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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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Showing 61-90 of 685 questions

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61 Easy
61 of 685
Understanding +4 | -1
The angle between the tangents drawn from a point on the director circle x2+y2=50x^2+y^2=50 to the circle x2+y2=25x^2+y^2=25, is
View answer and explanation
Correct option: C
90∘90^{\circ}
Option C (90∘90^{\circ}) is correct because the director circle is the locus of points from which tangents to the circle are perpendicular.
62 Easy
62 of 685
Application +4 | -1
If 3x+y+k=03x + y + k = 0 is a tangent to the circle x2+y2=10{x^2} + {y^2} = 10, the values of kk are
View answer and explanation
Correct option: C
±10\pm 10
Option C (±10\pm 10) is correct because it represents the values of kk for which the line 3x+y+k=03x + y + k = 0 is tangent to the circle x2+y2=10{x^2} + {y^2} = 10.
63 Medium
63 of 685
Application +4 | -1
A circle is described on the line joining the points (2,−3)(2,-3) and (−4,7)(-4,7) as a diameter. The circle
View answer and explanation
Correct option: D
origin lies inside the circle
The correct answer is that the origin lies inside the circle because the distance from the center of the circle to the origin is less than the radius of the circle.
64 Medium
64 of 685
Application +4 | -1
A tangent PTP T is drawn to the circle x2+y2=4x^2+y^2=4 at the point P(3,1)P(\sqrt{3}, 1). A straight line LL, perpendicular to PTP T is a tangent to the circle (x−3)2+y2=1(x-3)^2+y^2=1. A possible equation of LL is
View answer and explanation
Correct option: A
x−3y=1x-\sqrt{3} y=1
Option A (x+2y=0x+2 y=0) is correct because it is the locus of the centers of the circles passing through the intersection of the given circles.
65 Easy
65 of 685
Application +4 | -1
Equation of a circle which touches yy-axis at (0,2)(0,2) and cuts off an intercept of 3 units from the xx-axis is x2+y2−2αx−4y+4=0x^2+y^2-2 \alpha x-4 y+4=0 where α2=\alpha^2=
View answer and explanation
Correct option: B
25/425 / 4
Option B (25/425 / 4) is correct because it is the value of α2\alpha^2 that satisfies the given conditions for the circle.
66 Easy
66 of 685
Application +4 | -1
A point PP moves such that PA/PB=pP A / P B=p where AA and BB are two fixed points with AB=aA B=a, the locus of PP is a circle with radius
View answer and explanation
Correct option: C
ap/(1−p2)a p /\left(1-p^2\right)
Option C (ap/(1−p2)a p / (1-p^2)) is correct because it represents the radius of the circle that is the locus of P.
67 Easy
67 of 685
Application +4 | -1
If (1,  a),  (b,  2)\left( {1,\;a} \right),\;\left( {b,\;2} \right) are conjugate points with respect to the circle x2+y2=25{x^2} + {y^2} = 25, then 4a+2b4a + 2b is equal to
View answer and explanation
Correct option: B
50
Option B (50) is correct because it is the value of 4a + 2b when (1, a) and (b, 2) are conjugate points with respect to the given circle.
68 Easy
68 of 685
Application +4 | -1
The radius of the circle passing through the point P(6,2)P\left( {6,2} \right) 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 is the distance between the center of the circle and the point P(6,2), where the center is the intersection of the two diameters.
69 Easy
69 of 685
Application +4 | -1
ABCDABCD is a square whose side is a. The equation of the circle circumscribing the square, taking ABAB and ADAD as axes of reference, is
View answer and explanation
Correct option: C
x2+y2−ax−ay=0{x^2} + {y^2} - ax - ay = 0
Option C (x2+y2−ax−ay=0{x^2} + {y^2} - ax - ay = 0) is correct because the circle passes through the vertices of the square when AB and AD are the axes of reference.
70 Easy
70 of 685
Application +4 | -1
Equation of circle passing through (1,5)(1,5) and (4,1)(4,1) and touching yy-axis is x2+y2−5x−6y+9+λ(4x+3y−19)=0x^2+y^2-5 x-6 y+9+\lambda(4 x+3 y-19)=0 where λ\lambda is equal to
View answer and explanation
Correct option: A
0,−40/90,-40 / 9
Option A (0,−40/90, -40/9) is correct because these are the values of λ\lambda for which the circle passes through the given points and touches the y-axis.
71 Easy
71 of 685
Application +4 | -1
The area of square inscribed in a circle x2+y2−6x−8y=0{x^2} + {y^2} - 6x - 8y = 0 is
View answer and explanation
Correct option: B
50 sq units
Option B (50 sq units) is correct because the area of the square inscribed in the circle is half the square of the circle's diameter.
72 Medium
72 of 685
Application +4 | -1
The radius of inscribed circle in the quadrilateral formed by tangents drawn from (3,−1)(3,-1) to circle x2+y2−2x−2y−2=0x^2+y^2-2 x-2 y-2=0 and radius formed by point of contact of these tangents is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the point (3,-1) lies on the director circle, making the radius of the inscribed circle half the radius of the given circle.
73 Easy
73 of 685
Application +4 | -1
The line 4x+3y−4=04 x+3 y-4=0 divides the circumference of the circle centred at (5,3)(5,3) in the ratio 1:21: 2. Then the equation of the circle is
View answer and explanation
Correct option: A
x2+y2−10x−6y−66=0x^2+y^2-10 x-6 y-66=0
Option A (x2+y2−10x−6y−66=0x^2+y^2-10 x-6 y-66=0) is correct because the radius of the circle is calculated to be 10, and the equation is derived accordingly.
74 Easy
74 of 685
Application +4 | -1
The centres of those circles which touch the circle, x2+y2−8x−8y−4=0{x^2} + {y^2} - 8x - 8y - 4 = 0, externally and also touch the XX-axis, lie on
View answer and explanation
Correct option: D
A parabola
Option D is correct because the locus of the centers of the circles that touch the given circle externally and the x-axis is a parabola.
75 Medium
75 of 685
Application +4 | -1
The tangents to the circle x2+y2=169x^2+y^2=169 at the points (5,12)(5,12) and (−5,−12)(-5,-12) are
View answer and explanation
Correct option: A
parallel
Option A (parallel) is correct. The slopes of the tangents at the given points are equal, indicating that the tangents are parallel.

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76 Easy
76 of 685
Application +4 | -1
The coordinates of the middle point of the chord 2x−5y+18=02 x-5 y+18=0 cut off by the circle x2+y2−6x+2y−54=0x^2+y^2-6 x+2 y-54=0 is
View answer and explanation
Correct option: A
(1,4)(1,4)
Option C, (4,1), is correct because it represents the midpoint of the chord, which lies on the line perpendicular to the chord passing through the center of the circle.
77 Easy
77 of 685
Application +4 | -1
The number of common tangents to the circles
x2+y2−2x−4y+1=0{x^2} + {y^2} - 2x - 4y + 1 = 0 and x2+y2−12x−16y+91=0{x^2} + {y^2} - 12x - 16y + 91 = 0, is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the distance between the centers of the circles is greater than the sum of their radii, indicating four common tangents.
78 Easy
78 of 685
Application +4 | -1
If circumference of circle x2+y2+8x+8y−b=0x^2+y^2+8 x+8 y-b=0 is bisected by circle x2+y2−2x+4y+a=0x^2+y^2-2 x+4 y+a=0 then (a+b)(a+b) is equal to
View answer and explanation
Correct option: C
-56
The correct answer is C (-56) because the condition for a circle to bisect the circumference of another circle is that the radical axis passes through the center of the first circle.
79 Medium
79 of 685
Analyzing +4 | -1
The centre of the circle passing through (0,2)(0,2) and touching the parabola y=2x2y=2 x^2 at (2,2)(2,2) is
View answer and explanation
Correct option: B
(−2,52)\left(-2, \frac{5}{2}\right)
Option B ((-2, 5/2)) is correct because it represents the center of the circle that satisfies the given tangency and point conditions.
80 Medium
80 of 685
Application +4 | -1
In what ratio the line y−x+2=0y-x+2=0 divides the line joining the points (3,−1)(3,-1) and (8,9)(8,9) ?
View answer and explanation
Correct option: C
2:32: 3
Option C (2:3) is correct because the ratio is found by substituting the intersection point of the line and the line segment into the line's equation.
81 Easy
81 of 685
Application +4 | -1
The equations of the circles which touch both the axes and the line x=ax=a are
View answer and explanation
Correct option: C
x2+y2−ax±ay+a24=0x^2+y^2-a x \pm a y+\frac{a^2}{4}=0
Option C (x2+y2−ax±ay+a24=0x^2+y^2-a x \pm a y+\frac{a^2}{4}=0) is correct because it represents a circle touching both axes and the line x=a.
82 Medium
82 of 685
Application +4 | -1
A circle of radius 4 units touches the coordinate axes in the first quadrant. The equation of its image with respect to the line mirror y=0y=0 is
View answer and explanation
Correct option: A
x2+y2−8x+8y+16=0x^2+y^2-8 x+8 y+16=0
Option A (x2+y2−8x+8y+16=0x^2+y^2-8 x+8 y+16=0) is correct because it represents the equation of the circle obtained by reflecting the original circle about the line y=0.
83 Easy
83 of 685
Application +4 | -1
A variable circle passes through the fixed point A(p,q)A\left( {p,q} \right) and touches xx-axis. The locus of the other end of the diameter through AA is
View answer and explanation
Correct option: A
(x−p)2=4qy{\left( {x - p} \right)^2} = 4qy
Option A ((x−p)2=4qy{\left( {x - p} \right)^2} = 4qy) is correct because it represents the locus of the other end of the diameter through A.
84 Easy
84 of 685
Analyzing +4 | -1
If 2x+3y−6=02x + 3y - 6 = 0 and 9x+6y−18=09x + 6y - 18 = 0 cuts the axes in concyclic points, then the centre of the circle, is
View answer and explanation
Correct option: D
(5/2,5/2)\left( {5/2,5/2} \right)
Option D ((5/2,5/2)\left( {5/2,5/2} \right)) is correct because the center of the circle passing through the concyclic points is found by equating the equation of the circle to zero and finding the center.
85 Easy
85 of 685
Analyzing +4 | -1
The lengths of the tangents from any point on the circle 15x2+15y2−48x+64y=015 x^2+15 y^2-48 x+64 y=0 to the two circles 5x2+5y2−24x+32y+75=0,5 x^2+5 y^2-24 x+32 y+75=0,

5x2+5y2−48x+64y+300=05 x^2+5 y^2-48 x+64 y+300=0 are in the ratio
View answer and explanation
Correct option: A
1:21: 2
Option A (1:21:2) is correct because the ratio of the lengths of the tangents from a point on the circle to the other two circles is calculated to be 1:21:2.
86 Easy
86 of 685
Application +4 | -1
The line segment joining the points (4,7)\left( {4,7} \right) and (−2,  −1)\left( { - 2,\; - 1} \right) is a diameter of a circle. If the circle intersects the xx-axis at AA and BB, then ABAB is equal to
View answer and explanation
Correct option: D
8
Option D (8) is correct because the length of the chord AB, where the circle intersects the x-axis, is calculated using the formula 2g2−c2\sqrt{{g^2 - c}} after finding the equation of the circle.
87 Easy
87 of 685
Application +4 | -1
Coordinates of the mid-point of the segment cut by the circle x2+y2−6x+2y−54=0x^2+y^2-6 x+2 y-54=0 on the line 2x−5y+18=02 x-5 y+18=0, is
View answer and explanation
Correct option: C
(1,4)(1,4)
Option C (1,4) is correct because the midpoint of the chord is found by solving the equation of the line perpendicular to the chord and passing through the circle's center.
88 Easy
88 of 685
Application +4 | -1
The two circles x2+y2=axx^2+y^2 = ax and x2+y2=c2x^2+y^2=c^2 (with c>0c>0 ) touch each other, if
View answer and explanation
Correct option: A
c=∣a∣c=|a|
Option A (c=∣a∣c=|a|) is correct because the condition for touching circles (distance between centers equals sum or difference of radii) leads to this relationship.
89 Easy
89 of 685
Application +4 | -1
The circle x2+y2+4  x−7  y+12=0{x^2} + {y^2} + 4\;x - 7\;y + 12 = 0 cuts an intercept on yy-axis of length
View answer and explanation
Correct option: D
1
Option D (1) is correct because the y-intercept is calculated using the formula 2*sqrt(f2 - c) where f and c are coefficients from the circle's equation.
90 Medium
90 of 685
Application +4 | -1
The locus of the middle point of chords of the circle x2+y2=a2{x^2} + {y^2} = {a^2} which pass through the fixed point (h,k)\left( {h,k} \right) is
View answer and explanation
Correct option: A
x2+y2−hx−ky=0{x^2} + {y^2} - hx - ky = 0
Option A (x2+y2−hx−ky=0{x^2} + {y^2} - hx - ky = 0) is correct because it is the locus of the middle point of the chords, derived by using the equation of the chord and the given fixed point.
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