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JEE • MATHEMATICS • CIRCLES

Circles practice · Easy

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331 Easy
331 of 480
Application +4 | -1
The point (1,2)(1,2) lies inside and (3,4)(3,4) outside the circle x2+y2−7x+15y−c=0x^2+y^2-7 x+15 y-c=0, if
View answer and explanation
Correct option: B
c=35c=35
Option B (c=35c=35) is correct because the condition for a point to lie inside and outside the circle gives a range for c, and 35 is the only value that satisfies both conditions.
332 Easy
332 of 480
Application +4 | -1
A straight rod of length 9 units with its ends A,BA,B always on xx and yy axes respectively. then, the locus of the centroid of ΔOAB,\Delta OAB,, is
View answer and explanation
Correct option: B
x2+y2=9{x^2} + {y^2} = 9
Option B (x2+y2=9{x^2} + {y^2} = 9) is correct because it represents the locus of the centroid of the triangle formed by the rod and the axes.
333 Easy
333 of 480
Application +4 | -1
If g2+f2=cg^2+f^2=c, then the equation x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 will represent
View answer and explanation
Correct option: D
a circle of radius 0
Option D is correct because the radius of the circle is calculated to be 0, representing a point circle.
334 Easy
334 of 480
Application +4 | -1
If the line ycos⁡α=xsin⁡α+acos⁡αy \cos \alpha=x \sin \alpha+a \cos \alpha be aa tangent to the circle x2+y2=a2x^2+y^2=a^2, then
View answer and explanation
Correct option: B
cos⁡2α=1\cos ^2 \alpha=1
Option B (cos⁡2α=1\cos ^2 \alpha=1) is correct because the condition for tangency (distance from center to the line equals the radius) leads to this result.
335 Easy
335 of 480
Application +4 | -1
The equation of circle touches the line x=yx = y at origin and passes through the point (2,1)\left( {2,1} \right) is x2+y2+px+qy=0{x^2} + {y^2} + px + qy = 0. Then p,qp,q are
View answer and explanation
Correct option: B
−5,5- 5,5
The correct answer is B (−5,5- 5,5) because these values of p and q satisfy the conditions for the circle touching the line x=y at the origin and passing through (2,1).
336 Easy
336 of 480
Application +4 | -1
If the polar of a point (p,q)(p, q) with respect to the circle x2+y2=a2x^2+y^2=a^2 touches the circle (x−c)2+(y−d)2=b2(x-c)^2+(y-d)^2=b^2, then
View answer and explanation
Correct option: A
b2(p2+q2)=(a2−cp−dq)2b^2\left(p^2+q^2\right)=\left(a^2-c p-d q\right)^2
Option A (b2(p2+q2)=(a2−cp−dq)2b^2(p^2+q^2)=(a^2-c p-d q)^2) is correct because it represents the condition where the polar of a point with respect to one circle touches another circle, derived using the distance from the center to the line.
337 Easy
337 of 480
Application +4 | -1
A circle passes through the points (2,3)(2,3) and (4,5)(4,5). If its centre lies on the line, y−4x+3=0y-4 x+3=0, then its radius is equal to
View answer and explanation
Correct option: C
2\sqrt{2}
Option C (2\sqrt{2}) is correct because the radius is calculated using the distance formula between the center of the circle and a point on the circle.
338 Easy
338 of 480
Application +4 | -1
The equations of the tangents drawn from (7,1)(7,1) to the circle x2+y2=25x^2+y^2=25, are
View answer and explanation
Correct option: A
3x+4y−25=0,4x−3y−25=03 x+4 y-25=0,4 x-3 y-25=0
Option A (3x+4y−25=0,4x−3y−25=03 x+4 y-25=0,4 x-3 y-25=0) is correct because these are the equations of the tangents derived by finding the slopes of the lines passing through the given point and tangent to the circle.
339 Easy
339 of 480
Application +4 | -1
The equation of the chord of the circle x2+y2−6x+8y=0{x^2} + {y^2} - 6x + 8y = 0 which is bisected at the point (5,−3),(5, -3), is
View answer and explanation
Correct option: A
2x+y−7=02 x+y-7=0
Option A (2x+y−7=02 x+y-7=0) is correct because it is the equation of the chord bisected at the given point, calculated using T = S'.
340 Easy
340 of 480
Application +4 | -1
Find the equation of the pair of tangents drawn to the circle x2+y2−2x+4y=0x^2+y^2-2 x+4 y=0 from the point (0,1)(0,1).
View answer and explanation
Correct option: D
x−2y+2=0&2x+y−1=0x-2 y+2=0 \& 2 x+y-1=0
Option D (x−2y+2=0&2x+y−1=0x-2 y+2=0 \& 2 x+y-1=0) is correct because it represents the pair of tangents to the given circle from the specified point, derived using the equation SS1=T2SS_1 = T^2.
341 Easy
341 of 480
Application +4 | -1
If ABA B is a chord of the circle x2+y2=r2x^2+y^2=r^2 subtending a right angle at the centre. Then, the locus of the centroid of △PAB\triangle P A B as PP moves on the circle, is
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the locus of the centroid forms a circle with a radius related to the original circle's radius.
342 Easy
342 of 480
Application +4 | -1
If 3x+y=03x + y = 0 is tangent to the circle having its centre at (2,−1),\left( {2, - 1} \right), then the equation of other tangent to the circle from the origin is
View answer and explanation
Correct option: A
x−3y=0x - 3y = 0
Option A (x−3y=0x - 3y = 0) is correct because it represents the other tangent to the circle from the origin, found by using the distance from the center to the tangent.
343 Easy
343 of 480
Application +4 | -1
The length of the tangent to the circle x2+y2−2x−y−7=0x^2+y^2-2 x-y-7=0 from (−1,−3)(-1,-3), is
View answer and explanation
Correct option: B
222 \sqrt{2}
Option B (222 \sqrt{2}) is correct because it is the length of the tangent calculated using the formula S1\sqrt{S_1}.
344 Easy
344 of 480
Application +4 | -1
Equation of the circle passing through the point (3,4)\left( {3,4} \right) and concentric with the circle x2+y2−2x−4y+1=0{x^2} + {y^2} - 2x - 4y + 1 = 0 is
View answer and explanation
Correct option: C
x2+y2−2x−4y−3=0{x^2} + {y^2} - 2x - 4y - 3 = 0
Option C (x2+y2−2x−4y−3=0{x^2} + {y^2} - 2x - 4y - 3 = 0) is correct because it represents a circle concentric with the given circle and passing through the point (3,4).
345 Easy
345 of 480
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.

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346 Easy
346 of 480
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.
347 Easy
347 of 480
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.
348 Easy
348 of 480
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}.
349 Easy
349 of 480
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.
350 Easy
350 of 480
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.
351 Easy
351 of 480
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.
352 Easy
352 of 480
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.
353 Easy
353 of 480
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.
354 Easy
354 of 480
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).
355 Easy
355 of 480
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.
356 Easy
356 of 480
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.
357 Easy
357 of 480
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.
358 Easy
358 of 480
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.
359 Easy
359 of 480
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}.
360 Easy
360 of 480
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.
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