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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 601-630 of 685 questions

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601 Medium
601 of 685
Analyzing +4 | -1
If the point (−5+λ2,−3+λ2)\left( { - 5 + \frac{\lambda }{{\sqrt 2 }}, - 3 + \frac{\lambda }{{\sqrt 2 }}} \right) is an interior point of the larger segment of the circle x2+y2=16{x^2} + {y^2} = 16 cut off by the line x+y=2,x + y = 2, then
View answer and explanation
Correct option: B
λ∈(42−14,52)\lambda \in \left( {4\sqrt 2 - \sqrt {14} ,5\sqrt 2 } \right)
Option B (λ∈(42−14,52)\lambda \in \left( {4\sqrt 2 - \sqrt {14} ,5\sqrt 2 } \right)) is correct because it satisfies both conditions: the point lies inside the circle and on the correct side of the line.
602 Easy
602 of 685
Analyzing +4 | -1
Suppose f(x,y)=0f(x, y)=0 is the equation of a circle such that f(x,1)=0f(x, 1)=0 has equal roots (each equal to 2) and f(1,x)=0f(1, x)=0 also has equal roots (each equal to zero). The equation of circle is
View answer and explanation
Correct option: D
x2+y2−4x+3=0x^2+y^2-4 x+3=0
Option D (x2+y2−4x+3=0x^2+y^2-4 x+3=0) is correct because it satisfies the given conditions of equal roots for both f(x,1)=0f(x, 1)=0 and f(1,x)=0f(1, x)=0.
603 Easy
603 of 685
Application +4 | -1
An equation of the circle described on the chord xcos⁡α+ysin⁡α−p=0x \cos \alpha+y \sin \alpha-p=0 of the circle x2+y2=a2x^2+y^2=a^2 as diameter is
View answer and explanation
Correct option: B
x2+y2−2pxcos⁡α−2pysin⁡α+2p2−a2=0x^2+y^2-2 p x \cos \alpha-2 p y \sin \alpha+2 p^2-a^2=0
Option B (x2+y2−2pxcos⁡α−2pysin⁡α+2p2−a2=0x^2+y^2-2 p x \cos \alpha-2 p y \sin \alpha+2 p^2-a^2=0) is correct because the equation of the circle is found by considering the given chord as the diameter and using the condition that the center lies on the chord.
604 Medium
604 of 685
Application +4 | -1
A tangent is drawn to the circle 2(x2+y2)−3x+4y=02\left( {{x^2} + {y^2}} \right) - 3x + 4y = 0 and it touches the circle at point AA. If the tangent passes through the point P(2,1)P\left( {2,1} \right) then PA=PA =
View answer and explanation
Correct option: B
2
Option B (2) is correct because it is the length of the tangent calculated using the formula for the length of a tangent from a point to a circle.
605 Easy
605 of 685
Application +4 | -1
The smallest circle with centre on y-axis and passing through the point(7,3)has radius
View answer and explanation
Correct option: B
7
Option B (7) is correct because the smallest circle centered on the y-axis and passing through (7,3) will have its center at (0,3), making the radius 7.
606 Easy
606 of 685
Application +4 | -1
Consider four circles (x±1)2+(y±1)2=1(x \pm 1)^2+(y \pm 1)^2=1. Equation of smaller circle touching these four circles is
View answer and explanation
Correct option: D
x2+y2=3−22x^2+y^2=3-2 \sqrt{2}
Option D (x2+y2=3−22x^2+y^2=3-2 \sqrt{2}) is correct because it represents the equation of a circle centered at the origin with a radius equal to the distance from the origin to the point where the smaller circle touches the four given circles.
607 Hard
607 of 685
Application +4 | -1
Consider a family of circles, which are passing through the point (−1,  1)\left( { - 1,\;1} \right) and are tangent to xx-axis. If (h,  k)\left( {h,\;k} \right) are the coordinates of the centre of the circles, then the set of values of kk is given by the interval
View answer and explanation
Correct option: B
k≥12k \ge \frac{1}{2}
Option B (k≥12k \ge \frac{1}{2}) is correct because it represents the range of possible y-coordinates for the center of the circle, derived from the condition that the circle is tangent to the x-axis and passes through (-1, 1).
608 Easy
608 of 685
Analyzing +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: D
−35<m<15-35 < m < 15
Option D (-35 < m < 15) is correct because the condition for the line to intersect the circle at two distinct points is satisfied within this range of m values.
609 Medium
609 of 685
Application +4 | -1
The area of a circle and the area of a regular polygon of nn sides and of perimeter equal to that of the circle are in the ratio
View answer and explanation
Correct option: A
tan⁡πn:πn\tan \frac{\pi}{n}: \frac{\pi}{n}
Option A (tan(π/n):π/n) is correct because it represents the ratio of the area of a circle to the area of a regular polygon with the same perimeter.
610 Medium
610 of 685
Application +4 | -1
A cone is made from a circular sheet of radius 3\sqrt{3} by cutting out a sector and gluing the cut edges of the remaining piece together. The maximum volume attainable for the cone is
View answer and explanation
Correct option: C
2π/32 \pi / 3
Option C (2π/32 \pi / 3) is correct because it represents the maximum volume attainable for the cone formed from the circular sheet.
611 Easy
611 of 685
Application +4 | -1
Equation of a circle with centre (−4,3)(-4,3) touching internally and containing the circle x2+y2=1x^2+y^2=1 is
View answer and explanation
Correct option: C
x2+y2+8x−6y−11=0x^2+y^2+8 x-6 y-11=0
Option C (x2+y2+8x−6y−11=0x^2+y^2+8 x-6 y-11=0) is correct because it represents a circle with center (-4, 3) and radius 6, which touches the circle x2+y2=1x^2 + y^2 = 1 internally.
612 Easy
612 of 685
Analyzing +4 | -1
The number of common tangents to the circles x2+y2=4x^2+y^2=4 and x2+y2−6x−8y=24x^2+y^2-6 x-8 y=24, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the distance between the centers equals the difference of the radii, indicating the circles touch internally and have one common tangent.
613 Easy
613 of 685
Analyzing +4 | -1
A circle is passing through the points A(1,1)A(1,1) and B(1,3)B(1,3) and the bisector of first and third quadrant is normal to it, then its area, in square units, is
View answer and explanation
Correct option: A
2π2 \pi
Option A (2π2 \pi) is correct because it is the area of the circle passing through the given points with the specified normal.
614 Easy
614 of 685
Application +4 | -1
If tangents PAP A and PBP B are drawn to x2+y2=9x^2+y^2=9 from the point P(4,4)P(4,4), then area of △OAB,O\triangle O A B, O being the origin, is equal to
View answer and explanation
Correct option: A
932207\frac{9}{32} \sqrt{207} sq units
Option A (932207\frac{9}{32} \sqrt{207} sq units) is correct because it is calculated using the length of the perpendicular from the origin to the chord and the length of the chord.
615 Medium
615 of 685
Application +4 | -1
The angle between the pair of tangents from the point (1,12)\left(1, \frac{1}{2}\right) to the circle x2+y2+4x+2y−4=0x^2+y^2+4 x+2 y-4=0, is
View answer and explanation
Correct option: B
sin⁡−1(45)\sin ^{-1}\left(\frac{4}{5}\right)
Option B (sin⁡−1(45)\sin ^{-1}(\frac{{4}}{{5}})) is correct because it represents the angle between the pair of tangents from the given point to the given circle, calculated using the formula involving the radius and distance from the center.
616 Easy
616 of 685
Application +4 | -1
The circles x2+y2−10x+16=0x^2+y^2-10 x+16=0 and x2+y2=r2x^2+y^2=r^2 intersect each other in two distinct points if
View answer and explanation
Correct option: C
2<r<82 < r < 8
Option C (2<r<82 < r < 8) is correct because the condition for two circles to intersect at two distinct points is that the distance between their centers must be between the difference and the sum of their radii.
617 Easy
617 of 685
Application +4 | -1
Find the intercept made by the circle x2+y2−14x−10y+23=0x^2+y^2-14 x-10 y+23=0 on the line 7x+y−4=07 x+y-4=0.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the length of the intercept made by the circle on the line is calculated to be 2.
618 Easy
618 of 685
Application +4 | -1
The locus of pole of any tangent to the circle x2+y2x^2+y^2 =4=4 w.r.t. the hyperbola x2−y2=4x^2-y^2=4 is the circle
View answer and explanation
Correct option: C
x2+y2=4x^2+y^2=4
Option C is correct because the locus of the pole is found by comparing the tangent to the circle and the polar of the hyperbola.
619 Easy
619 of 685
Analyzing +4 | -1
If the circles x2+y2+2gx+2fy=0{x^2} + {y^2} + 2gx + 2fy = 0 and x2+y2+2g′ x+2f′ y=0{x^2} + {y^2} + 2g'\,x + 2f'\,y = 0 touch each other, then
View answer and explanation
Correct option: D
fg′=f′gf\mathrm{g'} = f'\mathrm{g}
Option D (fg′=f′gf\mathrm{g'} = f'\mathrm{g}) is correct because the circles touch each other if they have the same tangent at the origin.
620 Medium
620 of 685
Application +4 | -1
The range of values of aa such that the angle θ\theta between the pair of tangents drawn from (a,  0)\left( {a,\;0} \right) to the circle x2+y2=1{x^2} + {y^2} = 1 satisfies π2<θ<π\frac{\pi }{2} < \theta < \pi, is
View answer and explanation
Correct option: D
(−2,  −1)∪(1,  2)\left( { - \sqrt 2 ,\; - 1} \right) \cup \left( {1,\;\sqrt 2 } \right)
Option D (−2,  −1)∪(1,  2)\left( { - \sqrt 2 ,\; - 1} \right) \cup \left( {1,\;\sqrt 2 } \right) is correct because it represents the range of 'a' for which the angle between the tangents from (a,0) to the circle is between π2\frac{\pi}{2} and π\pi.
621 Medium
621 of 685
Analyzing +4 | -1
If y+3x=0y+3 x=0 is the equation of a chord of the circle, x2+y2−30x=0x^{2}+y^{2}-30 x=0, then the equation of the circle with this chord as diameter is :
View answer and explanation
Correct option: D
x2+y2−3x+9y=0x^{2}+y^{2}-3 x+9 y=0
Option D (x2+y2−3x+9y=0x^{2}+y^{2}-3 x+9 y=0) is correct because it represents the circle with the given chord as a diameter, found by determining the intersection points of the line and circle.
622 Easy
622 of 685
Application +4 | -1
The circle described on the line joining the points (0,1),(a,b)(0,1),(a, b) as diameter cuts the xx-axis in points whose abscissae are roots of the equation
View answer and explanation
Correct option: B
x2−ax+b=0x^2-a x+b=0
Option A (x2+y2−6x±8y+9=0x^2+y^2-6 x \pm 8 y+9=0) is correct because these are the equations of the circles that touch the x-axis at a distance of 3 from the origin and have an intercept of length 272\sqrt{7} on the y-axis.
623 Hard
623 of 685
Analyzing +4 | -1
Two concentric circles of which smallest is x2+y2=4x^2+y^2=4, have the difference in radii as dd, if line y=x+1y=x+1 cuts the circles in real points, then dd lies in the interval
View answer and explanation
Correct option: A
(−∞,−2−12)∪(−2+12,∞)\left(-\infty,-2-\frac{1}{\sqrt{2}}\right) \cup\left(-2+\frac{1}{\sqrt{2}}, \infty\right)
The correct answer is A (−∞,−2−12)∪(−2+12,∞)\left(-\infty,-2-\frac{{1}}{{\sqrt{{2}}}}\right) \cup\left(-2+\frac{{1}}{{\sqrt{{2}}}}, \infty\right) because it represents the interval for the difference in radii such that the line intersects both circles in real points.
624 Easy
624 of 685
Application +4 | -1
The tangents to the circle x2+y2=48x^2+y^2=48, which are inclined at angle of 60∘60^{\circ} with the axis of xx form a rhombus, the length of whose sides is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the length of the side of the rhombus is found by calculating the length of the tangent from the point where the tangents intersect the x-axis.
625 Easy
625 of 685
Application +4 | -1
The number of common tangents to the circles x2+y2−y=0{x^2} + {y^2} - y = 0 and x2+y2+y=0{x^2} + {y^2} + y = 0 is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the two circles touch each other externally, resulting in three common tangents.
626 Easy
626 of 685
Analyzing +4 | -1
If the chords of contact of the tangents from a point on the circle x2+y2=a2{x^2} + {y^2} = {a^2} to the circle x2+y2=b2{x^2} + {y^2} = {b^2} touch the circle x2+y2=c2,{x^2} + {y^2} = {c^2}, then the roots of the equation ax2+2bx+c=0a{x^2} + 2bx + c = 0, are
View answer and explanation
Correct option: B
Real and equal
The roots are real and equal because the condition for the chord of contact to touch the inner circle leads to b2=acb^2 = ac, which implies the discriminant of the quadratic equation is zero.
627 Easy
627 of 685
Application +4 | -1
A line through P(1,4)P\left( {1,4} \right) intersect a circle x2+y2=16{x^2} + {y^2} = 16 at AA and BB, then PA.PBPA.PB is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because PA.PBPA.PB is equal to the square of the length of the tangent from point PP to the circle, which can be calculated using PT2=S1PT^2 = S_1.
628 Medium
628 of 685
Analyzing +4 | -1
Tangent to circle x2+y2=5x^2+y^2=5 at (1,−2)(1,-2), also touches the circle x2+y2−8x+6y+20=0x^2+y^2-8 x+6 y+20=0. The coordinates of the corresponding point of contact is
View answer and explanation
Correct option: B
(3,−1)(3,-1)
Option B ((3,-1)) is correct. By finding the equation of the tangent to the first circle and substituting it into the equation of the second circle, we can solve for the point of contact.
629 Easy
629 of 685
Application +4 | -1
If the base of a triangle and the ratio of the lengths of the other two unequal sides are given, then the vertex lies on a/an
View answer and explanation
Correct option: B
circle
Option B (circle) is correct because the locus of the vertex is a circle according to Apollonius's circle theorem.
630 Easy
630 of 685
Application +4 | -1
The value of cc, for which the line y=2x+cy = 2x + c is a tangent to the circle x2+y2=16{x^2} + {y^2} = 16, is
View answer and explanation
Correct option: B
454\sqrt 5
Option B (454\sqrt 5) is correct because it represents the value of c for which the line is tangent to the circle, derived using the tangency condition.
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