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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 661-685 of 685 questions

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661 Easy
661 of 685
Application +4 | -1
Equation of a circle which passes through (3,6)(3,6) and touches the axes, is
View answer and explanation
Correct option: C
x2+y2−6x−6y+9=0x^2+y^2-6 x-6 y+9=0
Option C (x2+y2−6x−6y+9=0x^2+y^2-6 x-6 y+9=0) is correct because it represents a circle touching both axes and passing through the point (3,6).
662 Easy
662 of 685
Application +4 | -1
The equation of a tangent to the circle x2+y2=25x^2+y^2=25 passing through (−2,11)(-2,11), is
View answer and explanation
Correct option: A
4x+3y=254 x+3 y=25
The correct answer is 4x+3y=254 x+3 y=25 because the tangent equation is found by substituting the point into the general tangent equation and solving for the slope.
663 Easy
663 of 685
Application +4 | -1
The length of the tangent drawn to the circle x2+y2−2x+4y−11=0{x^2} + {y^2} - 2x + 4y - 11 = 0 from the point (1,3)\left( {1,3} \right) is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the length of the tangent is calculated by substituting the point's coordinates into the circle's equation and taking the square root.
664 Medium
664 of 685
Application +4 | -1
Find the equation of the chord of contact of the tangents drawn from (1,2)(1,2) to the circle x2+y2−2x+4y+7=0x^2+y^2-2 x+4 y+7=0.
View answer and explanation
Correct option: A
2y+5=02 y+5=0
Option A (2y+5=02 y+5=0) is correct because the equation of the chord of contact is found by substituting the point into the equation T=0.
665 Easy
665 of 685
Application +4 | -1
The chord of contact of tangents from a point PP to a circle passes through QQ. If l1l_1 and l2l_2 are the lengths of the tangents from PP and QQ to the circle, then PQP Q is equal to
View answer and explanation
Correct option: C
l12+l22\sqrt{l_1^2+l_2^2}
Option C (l12+l22\sqrt{l_1^2+l_2^2}) is correct because it represents the distance PQ, derived using the chord of contact and the lengths of tangents from P and Q.
666 Easy
666 of 685
Application +4 | -1
The diameter of a circle are along 2x+y−7=02x + y - 7 = 0 and x+3y−11=0x + 3y - 11 = 0. Then, the equation of this circle, which also passes through (5,7)\left( {5,7} \right), is
View answer and explanation
Correct option: C
x2+y2−4x−6y−12=0{x^2} + {y^2} - 4x - 6y - 12 = 0
Option C (x2+y2−4x−6y−12=0{x^2} + {y^2} - 4x - 6y - 12 = 0) is correct because it is the equation of the circle that satisfies the given conditions.
667 Easy
667 of 685
Application +4 | -1
The equation of a circle which has a tangent 3x+4y=63 x+4 y=6 and two normals given by (x−1)(y−2)=0(x-1)(y-2)=0 is
View answer and explanation
Correct option: C
x2+y2−2x−4y+4=0x^2+y^2-2 x-4 y+4=0
Option C (x2+y2−2x−4y+4=0x^2+y^2-2 x-4 y+4=0) is correct because the circle's center lies at the intersection of the normals and the radius is the distance from the center to the tangent.
668 Medium
668 of 685
Application +4 | -1
Find the equation of the chord of the circle x2+y2+6x+8y−x^2+y^2+6 x+8 y- 11=011=0, whose middle point is (1,−1)(1,-1).
View answer and explanation
Correct option: D
4x+3y−12=0;4x+3y−1=04 x+3 y-12=0 ; 4 x+3 y-1=0
Option D (4x+3y−12=0;4x+3y−1=04 x+3 y-12=0 ; 4 x+3 y-1=0) is correct because it represents the equation of the chord of the circle with the given midpoint, derived using the formula T = S1.
669 Easy
669 of 685
Application +4 | -1
If the equation of tangent to the circle x2+y2−2x+6y−6=0{x^2} + {y^2} - 2x + 6y - 6 = 0 parallel to 3x−4y+7=03x - 4y + 7 = 0 is 3x−4y+k=03x - 4y + k = 0, then the value of kk are
View answer and explanation
Correct option: A
5,−355, - 35
Option A (5,−355, - 35) is correct because these are the values of k for which the line 3x - 4y + k = 0 is tangent to the given circle.
670 Medium
670 of 685
Analyzing +4 | -1
Three distinct points A, B and C are given in the two-dimensional coordinate plane
such that the ratio of the distance of any one of them from (2, –1) to its distance from
(–1, 5) is 1 : 2. Then the centre of the circle passing through A, B and C is
View answer and explanation
Correct option: C
(3, –3)
Option C ((3, –3)) is correct because the center of the circle is the midpoint of the line segment joining the two given points.
671 Easy
671 of 685
Application +4 | -1
The shortest distance from the point (2,−7)(2,-7) to the circle x2+y2−14x−10y−151=0x^2+y^2-14 x-10 y-151=0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the shortest distance is the absolute difference between the distance from the point to the center and the radius of the circle.
672 Easy
672 of 685
Understanding +4 | -1
the equation ax2+2hxy+by2+2gx+2fy+c=0a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 represents a circle, the condition will be
View answer and explanation
Correct option: C
a=ba = b and h=0h = 0
Option C (a=ba = b and h=0h = 0) is correct because for a general second-degree equation to represent a circle, the coefficients of x2x^2 and y2y^2 must be equal, and the coefficient of xyxy must be zero.
673 Medium
673 of 685
Analyzing +4 | -1
The number of integral points on the circle, touching the parabola y2=8xy^2 = 8x at (2, 4) and passing through (0, 4), are equal to
View answer and explanation
Correct option: C
4
Option C (4) is correct because by finding the equation of the circle and checking for integer coordinates, we find there are 4 such points.
674 Easy
674 of 685
Application +4 | -1
The circle whose equations are x2+y2+c2=2ax{x^2} + {y^2} + {c^2} = 2ax and x2+y2+c2−2by=0{x^2} + {y^2} + {c^2} - 2by = 0 will touch one another externally if
View answer and explanation
Correct option: C
1a2+1b2=1c2\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{c^2}}}
The correct answer is C (1a2+1b2=1c2\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{c^2}}}) because it represents the condition for the two circles to touch each other externally, derived from the distance between their centers and the sum of their radii.
675 Easy
675 of 685
Application +4 | -1
A variable circle passes through the fixed point (2,0)\left( {2,0} \right) and touches yy-axis. Then, the locus of its centre is
View answer and explanation
Correct option: A
A parabola
Option A (A parabola) is correct because the locus of the center of a circle passing through a fixed point and touching the y-axis is a parabola.
676 Easy
676 of 685
Application +4 | -1
The equation of normal to the circle 2x2+2y2−2x−5y+3=02{x^2} + 2{y^2} - 2x - 5y + 3 = 0 at (1,1)\left( {1,1} \right) is
View answer and explanation
Correct option: C
x+2y=3x + 2y = 3
Option C (x+2y=3x + 2y = 3) is correct because it represents the equation of the normal to the given circle at the point (1,1).
677 Easy
677 of 685
Application +4 | -1
The pole of the straight line 9x+y−28=09x + y - 28 = 0 with respect to the circle 2x2+2y2−3x+5y−7=02{x^2} + 2{y^2} - 3x + 5y - 7 = 0 is
View answer and explanation
Correct option: C
(3,−1)\left( {3, - 1} \right)
Option C ((3, -1)) is correct because it is the pole of the given line with respect to the given circle, calculated using the polar equation.
678 Medium
678 of 685
Application +4 | -1
If f(x+y)=f(x)⋅f(y)f(x+y)=f(x) \cdot f(y) for all xx and y,f(1)=2y, f(1)=2 and αn=f(n),n∈N\alpha_n=f(n), n \in N, then the equation of the circle having (α1,α2)\left(\alpha_1, \alpha_2\right) and (α3,α4)\left(\alpha_3, \alpha_4\right) as the ends of its one diameter is
View answer and explanation
Correct option: A
(x−2)(x−8)+(y−4)(y−16)=0(x-2)(x-8)+(y-4)(y-16)=0
Option A ((x-2)(x-8)+(y-4)(y-16)=0) is correct because it uses the diametric form of a circle's equation with the correct coordinates derived from the given functional equation.
679 Easy
679 of 685
Application +4 | -1
Lines are drawn through the point P(−2,−3)P(-2,-3) to meet the circle x2+y2−2x−10y+1=0x^2+y^2-2 x-10 y+1=0. The length of the line segment PA,AP A, A being the point on the circle, where the line meets the circle at coincident points, is
View answer and explanation
Correct option: B
434 \sqrt{3} units
Option B (434 \sqrt{3} units) is correct because the length of the tangent from a point to a circle is calculated using the formula S1\sqrt{S_1}, where S_1 is obtained by substituting the point's coordinates into the circle's equation.
680 Easy
680 of 685
Analyzing +4 | -1
Which of the following statement(s) is true regarding the following circles?

x2+y2−4x−8y=0, x2+y2−8x−6y+20=0x^2+y^2-4 x-8 y=0,\, x^2+y^2-8 x-6 y+20=0
View answer and explanation
Correct option: B
These circles touch each other internally
Option B (These circles touch each other internally) is correct because the distance between their centers is equal to the difference of their radii.
681 Easy
681 of 685
Application +4 | -1
The number of common tangents to circles x2+y2+2x+8y−23=0{x^2} + {y^2} + 2x + 8y - 23 = 0 and x2+y2−4x−10y+9=0{x^2} + {y^2} - 4x - 10y + 9 = 0, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the distance between the centers is less than the sum of the radii, indicating two common tangents can be drawn.
682 Easy
682 of 685
Application +4 | -1
The lengths of the intercepts made by any circle on the coordinates axes are equal if the centre lies on the line (s) represented by
View answer and explanation
Correct option: C
x2−y2=0x^2-y^2=0
Option C (x2−y2=0x^2-y^2=0) is correct because equal intercepts on the axes imply that the center of the circle lies on the lines x=yx = y or x=−yx = -y.
683 Medium
683 of 685
Analyzing +4 | -1
If O{O} is the origin and OP,OQO P, O Q are distinct tangents to the circle x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0, the circumcenter of the triangle OPQO P Q is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the circumcenter of the triangle OPQ is (-g/2, -f/2), not (-g, -f) or (g, f) or (-f, -g).
684 Easy
684 of 685
Application +4 | -1
Locus of mid-point of chords of x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 that pass through the origin, is
View answer and explanation
Correct option: C
x2+y2+gx+fy=0x^2+y^2+g x+f y=0
Option C (x2+y2+gx+fy=0x^2+y^2+gx+fy=0) is correct because it represents the locus of the midpoint of the chords passing through the origin.
685 Easy
685 of 685
Analyzing +4 | -1
If y=mx+cy=m x+c is a tangent to the circle (x−3)2+y2=1(x-3)^2+y^2=1 and is also perpendicular to the tangent to the circle x2+y2=1x^2+y^2=1 at (1/2,/2)(1 / \sqrt{2}, / \sqrt{2}), then
View answer and explanation
Correct option: A
c2+6c+7=0c^2+6 c+7=0
Option A (c2+6c+7=0c^2+6 c+7=0) is correct because it represents the condition for the line to be tangent to the circle and perpendicular to the tangent of the other circle.
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