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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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511 Easy
511 of 685
Application +4 | -1
The range of values of a for which the point (a,4)\left( {a,4} \right) is outside the circles x2+y2+10x=0{x^2} + {y^2} + 10x = 0 and x2+y2−12x+20=0,{x^2} + {y^2} - 12x + 20 = 0, is
View answer and explanation
Correct option: A
(−∞,−8)∪(−2,6)∪(6,∞)\left( { - \infty , - 8} \right) \cup \left( { - 2,6} \right) \cup \left( {6,\infty } \right)
Option A ((−∞,−8)∪(−2,6)∪(6,∞)\left( { - \infty , - 8} \right) \cup \left( { - 2,6} \right) \cup \left( {6,\infty } \right)) is correct because it satisfies the conditions for the point to be outside both circles.
512 Easy
512 of 685
Application +4 | -1
The coordinates of the centre and radius of the circle represented by the equation (3−2λ)x2+λy2−4x+2y−4=0  are(3 - 2\lambda ){x^2} + \lambda {y^2} - 4x + 2y - 4 = 0\,\,\mathrm{ are }
View answer and explanation
Correct option: C
(2,−1),3(2,-1), 3
The correct answer is C ((2,-1), 3) because when λ\lambda is 1, the equation represents a circle with center (2, -1) and radius 3.
513 Easy
513 of 685
Application +4 | -1
A circle SS of radius ' aa ' is the director circle of another circle S1S_1. S1S_1 is the director circle of circle S2S_2 and so on. If the sum of the radii of all these circles is 2 , then the value of aa is
View answer and explanation
Correct option: B
2−22-\sqrt{2}
Option B (2−22-\sqrt{2}) is correct because the sum of the radii of the director circles forms an infinite geometric progression.
514 Easy
514 of 685
Application +4 | -1
The radius of the circle 3x2+by2+4bx−6by+b2=03 x^2+b y^2+4 b x-6 b y+b^2=0 is
View answer and explanation
Correct option: C
10\sqrt{10}
Option C (10\sqrt{10}) is correct because after normalizing the equation of the circle, the radius can be calculated using the formula g2+f2−c\sqrt{g^2 + f^2 - c}.
515 Easy
515 of 685
Application +4 | -1
The locus of the mid-points of the chords of the circle x2+y2=16{x^2} + {y^2} = 16 which are tangents to the hyperbola 9x2−16y2=1449{x^2} - 16{y^2} = 144 is
View answer and explanation
Correct option: A
(x2+y2)2=16x2−9y2{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}
Option A ((x2+y2)2=16x2−9y2{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}) is correct because it is the locus of the mid-points of the chords, derived by using the tangent condition of the hyperbola.
516 Easy
516 of 685
Application +4 | -1
The radius of the circle 2x2+2y2−4xcos⁡θ+4ysin⁡θ−1−4cos⁡θ−cos⁡2θ=02 x^2+2 y^2-4 x \cos \theta+4 y \sin \theta-1-4 \cos \theta-\cos 2 \theta=0 is
View answer and explanation
Correct option: B
1+cos⁡θ1+\cos \theta
Option B (1+cos⁡θ1+\cos \theta) is correct because after completing the square and simplifying, the radius of the circle is found to be 1+cos⁡θ1+\cos \theta.
517 Easy
517 of 685
Analyzing +4 | -1
Consider the two curves C1:y2=4x{C_1}:{y^2} = 4x
C2:x2+y2−6x+1=0,{C_2}:{x^2} + {y^2} - 6x + 1 = 0, then
View answer and explanation
Correct option: B
C1{C_1} and C2{C_2} touch each other exactly at two points
Option B (C1{C_1} and C2{C_2} touch each other exactly at two points) is correct because solving the equations simultaneously yields two identical x-values, indicating tangency at two points.
518 Medium
518 of 685
Application +4 | -1
Consider the circles x2+y2=9x^2+y^2=9 and x2+y2−10x+9=0x^2+y^2-10 x+9=0, then find the length of the common chord of circle is
View answer and explanation
Correct option: C
245\frac{24}{5}
Option C (245\frac{{24}}{{5}}) is correct because the length of the common chord is calculated using the radical axis and the distance from the center to the radical axis.
519 Easy
519 of 685
Application +4 | -1
The intercepts on the line y=xy = x by the circle x2+y2−2x=0{x^2} + {y^2} - 2x = 0 is ABAB. Equation of the circle on ABAB as a diameter is
View answer and explanation
Correct option: A
x2+y2−x−y=0{x^2} + {y^2} - x - y = 0
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 a diameter, where A and B are the intersection points of the given line and circle.
520 Easy
520 of 685
Understanding +4 | -1
A point P moves in such a way that the ratio of its distance from two coplanar points is always a fixed number (≠1)\left( { \ne 1} \right). Then, its locus is a
View answer and explanation
Correct option: B
Circle
Option B (Circle) is correct because the locus of a point with a constant ratio of distances from two fixed points (not equal to 1) is a circle.
521 Easy
521 of 685
Application +4 | -1
The radius of the circle passing through the point (6, 2) and two of whose diameters are x+y=6x + y = 6 and x+2y=4x + 2y = 4, is
View answer and explanation
Correct option: D
20\sqrt {20}
Option D (20\sqrt {20}) is correct because the radius of the circle is the distance between the center and the given point (6, 2).
522 Easy
522 of 685
Application +4 | -1
The equation of the smallest circle passing from points (1,1)(1,1) and (2,2)(2,2) and always in the first quadrant, is
View answer and explanation
Correct option: C
x2+y2−3x−3y+4=0x^2+y^2-3 x-3 y+4=0
Option D (x2+y2−3x−3y+4=0x^2+y^2-3 x-3 y+4=0) is correct because the smallest circle passing through (1,1) and (2,2) and always in the first quadrant has the given equation.
523 Medium
523 of 685
Analyzing +4 | -1
If a line, y=mx+cy = mx + c is a tangent to the circle, (x−3)2+y2=1{(x - 3)^2} + {y^2} = 1 and it is perpendicular to a line L1,{L_1}, where L1{L_1} is the tangent to the circle, x2+y2=1{x^2} + {y^2} = 1 at the point (12, 12)\left( {\frac{1}{{\sqrt 2 }},\,\frac{1}{{\sqrt 2 }}} \right); then
View answer and explanation
Correct option: D
c2+6c+7=0{c^2} + 6c + 7 = 0
Option D (c2+6c+7=0{c^2} + 6c + 7 = 0) is correct because it represents the equation derived from the tangency condition of the line to the given circle.
524 Easy
524 of 685
Analyzing +4 | -1
The circles x2+y2−2x−15=0x^2+y^2-2 x-15=0 and x2+y2+4y+3=0x^2+y^2+4 y+3=0 have
View answer and explanation
Correct option: A
no common tangent
Option A (no common tangent) is correct because the distance between the centers of the circles is less than the difference of their radii, meaning one circle lies entirely inside the other.
525 Easy
525 of 685
Application +4 | -1
The locus of the point, the sum of the squares of whose distances from nn fixed points Ai(xi,yi)A_i\left(x_i, y_i\right), i=1,2,…,ni=1,2, \ldots, n is equal to k2k^2 is a circle
View answer and explanation
Correct option: C
with centre at the point of mean position of the given points
Option C (with centre at the point of mean position of the given points) is correct because the locus represents a circle whose center is at the average of the coordinates of the fixed points.

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526 Easy
526 of 685
Analyzing +4 | -1
A point Plies on the circle (x−1)2+(y−1)2=1(x-1)^2+(y-1)^2=1 such that PA2+PB2P A^2+P B^2 is maximum where A(1,4)A(1,4) and B(1,−5)B(1,-5), then P,AP, A, B lie on a
View answer and explanation
Correct option: A
straight line
Option A (straight line) is correct because the condition PA2+PB2P A^2+P B^2 being maximum implies that P, A, and B are collinear on the line x=1.
527 Easy
527 of 685
Application +4 | -1
If the circle x2+y2−10x−14y+24=0{x^2} + {y^2} - 10x - 14y + 24 = 0 cuts an intercepts on yy-axis of length
View answer and explanation
Correct option: B
10
Option B (10) is correct because the length of the intercept on the y-axis is calculated using the formula 2f2−c2\sqrt{{f^2} - c}.
528 Easy
528 of 685
Application +4 | -1
A circle touches both the y-axis and the line x+y=0.x + y = 0. Then the locus of its center is
View answer and explanation
Correct option: D
x2−y2=2xy{x^2} - {y^2} = 2xy
Option D (x2−y2=2xy{x^2} - {y^2} = 2xy) is correct because it represents the locus of the center of a circle that touches both the y-axis and the line x + y = 0, derived by equating the radius to the distance from the center to the line.
529 Easy
529 of 685
Application +4 | -1
The coordinates of the point on the circle x2+y2−12x−4y+30=0,{x^2} + {y^2} - 12x - 4y + 30 = 0, which is farthest from the origin are
View answer and explanation
Correct option: A
(9,3)\left( {9,3} \right)
Option A ((9,3)) is correct because it is the only point that lies on the circle and is farthest from the origin.
530 Easy
530 of 685
Application +4 | -1
Length of the tangents from the point (1,2)\left( {1,2} \right) to the circles x2+y2+x+y−4=0{x^2} + {y^2} + x + y - 4 = 0 and 3x2+3y2−x−y−k=03{x^2} + 3{y^2} - x - y - k = 0 are in the ration 4:34:3, then kk is equal to
View answer and explanation
Correct option: D
394\frac{{39}}{4}
Option D (394\frac{{39}}{4}) is correct because it is the value of k that satisfies the given ratio of tangent lengths from the point (1,2) to the two circles.
531 Easy
531 of 685
Application +4 | -1
If the common chord of the circles x2+(y−b)2=16x^2+(y-b)^2=16 and x2+y2=16x^2+y^2=16 subtends a right angle at the origin, then b=b=
View answer and explanation
Correct option: C
±42\pm 4 \sqrt{2}
Option C (±42\pm 4 \sqrt{2}) is correct because it is the value of b for which the common chord subtends a right angle at the origin.
532 Easy
532 of 685
Understanding +4 | -1
Find the centre and the radius of the circle x=a+c⋅cos⁡θ,y=x=a+c \cdot \cos \theta, y= b+c⋅sin⁡θb+c \cdot \sin \theta.
View answer and explanation
Correct option: C
(a,b),c(a, b), c
Option C ((a,b),c(a, b), c) is correct because the given parametric equations represent a circle with center (a, b) and radius c.
533 Medium
533 of 685
Application +4 | -1
The line joining (5,0)(5,0) to (10cos⁡θ,10sin⁡θ)(10 \cos \theta, 10 \sin \theta) is divided internally in the ratio 2:32: 3 at PP. If θ\theta varies, then the locus of PP is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the locus of point P is a circle derived from the section formula and trigonometric identities.
534 Medium
534 of 685
Application +4 | -1
How many common tangents can be drawn to the following circles x2+y2=6xx^2+y^2=6 x and x2+y2+6x+2y+1=0?x^2+y^2+6 x+2 y+1=0?
View answer and explanation
Correct option: A
4
Option A (4) is correct because the distance between the centers of the circles is greater than the sum of their radii, allowing for four common tangents.
535 Easy
535 of 685
Application +4 | -1
If PP is a point such that the ratio of the square of the lengths of the tangents from PP to the circles x2+y2+2x−4y−20=0{x^2} + {y^2} + 2x - 4y - 20 = 0 and x2+y2−4x+2y−44=0{x^2} + {y^2} - 4x + 2y - 44 = 0 is 2:32:3, then the locus of pp is a circle with centre
View answer and explanation
Correct option: B
(−7,8)\left( { - 7,8} \right)
Option B ((−7,8-7, 8)) is correct because it represents the center of the circle formed by the locus of point P, derived from the given ratio of tangent lengths.
536 Medium
536 of 685
Application +4 | -1
Let O(0, 0)O(0,\,0) and A(0, 1)A(0,\,1) be two fixed points, then the locus of a point PP such that the perimeter of ΔAOP\Delta AOP is 4, is
View answer and explanation
Correct option: C
9x2+8y2−8y=16 9{x^2} + 8{y^2} - 8y = 16\,
Option C (9x2+8y2−8y=169{x^2} + 8{y^2} - 8y = 16) is correct because it represents the locus of point P satisfying the given perimeter condition.
537 Easy
537 of 685
Application +4 | -1
The inverse point of (1,−1)\left( {1, - 1} \right) with respect to x2+y2=4{x^2} + {y^2} = 4 is
View answer and explanation
Correct option: D
(2,−2)\left( {2, - 2} \right)
Option D ((2,−2)(2, - 2)) is correct because it lies on the polar of (1, -1) with respect to the circle x2+y2=4{x^2} + {y^2} = 4.
538 Easy
538 of 685
Application +4 | -1
The equation of tangent to the circle x2+y2−2ax=0x^2+y^2-2 a x=0 at the point [a(1+cos⁡α),asin⁡α][a(1+\cos \alpha), a \sin \alpha] is
View answer and explanation
Correct option: C
xcos⁡α+ysin⁡α=a(1+cos⁡α)x \cos \alpha+y \sin \alpha=a(1+\cos \alpha)
Option C (xcos⁡α+ysin⁡α=a(1+cos⁡α)x \cos \alpha+y \sin \alpha=a(1+\cos \alpha)) is correct because it represents the equation of the tangent to the given circle at the specified point, derived using the tangent equation formula.
539 Easy
539 of 685
Application +4 | -1
Let the abscissa of the two points P\mathrm{P} and Q\mathrm{Q} on a circle be the roots of x2−4x−6=0x^2-4 x-6=0 and the ordinates of P\mathrm{P} and Q\mathrm{Q} be the roots of y2+2y−7=0y^2+2 y-7=0. If PQ\mathrm{PQ} is diameter of the circle x2+y2+2ax+2by+c=0x^2+y^2+2 a x+2 b y+c=0, then the value of (a+b−c)(a+b-c) is
View answer and explanation
Correct option: A
12
Option A (12) is correct because it is the value of (a+b-c) after finding the equation of the circle using the diameter form and comparing it with the given equation.
540 Easy
540 of 685
Application +4 | -1
A circle touches a given straight line and cuts of a constant length 2d2 d from another straight line perpendicular to the first straight line. The locus of the centre of the circle is
View answer and explanation
Correct option: A
y2−x2=d2y^2-x^2=d^2
Option A (y2−x2=d2y^2-x^2=d^2) is the correct locus because it is derived from the geometric conditions that the circle touches one axis and cuts a constant length from the other.
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