of the intercept cut by the circle C on the x-axis is calculated to be 252 \sqrt{5} after finding the equation of the circle.
257 Easy
257 of 685
Application +4 | -1
If the length of the tangent from any point on the circle (x−3)2+(y−2)2=5r2{\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} = 5{r^2} to the circle (x−3)2+(y+2)2=r2{\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} = {r^2} is 16 units, then the area between the two circles in sq units is
View answer and explanation
Correct option: D
256π256\pi
Option D (256π256\pi) is correct because it is the area between the two circles calculated using the given tangent length.
258 Medium
258 of 685
Application +4 | -1
If two vertices of an equilateral triangle are A(−a,0)\mathrm{A}(-a, 0) and B(a,0),a>0B(a, 0), a>0, and the third vertex C\mathrm{C} lies above xx-axis then the equation of the circumcircle of △ABC\triangle \mathrm{ABC} is :
View answer and explanation
Correct option: A
3x2+3y2−23ay=3a23 x^{2}+3 y^{2}-2 \sqrt{3} a y=3 a^{2}
Option A (3x2+3y2−23ay=3a23 x^{2}+3 y^{2}-2 \sqrt{3} a y=3 a^{2}) is correct because it represents the circumcircle equation derived from the properties of an equilateral triangle and the general circle equation.
259 Easy
259 of 685
Application +4 | -1
If the lines 2x−3y=52x - 3y = 5 and 3x−4y=73x - 4y = 7 are two diameters of a circle of radius 7, then 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 equation is found using the center and radius.
260 Easy
260 of 685
Application +4 | -1
If the angle between tangents drawn to x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 from (0,0)(0,0) is π2\frac{\pi}{2}, then
View answer and explanation
Correct option: B
g2+f2=2cg^2+f^2=2 c
Option B (g2+f2=2cg^2+f^2=2 c) is correct because the origin lies on the director circle of the given circle, which leads to this condition.
261 Easy
261 of 685
Application +4 | -1
Two circles of equal radius of 5 units have their centres at the origin and the point (2,−3)(2,-3). Equation of the common chord of these circles is
View answer and explanation
Correct option: A
4x−6y−13=04 x-6 y-13=0
Option A (4x−6y−13=04 x-6 y-13=0) is correct because it represents the equation of the common chord obtained by subtracting the equations of the two circles.
262 Easy
262 of 685
Application +4 | -1
The line (x−1)cos⁡θ+(y−1)sin⁡θ=1(x-1) \cos \theta+(y-1) \sin \theta=1, for all values of θ\theta touches the circle
View answer and explanation
Correct option: D
x2+y2−2x−2y+1=0x^2+y^2-2 x-2 y+1=0
Option D (x2+y2−2x−2y+1=0x^2+y^2-2 x-2 y+1=0) is correct because the given line is tangent to the circle (x−1)2+(y−1)2=1(x-1)^2 + (y-1)^2 = 1, which simplifies to the given equation.
263 Easy
263 of 685
Application +4 | -1
The circle passing through (1,−2)(1,-2) and touching the axis of xx at (3,0)(3,0) also passes through the point
View answer and explanation
Correct option: C
(5,−2)(5,-2)
Option C ((5,-2)) is correct because the equation of the circle is found using the given conditions, and substituting (5,-2) into the equation satisfies it.
264 Medium
264 of 685
Application +4 | -1
The number of tangents to the circle x2+y2−2x+4y=0x^2+y^2-2 x+4 y=0 through the point (−1,2)(-1,2) is
View answer and explanation
Correct option: B
22
Option D (\a=±2ba= \pm \sqrt{{2}} b\)iscorrect.Byfindingtheslopesofthemediansandsettingtheirproductto−1(sincetheyareperpendicular),therelationshipbetweenaandbisfoundtobe) is correct. By finding the slopes of the medians and setting their product to -1 (since they are perpendicular), the relationship between a and b is found to bea= ± 2\sqrt{{2}} b.
265 Medium
265 of 685
Application +4 | -1
PQPQ is a chord of the circle x2+y2−2x−8=0{x^2} + {y^2} - 2x - 8 = 0 whose midpoint is (2,2)\left( {2,2} \right). The circle passing through P,QP,Q and (1,2)\left( {1,2} \right) is
View answer and explanation
Correct option: B
x2+y2−7x−10y+22=0{x^2} + {y^2} - 7x - 10y + 22 = 0
Option B (x2+y2−7x−10y+22=0{x^2} + {y^2} - 7x - 10y + 22 = 0) is correct because it represents a circle passing through P, Q, and (1, 2), derived using the chord equation and circle family concept.
266 Easy
266 of 685
Application +4 | -1
If the circle x2+y2−2gx+6y−19c=0,g,c∈Rx^2+y^2-2 g x+6 y-19 c=0, g, c \in \mathbb{R} passes through the point (6,1)(6,1) and its centre lies on the line x−2cy=8x-2 c y=8, then the length of intercept made by the circle on xx-axis is
View answer and explanation
Correct option: D
2232 \sqrt{23}
Option D (2232 \sqrt{23}) is correct because it is the length of the intercept made by the circle on the x-axis, calculated after finding the values of g and c.
267 Easy
267 of 685
Analyzing +4 | -1
The centre of a set of circles, each of radius 3 , lies on the circles x2+y2=25x^2+y^2=25. The locus of any point in the set is
View answer and explanation
Correct option: A
4≤x2+y2≤644 \leq x^2+y^2 \leq 64
Option A (4≤x2+y2≤644 \leq x^2+y^2 \leq 64) is correct because the locus represents the region between two circles centered at the origin with radii 2 and 8, respectively, due to the radius of the moving circle being 3 and the fixed circle having a radius of 5.
268 Easy
268 of 685
Application +4 | -1
Let C1C_1 be the graph of the curve represented by the equation x2−13x+4y=1x^2-13 x+4 y=1. If C2C_2 is the new curve obtained when C1C_1 is reflected in the origin, then the equation of C2C_2 is
View answer and explanation
Correct option: D
x2+13x−4y=1x^2+13 x-4 y=1
Option D (x2+13x−4y=1x^2+13 x-4 y=1) is correct. Reflecting a curve in the origin involves replacing x with -x and y with -y in the original equation.
269 Easy
269 of 685
Application +4 | -1
If a circle touches yy-axis at (0,4)(0,4) and passes through (2,0)(2,0), then which of the following can be the tangent to the circle?
View answer and explanation
Correct option: A
3x+4y−6=03 x+4 y-6=0
Option A (3x+4y−6=03x + 4y - 6 = 0) is correct because the perpendicular distance from the center of the circle to this line equals the radius, indicating it's a tangent.
270 Medium
270 of 685
Application +4 | -1
The equation of the circle which passes through the points of intersection of the circles x2+y2−6x=0{x^2} + {y^2} - 6x = 0 and x2+y2−6y=0{x^2} + {y^2} - 6y = 0 and has its centre at (32,32)\left( {\frac{3}{2},\frac{3}{2}} \right), is
View answer and explanation
Correct option: C
x2+y2−3x−3y=0{x^2} + {y^2} - 3x - 3y = 0
Option C (x2+y2−3x−3y=0{x^2} + {y^2} - 3x - 3y = 0) is correct because it represents the equation of the circle passing through the intersection points of the given circles and having its center at (3/2, 3/2).
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