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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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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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Showing 631-660 of 685 questions

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631 Easy
631 of 685
Application +4 | -1
Two circles x2+y2−2x−3=0{x^2} + {y^2} - 2x - 3 = 0 and x2+y2−4x−6y−8=0{x^2} + {y^2} - 4x - 6y - 8 = 0 are such that
View answer and explanation
Correct option: C
They intersect at two points
Option C (They intersect at two points) is correct because the distance between the centers of the circles is less than the sum of their radii.
632 Easy
632 of 685
Application +4 | -1
If the area of the circle 4x2+4y2−8x+16y+k=04{x^2} + 4{y^2} - 8x + 16y + k = 0 is 9π9\pi sq unit, then the value of kk is
View answer and explanation
Correct option: C
−16- 16 units
Option C (−16- 16 units) is correct because it is the value of k that makes the area of the circle equal to 9π9\pi sq units.
633 Easy
633 of 685
Application +4 | -1
If the abscissa and the ordinates of two points AA and BB are the roots of the equations x2+2ax−b2=0x^2+2 a x-b^2=0 and y2+2py−q2=0y^2+2 p y-q^2=0 respectively.
Then, the equation of circle with ABA B as diameter is
View answer and explanation
Correct option: A
x2+y2+2ax+2py−b2−q2=0x^2+y^2+2 a x+2 p y-b^2-q^2=0
Option A (x2+y2+2ax+2py−b2−q2=0x^2+y^2+2 a x+2 p y-b^2-q^2=0) is correct because it represents the equation of a circle with AB as diameter, derived using the roots of the given quadratic equations and the diameter form of the circle equation.
634 Medium
634 of 685
Analyzing +4 | -1
Suppose a circle passes through (2,2)\left( {2,2} \right) and (9,9)\left( {9,9} \right) and touches the xx-axis at P. If OO is the origin, then OPOP is equal to
View answer and explanation
Correct option: C
6
Option C (6) is correct because the x-coordinate of the point where the circle touches the x-axis is the geometric mean of the x-coordinates of the other two points on the circle.
635 Easy
635 of 685
Application +4 | -1
Find the radius of the circle whose centre is in fourth quadrant and which touches as each of the three lines x=0x=0, y=0y=0 and 3x+4y−12=0.3 x+4 y-12=0.
View answer and explanation
Correct option: A
3
Option A (3) is correct because the radius of the circle is found by using the distance from the center to the line and solving the resulting equation.
636 Easy
636 of 685
Application +4 | -1
Find the parametric equations of the circle x2+y2−4x−2y+x^2+y^2-4 x-2 y+ 1=01=0.
View answer and explanation
Correct option: A
x=2+2cos⁡θ;y=1+2sin⁡θx=2+2 \cos \theta ; y=1+2 \sin \theta
Option A (x=2+2cos⁡θ;y=1+2sin⁡θx=2+2 \cos \theta ; y=1+2 \sin \theta) is correct because it represents the parametric equations of the circle with center (2, 1) and radius 2.
637 Easy
637 of 685
Application +4 | -1
The equation of the tangent to the circle x2+y2=4{x^2} + {y^2} = 4, which are parallel to x+2y+3=0x + 2y + 3 = 0, are
View answer and explanation
Correct option: C
x+2y=±25x + 2y = \pm 2\sqrt 5
Option C (x + 2y = ±25\pm 2\sqrt 5) is correct because it is the result of applying the condition for tangency (c2 = a2(1+m2)) to find the correct tangent lines parallel to the given line.
638 Medium
638 of 685
Analyzing +4 | -1
The distinct points A(0,  0),  B(0,  1),  C(1,  0)A\left( {0,\;0} \right),\;B\left( {0,\;1} \right),\;C\left( {1,\;0} \right) and D(2a,  3a)D\left( {2a,\;3a} \right) are concyclic, then
View answer and explanation
Correct option: A
aa' can attain only rational values
The correct answer is A. Since the points A, B, C, and D are concyclic and A, B, and C form a right-angled triangle, the circle's equation can be found, and the condition for D to lie on it leads to rational values for 'a'.
639 Easy
639 of 685
Application +4 | -1
If the equation of the circle is x2+y2−8x+10y−12=0x^2+y^2-8 x+10 y-12=0, then

I. Centre of the circle is (4,−5)(4,-5).

II. Radius of the circle is 53\sqrt{53}.
View answer and explanation
Correct option: C
Both are true
Option C is correct because both statements are true. The center and radius are correctly derived from the given circle equation by completing the square.
640 Easy
640 of 685
Analyzing +4 | -1
Let a circle C1C_{1} be obtained on rolling the circle x2+x^{2}+ y2−4x−6y+11=0y^{2}-4 x-6 y+11=0 upwards 4 units on the tangent TT to it at the point (3,2)(3,2). Let C2C_{2} be the image of C1C_{1} in TT. Let AA and BB be the centers of circles C1C_{1} and C2C_{2} respectively, and MM and NN be respectively the feet of perpendiculars drawn from AA and BB on the xx-axis. Then the area of the trapezium AMNB is:
View answer and explanation
Correct option: C
4(1+2)4(1+\sqrt{2})
Option C (4(1+2)4(1+\sqrt{2})) is correct because the area of the trapezium is calculated by finding the coordinates of the centers of the circles and using the formula for the area of a trapezium.
641 Easy
641 of 685
Application +4 | -1
The locus of a point which moves so that the ratio of the length of the tangents to the circles x2+y2+4x+3=0{x^2} + {y^2} + 4x + 3 = 0 and x2+y2−6x+5=0{x^2} + {y^2} - 6x + 5 = 0 is 2:32:3, is
View answer and explanation
Correct option: C
5x2+5y2+60x+7=05{x^2} + 5{y^2} + 60x + 7 = 0
Option C (5x2+5y2+60x+7=05{x^2} + 5{y^2} + 60x + 7 = 0) is correct because it represents the locus of the point where the ratio of the lengths of tangents to the two circles is 2:3.
642 Easy
642 of 685
Application +4 | -1
Discuss the position of the points (1,2)(1,2) and (6,0)(6,0) with respect to the circle x2+y2−4x+2y−11=0x^2+y^2-4 x+2 y-11=0.
View answer and explanation
Correct option: D
(1,2)(1,2) inside; (6,0)(6,0) outside
Option D ((1,2) inside; (6,0) outside) is correct because substituting the points into the circle's equation shows that (1,2) results in a negative value (inside), while (6,0) results in a positive value (outside).
643 Easy
643 of 685
Application +4 | -1
Equation of a circle with centre C(h,k)C(h, k) and radius 5 such that h2−3h+2=0h^2-3 h+2=0 and k2+5k−6=0k^2+5 k-6=0 is
View answer and explanation
Correct option: C
x2+y2−4x−2y−20=0x^2+y^2-4 x-2 y-20=0
Option D (x2+y2−4x−6y−12=0x^2+y^2-4 x-6 y-12=0) is correct because it represents a circle with radius 5 and a center (h,k) that satisfies the given conditions.
644 Easy
644 of 685
Application +4 | -1
The area of the circle whose centre is at (2,3)\left( {2,3} \right) and passing through (4,6)\left( {4,6} \right), is
View answer and explanation
Correct option: C
13π13\mathrm{\pi } sq units
Option C (13π13{\pi } sq units) is correct because it represents the area of the circle calculated using the distance between the center and the point on the circle as the radius.
645 Easy
645 of 685
Application +4 | -1
The locus of the centres of the circles for which one end of a diameter is (1,1)(1,1) while the other end is on the line x+y=3x+y=3, is
View answer and explanation
Correct option: C
2x+2y=52 x+2 y=5
Option C (2x+2y=52 x+2 y=5) is correct because it represents the locus of the centers of the circles that satisfy the given conditions, derived by eliminating the parameter 't'.
646 Easy
646 of 685
Application +4 | -1
The line 3x−2y=k3x - 2y = k meets the circle x2+y2=4r2{x^2} + {y^2} = 4{r^2} at only one point if k2{k^2} is equal to
View answer and explanation
Correct option: A
52r252{r^2}
Option A (52r252r^2) is correct because it satisfies the tangency condition for a line and a circle, where the distance from the center of the circle to the line equals the radius.
647 Easy
647 of 685
Application +4 | -1
The equation of the circle which touches the axes of the coordinates and the line x3+y4=1\frac{x}{3} + \frac{y}{4} = 1 and whose centre lies in the first quadrant is x2+y2−2cx−2cy+c2=0,{x^2} + {y^2} - 2cx - 2cy + {c^2} = 0, where cc is
View answer and explanation
Correct option: A
1,61,6
Option A (1,61,6) is correct because these are the two possible values of the radius of the circle that touches both axes and the given line.
648 Easy
648 of 685
Application +4 | -1
Find the equation of the circle which passes through the point of intersection of the lines 3x−2y−1=03 x-2 y-1=0 and 4x+y−27=04 x+y-27=0 and whose centre is (2,−3)(2,-3).
View answer and explanation
Correct option: C
109\sqrt{109}
Option C (109\sqrt{109}) is correct because it is the distance between the center of the circle and the point of intersection of the given lines, which equals the radius.
649 Easy
649 of 685
Application +4 | -1
The intercept on the line y=xy=x by the circle x2+y2−2x=0x^2+y^2-2 x=0 is ABA B. The equation of the circle with ABA B as a diameter, is
View answer and explanation
Correct option: B
x2+y2=x+yx^2+y^2=x+y
Option B (x2+y2=x+yx^2+y^2=x+y) is correct because it represents the equation of the circle with AB as a diameter, where A and B are the intersection points of the given circle and the line y=x.
650 Easy
650 of 685
Analyzing +4 | -1
If the straight line ax+by=2;a,b≠0a x+b y=2 ; a, b \neq 0 touches the circle x2+y2−2x=3x^2+y^2-2 x=3 and is normal to circle x2+y2−4y=6x^2+y^2-4 y=6, then the values of aa and bb are
View answer and explanation
Correct option: C
a=−43,b=1a=\frac{-4}{3}, b=1
Option C (a=−43,b=1a=\frac{{-4}}{{3}}, b=1) is correct because by using the conditions of tangency and normality, we can solve for a and b.
651 Easy
651 of 685
Analyzing +4 | -1
The equation of a circle CC is x2+y2−6c−8y−11=0.{x^2} + {y^2} - 6c - 8y - 11 = 0. The number of real points at which the circle drawn with points (1,8)\left( {1,8} \right) and (0,0)\left( {0,0} \right) as the ends of a diameter cuts the circle, CC is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the condition r1−r2<C1C2<r1+r2{r_1} - {r_2} < {C_1}{C_2} < {r_1} + {r_2} is satisfied, indicating that the circles intersect at two points.
652 Easy
652 of 685
Analyzing +4 | -1
If the line y=mx−(m−1)y=m x-(m-1) cuts the circle x2+y2=4x^2+y^2=4 at two real and distinct points, then
View answer and explanation
Correct option: D
m∈Rm \in R
Option D (m∈Rm \in R) is correct because the condition for the line to intersect the circle at two distinct points holds for all real values of m.
653 Easy
653 of 685
Application +4 | -1
Two parallel chords of a circle SS have length 10 and 14 and are 6 units apart. If a regular polygon of 12 sides is inscribed in a circle SS, then find the area of regular polygon.
View answer and explanation
Correct option: A
150
Option A (150) is correct because the area of the regular polygon is calculated using the given chord lengths and their distance apart to find the circle's radius, then applying the area formula for a regular polygon.
654 Easy
654 of 685
Application +4 | -1
If a>2b>0a > 2b > 0, then the positive value of mm for which y=mx−b1+m2y = mx - b\sqrt {1 + {m^2}} is a common tangent to x2+y2=b2{x^2} + {y^2} = {b^2} and (x−a)2+y2=b2{\left( {x - a} \right)^2} + {y^2} = {b^2}, is
View answer and explanation
Correct option: A
2ba2−4b2\frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}
Option A (2ba2−4b2\frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}) is correct because it is derived from the tangency condition of a line to a circle.
655 Easy
655 of 685
Application +4 | -1
The circle x2+y2=4x+8y+5x^2+y^{2}=4 x+8 y+5 intersects the line 3x−4y=m3 x-4 y=m at two distinct points if
View answer and explanation
Correct option: A
−35<m<15- 35 < m < 15
Option A (−35<m<15- 35 < m < 15) is correct because it satisfies the condition for the line to intersect the circle at two distinct points.
656 Easy
656 of 685
Analyzing +4 | -1
The number of common tangents to the two circles x2+y2−8x+2y=0{x^2} + {y^2} - 8x + 2y = 0 and x2+y2−2x−16y+25=0{x^2} + {y^2} - 2x - 16y + 25 = 0 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the distance between the centers of the two circles is less than the sum of their radii, indicating that they intersect, and thus have two common tangents.
657 Easy
657 of 685
Application +4 | -1
The equation of the circle with centre on the xx-axis and touching the line 3x+4y−11=03 x+4 y-11=0 at point (1,2)(1,2) is
View answer and explanation
Correct option: C
x2+y2+x−6=0x^2+y^2+x-6=0
Option D (None of these) is incorrect; the correct answer is C (x2+y2+x−6=0x^2+y^2+x-6=0) which is found by using the given point and the condition that the center lies on the x-axis to solve for the circle's equation.
658 Medium
658 of 685
Application +4 | -1
A circle has two of its diameters along the lines 2x+2 x+ 3y−18=03 y-18=0 and 3x−y−5=03 x-y-5=0 and touches the line x+2y+4=0x+2 y+4=0. Equation of the circle is
View answer and explanation
Correct option: D
x2+y2−6x−8y+16=0x^2+y^2-6 x-8 y+16=0
Option D (x2+y2−6x−8y+16=0x^2+y^2-6 x-8 y+16=0) is correct because it represents a circle with center (3,4) and radius 3, which satisfies the given conditions.
659 Easy
659 of 685
Application +4 | -1
If 5x−12y+10=05 x-12 y+10=0 and 12y−5x+16=012 y-5 x+16=0 are two tangents to a circle, then the radius the circle, is
View answer and explanation
Correct option: A
1
The correct answer is 1 because the radius is half the distance between the two parallel tangent lines.
660 Hard
660 of 685
Application +4 | -1
The equation of the circle which passes through the intersection of x2+y2+13x−3y=0{x^2} + {y^2} + 13x - 3y = 0 and 2x2+2y2+4x−7y−25=02{x^2} + 2{y^2} + 4x - 7y - 25 = 0 and whose centre lies on 13x+30y=013x + 30y = 0, is
View answer and explanation
Correct option: B
4x2+4y2+30x−13y−25=04{x^2} + 4{y^2} + 30x - 13y - 25 = 0
Option C (4x2+4y2+30x−13y−25=04{x^2} + 4{y^2} + 30x - 13y - 25 = 0) is correct because it represents the equation of a circle passing through the intersection of two given circles and whose center lies on the given line.
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