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JEE • MATHEMATICS • CIRCLES

Circles practice · Medium

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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 121-150 of 189 questions

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121 Medium
121 of 189
Application +4 | -1
The coordinates of the centre of the smallest circle passing through the origin and having y=x+1y = x + 1 as a diameter are
View answer and explanation
Correct option: D
(−12,12)\left( { - \frac{1}{2},\frac{1}{2}} \right)
Option D ((−12,12)\left( { - \frac{1}{2},\frac{1}{2}} \right)) is correct because the center of the smallest circle lies on the perpendicular bisector of the line segment joining the origin and the intersection of the line y = x + 1 with the perpendicular bisector.
122 Medium
122 of 189
Analyzing +4 | -1
If (2,4)(2,4) is a point interior to the circle x2+y2−6x−10y+λ=0x^2+y^2-6 x-10 y+\lambda=0 and the circle does not cut the axes at any point, then
View answer and explanation
Correct option: A
λ∈(25,32)\lambda \in(25,32)
Option A (λ∈(25,32)\lambda \in(25,32)) is correct because it satisfies both conditions: the point (2,4) lies inside the circle, and the circle does not cut the axes.
123 Medium
123 of 189
Application +4 | -1
In the two circles (x+7)2+(y−3)2=36{\left( {x + 7} \right)^2} + {\left( {y - 3} \right)^2} = 36 and (x−5)2+(y+2)2=49{\left( {x - 5} \right)^2} + {\left( {y + 2} \right)^2} = 49 touch each other externally, then the point of contact is
View answer and explanation
Correct option: B
(−1913,913)\left( {\frac{{ - 19}}{{13}},\frac{9}{{13}}} \right)
Option B (−1913,913)\left( {\frac{{ - 19}}{{13}},\frac{9}{{13}}} \right) is correct because it represents the point of contact when two circles touch each other externally, calculated using the section formula.
124 Medium
124 of 189
Application +4 | -1
The equation of the tangent to circle 5x2+5y2=15{x^2} + 5{y^2} = 1, parallel to line 3x+4y=13x + 4y = 1 are
View answer and explanation
Correct option: C
3x+4y=±53x + 4y = \pm \sqrt 5
Option C (3x+4y=±53x + 4y = \pm \sqrt 5) is correct because these are the equations of the tangents to the circle that are parallel to the given line.
125 Medium
125 of 189
Application +4 | -1
If the centroid of an equilateral triangle is (2,−2)(2,-2) and its one vertex is (−1,1)(-1,1), then the equation of its circumcircle is
View answer and explanation
Correct option: A
x2+y2−4x+4y−10=0x^2+y^2-4 x+4 y-10=0
Option A (x2+y2−4x+4y−10=0x^2+y^2-4 x+4 y-10=0) is correct because the circumcircle passes through the vertex and has its center at the centroid.
126 Medium
126 of 189
Analyzing +4 | -1
The number of points with integral coordinates (2a,a−1)\left( {2a,a - 1} \right) that fall in the interior of the larger segment of the circle x2+y2=25{x^2} + {y^2} = 25 cut off by the parabola x2+4y=0,{x^2} + 4y = 0, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because there are two integral values of 'a' that satisfy the given conditions for the point to lie inside the larger segment.
127 Medium
127 of 189
Application +4 | -1
Find the equation of the circle, the coordinates of the end points of whose diameter are (−1,2)(-1,2) and (4,−3)(4,-3).
View answer and explanation
Correct option: C
x2+y2−3x+y−10=0x^2+y^2-3 x+y-10=0
Option C (x2+y2−3x+y−10=0x^2+y^2-3 x+y-10=0) is correct because it is derived using the diameter form of the circle equation with the given endpoints.
128 Medium
128 of 189
Application +4 | -1
The range of values of λ\lambda for which the circles x2+y2=4x^2+y^2=4 and x2+y2−4λx+9=0x^2+y^2-4 \lambda x+9=0 have two common tangents, is
View answer and explanation
Correct option: B
(−∞,−13/8)∪(13/8,∞)(-\infty,-13 / 8) \cup(13 / 8, \infty)
Option B ((−∞,−13/8)∪(13/8,∞))((-\infty,-13 / 8) \cup(13 / 8, \infty)) is correct because it represents the range of lambda values for which the circles have two common tangents.
129 Medium
129 of 189
Analyzing +4 | -1
A line is at a constant distance cc from the origin and meets the coordinate axes in AA and BB. The locus of the centre of the circle passing through O,A,BO,A,B is
View answer and explanation
Correct option: D
x−2+y−2=4c−2{x^{ - 2}} + {y^{ - 2}} = 4{c^{ - 2}}
The correct answer is D (x−2+y−2=4c−2{x^{ - 2}} + {y^{ - 2}} = 4{c^{ - 2}}) because it represents the locus of the center of the circle passing through O, A, and B, given the line is at a constant distance c from the origin.
130 Medium
130 of 189
Analyzing +4 | -1
Let z∈C\mathrm{z} \in \mathrm{C}, the set of complex numbers. Then the equation, 2∣z+3i∣−∣z−i∣=02|z+3 i|-|z-i|=0 represents :
View answer and explanation
Correct option: A
a circle with radius 83\frac{8}{3}.
The correct answer is A (a circle with radius 83\frac{8}{3}) because the given equation simplifies to the equation of a circle, and the radius is calculated to be 83\frac{8}{3}.
131 Medium
131 of 189
Application +4 | -1
The circle x2+y2−8x+4y+4=0{x^2} + {y^2} - 8x + 4y + 4 = 0 touches
View answer and explanation
Correct option: B
y−axisy - axis
Option B (y−axisy - axis) is correct because the radius of the circle is equal to the x-coordinate of the center, indicating that the circle touches the y-axis.
132 Medium
132 of 189
Analyzing +4 | -1
Let PP be a point on the parabola, x2=4yx^2=4 y. If the distance of PP from the centre of the circle, x2+y2+6x+8=0x^2+y^2+6 x+8=0 is minimum, then the equation of the tangent to the parabola at P\mathrm{P}, is
View answer and explanation
Correct option: C
x+y+1=0x+y+1=0
Option C (x+y+1=0x+y+1=0) is correct. The tangent to the parabola at the point where the distance to the circle's center is minimized is found by using the normal to the parabola.
133 Medium
133 of 189
Application +4 | -1
The locus of the point of intersection of tangents to the circle x=acos⁡θx=a \cos \theta, y=asin⁡θy=a \sin \theta at the points, whose parametric angles differ by π2\frac{\pi}{2} is a
View answer and explanation
Correct option: B
circle
Option B (circle) is correct because the locus of the point of intersection of tangents is a circle.
134 Medium
134 of 189
Application +4 | -1
The points of contact of tangents to the circle x2+y2=25{x^2} + {y^2} = 25 which are inclined at an angle of 300{30^0} to the x−axisx - \mathrm{axis} are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the points of contact are calculated using the formula involving the slope and radius, and the result doesn't match any of the given options.
135 Medium
135 of 189
Application +4 | -1
The circle x2+y2=4x+8y+5{x^2} + {y^2} = 4x + 8y + 5 intersects the line 3x−4y=m3x - 4y = m at two distinct points, if
View answer and explanation
Correct option: B
−35<m<15- 35 < m < 15
Option B (−35<m<15- 35 < m < 15) is correct because the line intersects the circle at two distinct points if the distance of the center from the line is less than the radius.
136 Medium
136 of 189
Analyzing +4 | -1
The circle passing through the distinct points (1,t),(t,1)(1, t),(t, 1) and (t,t)(t, t) for all values of t{t}, passes through the point
View answer and explanation
Correct option: A
(1,1)(1,1)
Option A ((1,1)) is correct because the circle passing through the distinct points (1, t), (t, 1), and (t, t) for all values of t, passes through the point (1,1).
137 Medium
137 of 189
Application +4 | -1
The straight line xa+yb=1\frac{x}{a}+\frac{y}{b}=1 cuts the coordinate axes at AA and BB. The equation of the circle passing through O(0,0)O(0,0), AA and BB, is
View answer and explanation
Correct option: A
x2+y2−ax−by=0x^2+y^2-a x-b y=0
Option A (x2+y2−ax−by=0x^2+y^2-a x-b y=0) is correct because it represents the circle passing through the origin and the intercepts of the given line.
138 Medium
138 of 189
Analyzing +4 | -1
f(x,y)≡x2+y2+2ax+2by+c=0f(x, y) \equiv x^2+y^2+2 a x+2 b y+c=0 represent a circle. If f(x,0)=0f(x, 0)=0 has equal roots, each being 2 and f(0,y)=0f(0, y)=0 has 2 and 3 as its roots, then the centre of the circle is
View answer and explanation
Correct option: B
Data are not consistent
Option D (Data are inconsistent) is correct because the given conditions lead to conflicting values for the constant term 'c' in the circle's equation.
139 Medium
139 of 189
Analyzing +4 | -1
If in a △ABC\triangle A B C (whose circumcentre is origin), a≤sin⁡Aa \leq \sin A, then for any point (x,y)(x, y) inside the circumcircle of △ABC\triangle A B C
View answer and explanation
Correct option: A
∣xy∣<18|x y|<\frac{1}{8}
Option A (∣xy∣<18|xy| < \frac{{1}}{{8}}) is correct because it is derived from the given condition a≤sin⁡Aa \leq \sin A and applying AM-GM inequality inside the circumcircle.
140 Medium
140 of 189
Analyzing +4 | -1
The number of common tangents to two circles x2+y2=4{x^2} + {y^2} = 4 and x2+y2−8x+12=0{x^2} + {y^2} - 8x + 12 = 0 is
View answer and explanation
Correct option: D
3
Option D (3) is correct because the distance between the centers equals the sum of the radii, indicating the circles touch externally, allowing for three common tangents.
141 Medium
141 of 189
Application +4 | -1
A line is drawn through a fixed point P(α,β)P\left( {\alpha ,\beta } \right) to cut the circle x2+y2=r2{x^2} + {y^2} = {r^2} at AA and BB. Then, PA.PBPA.PB is equal to
View answer and explanation
Correct option: B
α2+β2−r2{\alpha ^2} + {\beta ^2} - {r^2}
Option B (α2+β2−r2{\alpha ^2} + {\beta ^2} - {r^2}) is correct because it represents the power of the point P with respect to the circle.
142 Medium
142 of 189
Application +4 | -1
The locus of the point from which mutually perpendicular tangents can be drawn to the circle x2+y2=36x^2+y^2=36 is
View answer and explanation
Correct option: D
x2+y2=72x^2+y^2=72.
Option D (x2+y2=72x^2+y^2=72) is correct because the locus of the point from which mutually perpendicular tangents can be drawn to a circle is the director circle, and its radius squared is twice the radius squared of the original circle.
143 Medium
143 of 189
Application +4 | -1
The equations of the tangents to the circle x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0 which are parallel to the line 4x+3y+5=04x + 3y + 5 = 0, are
View answer and explanation
Correct option: C
4x+3y+19=04x + 3y + 19 = 0 and 4x+3y−31=04x + 3y - 31 = 0
Option C (4x + 3y + 19 = 0 and 4x + 3y - 31 = 0) is correct because these are the tangent lines found by using the perpendicular distance from the center of the circle to the tangent equals the radius.
144 Medium
144 of 189
Application +4 | -1
If (α,β)\left( {\alpha ,\beta } \right) is a point on the chord PQPQ of the circle x2+y2=25,{x^2} + {y^2} = 25, where the coordinates of PP and QQ are (3,−4)\left( {3, - 4} \right) and (4,3)\left( {4,3} \right) respectively, then
View answer and explanation
Correct option: A
3≤α≤4 and−4≤β≤33 \le \alpha \le 4\,\mathrm{and} - 4 \le \beta \le 3
Option A is correct because if (α,β) is a point on the chord PQ, then α must lie between the x-coordinates of P and Q, and β must lie between the y-coordinates of P and Q.
145 Medium
145 of 189
Analyzing +4 | -1
The equation of the circle passing through (2,0)(2,0) and (0,4)(0,4) and having the minimum radius is
View answer and explanation
Correct option: B
x2+y2−2x−4y=0x^2+y^2-2 x-4 y=0
Option B (x2+y2−2x−4y=0x^2+y^2-2 x-4 y=0) is correct because it represents the circle passing through (2,0) and (0,4) with the minimum radius, which is the circle with the line segment joining the points as the diameter.
146 Medium
146 of 189
Application +4 | -1
If xα+yβ=1\frac{x}{\alpha } + \frac{y}{\beta } = 1 touches the circle x2+y2=a2{x^2} + {y^2} = {a^2}, the point (1/α,1/β)\left( {1/\alpha ,1/\beta } \right) lies on a/an
View answer and explanation
Correct option: B
Circle
Option B (Circle) is correct because the locus of the point (1/α, 1/β) is a circle, derived from the condition of tangency to the given circle.
147 Medium
147 of 189
Application +4 | -1
The equation of the common tangent to the circle (x−3)2+y2=9(x-3)^2+y^2=9 and the parabola y2=4xy^2=4 x above the xx-axis is
View answer and explanation
Correct option: C
3y=x+3\sqrt{3} y=x+3
Option C (3y=x+3\sqrt{3} y=x+3) is correct because it represents the common tangent to the given circle and parabola above the x-axis, found by equating the perpendicular distance from the circle's center to the tangent with the circle's radius.
148 Medium
148 of 189
Analyzing +4 | -1
If there does not exist any chord drawn from the point (22sin⁡θ,cos⁡θ3)\left(2 \sqrt{2} \sin \theta, \frac{\cos \theta}{\sqrt{3}}\right) to the circle 3(x2+y2)=26sin⁡θ+cos⁡θy\sqrt{3}\left(x^2+y^2\right)=2 \sqrt{6} \sin \theta+\cos \theta y which are bisected by xx-axis, then
View answer and explanation
Correct option: C
θ∈(nπ−π6,nπ+π6);n∈Z\theta \in\left(\mathrm{n} \pi-\frac{\pi}{6}, \mathrm{n} \pi+\frac{\pi}{6}\right) ; \mathrm{n} \in \mathrm{Z}
Option C (θ∈(nπ−π6,nπ+π6);n∈Z\theta \in\left(n \pi-\frac{\pi}{6}, n \pi+\frac{\pi}{6}\right) ; n \in Z) is correct because the condition for non-real roots of the quadratic equation derived from the midpoint of the chord leads to the specified interval for θ\theta.
149 Medium
149 of 189
Application +4 | -1
The locus of the point which moves so that the ratio of the lengths of the tangents to the circle x2+y2−4y+3=0x^2+y^2-4 y+3=0 and x2+y2+6y+5=0x^2+y^2+6 y+5=0 is
View answer and explanation
Correct option: C
5x2+5y2+70y+33=05 x^2+5 y^2+70 y+33=0
Option C (5x2+5y2+70y+33=05 x^2+5 y^2+70 y+33=0) is correct because it represents the locus of a point where the ratio of the lengths of tangents to the given circles is a constant.
150 Medium
150 of 189
Application +4 | -1
The circles x2+y2+2x−2y+1=0       and      x2+y2−2x−2y+1=0{x^2} + {y^2} + 2x - 2y + 1 = 0\,\,\,\,\,\,\,\mathrm{and}\,\,\,\,\,\,{x^2} + {y^2} - 2x - 2y + 1 = 0 touch each other
View answer and explanation
Correct option: A
externally at (0,1)(0,1)
Option A (externally at (0,1)) is correct because the distance between the centers equals the sum of the radii, and the common tangent equation gives the point of contact.
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