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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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361 Medium
361 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)(a, 0) to the circle x2+y2=1x^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)(-\sqrt{2},-1) \cup(1, \sqrt{2})
Option D ((−2,−1)∪(1,2)(-\sqrt{{2}},-1) \cup (1, \sqrt{{2}})) is correct because the condition π2<θ<π\frac{{\pi}}{{2}} < \theta < \pi implies that tan⁡θ<0\tan \theta < 0, which leads to the given range for 'a'.
362 Easy
362 of 685
Analyzing +4 | -1
If two circles a(x2+y2)+bx+cy=0a\left(x^2+y^2\right)+b x+c y=0 and A(x2+y2)+Bx+Cy=0A\left(x^2+y^2\right)+B x+C y=0 touch each other then
View answer and explanation
Correct option: B
bC=cBb C=c B
Option B (bC=cBb C=c B) is correct because the circles touch each other if they have a common tangent at the origin, leading to the given condition.
363 Easy
363 of 685
Application +4 | -1
The locus of a point which moves such that the tangents from it to the two circles x2+y2−5x−3=0x^2+y^2-5 x-3=0 and 3x2+3y2+2x+4y−6=03 x^2+3 y^2+2 x+4 y-6=0 are equal, is given by
View answer and explanation
Correct option: B
17x+4y+3=017 x+4 y+3=0
Option B (17x+4y+3=017 x+4 y+3=0) is correct because it represents the radical axis of the two circles, which is the locus of points with equal tangent lengths.
364 Easy
364 of 685
Analyzing +4 | -1
If the circles x2+y2+2ax+b=0x^2+y^2+2 a x+b=0 and x2+y2+2cx+b=0x^2+y^2+2 c x+b=0 touch each other, then
View answer and explanation
Correct option: C
b=0b=0
Option C (b=0b=0) is correct because the condition for the circles to touch each other requires the common tangent to have equal roots, leading to b=0.
365 Easy
365 of 685
Analyzing +4 | -1
The set of values of cc, so that the equations y=∣x∣+cy=|x|+c and x2+y2−8∣x∣−9=0x^2+y^2-8|x|-9=0 have no solution, is
View answer and explanation
Correct option: D
(52−4,∞)(5 \sqrt{2}-4, \infty)
Option D ((52−4,∞)(5 \sqrt{2}-4, \infty)) is correct because the condition for no solution is that the constant c in y=|x|+c must be greater than the y-intercept of the tangent to the circle parallel to y=x.
366 Medium
366 of 685
Application +4 | -1
If the circles x2+y2+4x+8y=0{x^2} + {y^2} + 4x + 8y = 0 and x2+y2+8x+2ky=0{x^2} + {y^2} + 8x + 2ky = 0 touch each other, then kk is equal to
View answer and explanation
Correct option: B
8
Option B (8) is correct because solving the equation derived from the condition that the circles touch each other yields k = 8.
367 Medium
367 of 685
Application +4 | -1
Tangents drawn from point P(1,8)P(1,8) to the circle x2+y2−6x−4y−11=0x^2+y^2-6 x-4 y-11=0 touch the circle at the points AA and BB. The equation of the circumcircle of the triangle PABP A B is
View answer and explanation
Correct option: B
x2+y2−4x−10y+19=0x^2+y^2-4 x-10 y+19=0
Option B (x2+y2−4x−10y+19=0x^2+y^2-4 x-10 y+19=0) is correct because CP is the diameter of the circumcircle of triangle PAB, and the equation is derived accordingly.
368 Easy
368 of 685
Understanding +4 | -1
The circle (x−r)2+(y−r)2=r2(x-r)^2+(y-r)^2=r^2 touches
View answer and explanation
Correct option: C
both the coordinate axes
Option C (both the coordinate axes) is correct because the distance of the center from both axes is equal to the radius.
369 Easy
369 of 685
Application +4 | -1
Equation of a circle with centre (4,3)(4,3) touching the circle x2+y2=1x^2+y^2=1 is
View answer and explanation
Correct option: D
x2+y2−8x−6y+9=0x^2+y^2-8 x-6 y+9=0
Option D (x2+y2−8x−6y+9=0x^2+y^2-8 x-6 y+9=0) is the correct answer because it represents a circle with center (4,3) that touches the circle x2+y2=1x^2 + y^2 = 1 externally.
370 Easy
370 of 685
Application +4 | -1
The locus of the point of intersection of the tangents to the circle x=rcos⁡θ,y=rsin⁡θx=r \cos \theta, y=r \sin \theta at points whose parametric angles differ by π3\frac{\pi}{3} is
View answer and explanation
Correct option: D
3(x2+y2)=4r23\left(x^2+y^2\right)=4 r^2
Option D (3(x2+y2)=4r23(x^2+y^2)=4 r^2) is correct because it represents the locus of the intersection point of tangents to the circle whose parametric angles differ by π3\frac{\pi}{3}.
371 Medium
371 of 685
Application +4 | -1
A circle passes through the origin OO and cuts the axes at A(a,0)A(a, 0) and B(0,b)B(0, b). The reflection of origin OO in the line ABA B is the point
View answer and explanation
Correct option: C
(2aba2+b2,2aba2+b2)\left(\frac{2 a b}{a^2+b^2}, \frac{2 a b}{a^2+b^2}\right)
The correct answer is B (2a2ba2+b2,2ab2a2+b2)\left(\frac{{2 a^2 b}}{{{{a^2+b^2}}}}, \frac{{2 a b^2}}{{{{a^2+b^2}}}}\right) because it represents the reflection of the origin in the line AB, calculated using perpendicularity and equidistance conditions.
372 Easy
372 of 685
Application +4 | -1
If (a,b)(a, b) is a point on the chord ABA B of the circle, where the ends of the chord are A=(2,−3)A=(2,-3) and B=(3,2)B=(3,2), then
View answer and explanation
Correct option: B
a∈(2,3),b∈(−3,2)a \in(2,3), b \in(-3,2)
Option B (a∈(2,3),b∈(−3,2)a \in(2,3), b \in(-3,2)) is correct because if (a, b) lies on the line segment AB, then 'a' must lie between the x-coordinates of A and B, and 'b' must lie between the y-coordinates of A and B.
373 Medium
373 of 685
Application +4 | -1
If two circles, each of radius 5 unit, touch each other at (1, 2) and the equation of their common tangent is 4x+3y=104x + 3y = 10, then equation of the circle a portion of which lies in all the quadrants, is
View answer and explanation
Correct option: B
x2+y2+6x+2y−15=0{x^2} + {y^2} + 6x + 2y - 15 = 0
Option B x2+y2+6x+2y−15=0{x^2} + {y^2} + 6x + 2y - 15 = 0 is correct because it represents a circle with radius 5, touching the given line at (1,2), and having a portion in all quadrants.
374 Easy
374 of 685
Analyzing +4 | -1
If a circle having centre at (α,β)(\alpha, \beta) radius rr completely lies with in two lines x+y=2x+y=2 and x+y=−2x+y=-2, then, min⁡.(∣α+β+2∣,∣α+β−2∣)\min .(|\alpha+\beta+2|,|\alpha+\beta-2|) is
View answer and explanation
Correct option: A
greater than 2r\sqrt{2} r
Option A (greater than 2r\sqrt{2} r) is correct because the minimum distance of the center from either line must be greater than the radius of the circle.
375 Medium
375 of 685
Application +4 | -1
If the tangent from a point PP to the circle x2+y2=1x^2+y^2=1 is perpendicular to the tangent from PP to the circle x2+y2=3x^2+y^2=3, then the locus of PP is a
View answer and explanation
Correct option: A
circle of radius 2
Option A (circle of radius 2) is correct because the locus of P is a circle with radius 2, derived from the perpendicular tangent condition.

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376 Easy
376 of 685
Application +4 | -1
The value of cc for which line y=2x+cy=2 x+c is tangent to circle x2+y2=(5)2x^2+y^2=(\sqrt{5})^2 is
View answer and explanation
Correct option: B
±5\pm 5
Option B is correct because the value of c for the line y = 2x + c to be tangent to the circle x2 + y2 = (sqrt(5))2 is ±5, which is found using the tangency condition.
377 Easy
377 of 685
Application +4 | -1
The centre of a circle is (2,−3)\left( {2, - 3} \right) and the circumference is 10π10\pi. Then, the equation of the circle 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 represents the equation of a circle with center (2, -3) and radius 5, derived from the given circumference.
378 Medium
378 of 685
Application +4 | -1
S:x2+y2–6x+4y–3=0S: x^2+y^2– 6x + 4y – 3=0 is a circle and L:4x+3y+19=0L : 4x + 3y + 19 = 0 is a straight line.
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the perpendicular distance from the center of the circle to the line is greater than the radius, meaning the line lies outside the circle.
379 Easy
379 of 685
Application +4 | -1
The equation of the circle passing through (4, 5) and having the centre (2, 2), is
View answer and explanation
Correct option: B
x2+y2−4x−4y−5=0{x^2} + {y^2} - 4x - 4y - 5 = 0
Option B (x2+y2−4x−4y−5=0{x^2} + {y^2} - 4x - 4y - 5 = 0) is correct because it represents the equation of the circle passing through (4, 5) and having the center (2, 2).
380 Easy
380 of 685
Application +4 | -1
Extremities of a diagonal of a rectangle are (0,0)\left( {0,0} \right) and (4,3)\left( {4,3} \right). The equations of the tangents to the circumcircle of the rectangle which are parallel to the diagonal are
View answer and explanation
Correct option: B
6x−8y±25=06x - 8y \pm 25 = 0
Option B (6x−8y±25=06x - 8y \pm 25 = 0) is correct because it represents the tangent lines to the circumcircle that are parallel to the given diagonal.
381 Medium
381 of 685
Analyzing +4 | -1
The circle x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2\mathrm{g}x + 2fy + c = 0 cuts the parabola x2=4ay{x^2} = 4ay at points (xi,yi),  i=1,  2,  3,  4,\left( {{x_i},{y_i}} \right),\;i = 1,\;2,\;3,\;4, then
View answer and explanation
Correct option: B
∑yi=−4(f+2a)\sum {y_i} = - 4\left( {f + 2a} \right)
Option B (∑yi=−4(f+2a)\sum {y_i} = - 4\left( {f + 2a} \right)) is correct because it is derived from Vieta's formulas applied to the quartic equation obtained by substituting the parabola equation into the circle equation.
382 Medium
382 of 685
Application +4 | -1
From the point P(2+32cos⁡θ,3+32sin⁡θ),0<θ<2π{P}(2+3 \sqrt{2} \cos \theta, 3+3 \sqrt{2} \sin \theta), 0 < \theta < 2 \pi, tangents are drawn to the circle x2+y2−4x−6y+4=0x^2+y^2-4 x-6 y+4=0, then the angle between them is
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{\pi}{2}) is correct because point P lies on the director circle of the given circle, implying tangents from P are perpendicular.
383 Medium
383 of 685
Analyzing +4 | -1
Let PP and QQ be any points on the curves (x−1)2+(x-1)^2+ (y+1)2=1(y+1)^2=1 and y=x2y=x^2, respectively. The distance between PP and QQ is minimum for some value of the abscissa of PP in the interval
View answer and explanation
Correct option: C
(14,12)\left(\frac{1}{4}, \frac{1}{2}\right)
Option C (14,12)\left(\frac{1}{4}, \frac{1}{2}\right) is correct because it involves finding the point on the circle closest to the parabola, which requires analyzing the normal to the parabola and its intersection with the circle.
384 Easy
384 of 685
Application +4 | -1
If P1,P2,P3{P_1},{P_2},{P_3} are the perimeter of the three circles x2+y2+8x−6y=0,  4x2+4y2−4x−12y−186=0{x^2} + {y^2} + 8x - 6y = 0,\;4{x^2} + 4{y^2} - 4x - 12y - 186 = 0 and x2+y2−6x+6y−9=0{x^2} + {y^2} - 6x + 6y - 9 = 0 respectively, then
View answer and explanation
Correct option: B
P1<P3<P2{P_1} < {P_3} < {P_2}
Option B (P1<P3<P2{P_1} < {P_3} < {P_2}) is correct because it accurately compares the perimeters of the three circles based on their calculated radii.
385 Easy
385 of 685
Application +4 | -1
If lx+my=1l x+m y=1 touches the circle x2+y2=a2x^2+y^2=a^2 then the point (l,m)(l, m) lies on the circle
View answer and explanation
Correct option: A
x2+y2=a−2x^2+y^2=a^{-2}
Option A (x2+y2=a−2x^2+y^2=a^{-2}) is correct because the condition for the line to touch the circle leads to the point (l, m) lying on this circle.
386 Medium
386 of 685
Application +4 | -1
If the polar of a point (α,β)(\alpha, \beta) with respect to any one of the circle x2+y2−2kx+3=0x^2+y^2-2 k x+3=0, where kk is a variable, always passes through a fixed point, whatever be the value of kk, then the fixed point is
View answer and explanation
Correct option: A
(−α,1β(α2−3))\left(-\alpha, \frac{1}{\beta}\left(\alpha^2-3\right)\right)
Option A ((−α,1β(α2−3))\left(-\alpha, \frac{{1}}{{\beta}}\left(\alpha^2-3\right)\right)) is correct because the polar equation must be independent of k, leading to two equations that can be solved for the fixed point.
387 Medium
387 of 685
Application +4 | -1
If tangents are drawn from any point on the line x+y=3x+y=3 to the circle x2+y2=9x^2+y^2=9 then the chord of contact passes through the point
View answer and explanation
Correct option: C
(3,3)(3,3)
Option C ((3,3)) is correct because the chord of contact for tangents drawn from a point on the line x+y=3 to the circle x2+y2=9 always passes through the point (3,3).
388 Easy
388 of 685
Application +4 | -1
The coordinates of the middle point of the chord intercepted on line lx+my+n=0l x+m y+n=0 by the circle x2+y2=a2x^2+y^2=a^2 are
View answer and explanation
Correct option: B
(−ln⁡l2+m2,−mnl2+m2)\left(\frac{-\ln }{l^2+m^2}, \frac{-m n}{l^2+m^2}\right)
Option B ((−ln⁡l2+m2,−mnl2+m2)\left(\frac{-\ln }{l^2+m^2}, \frac{-m n}{l^2+m^2}\right)) is correct because it represents the midpoint of the chord intercepted by the line on the circle.
389 Easy
389 of 685
Application +4 | -1
Equation of a circle S(x,y)=0,(S(2,3)=16)S(x, y)=0,(S(2,3)=16) which touches the line 3x+4y−7=03 x+4 y-7=0 at (1,1)(1,1) is given by
View answer and explanation
Correct option: A
x2+y2+x+2y−5=0x^2+y^2+x+2 y-5=0
Option A (x2+y2+x+2y−5=0x^2+y^2+x+2 y-5=0) is correct because it satisfies the given conditions of touching the line at (1,1) and S(2,3)=16.
390 Easy
390 of 685
Application +4 | -1
The equation of the circle touching x=0,y=0x = 0,y = 0 and x=4x = 4 is
View answer and explanation
Correct option: D
x2+y2−4x−4y+4=0{x^2} + {y^2} - 4x - 4y + 4 = 0
Option D (x2+y2−4x−4y+4=0{x^2} + {y^2} - 4x - 4y + 4 = 0) is correct because it represents a circle touching x = 0, y = 0, and x = 4, which means its center is at (2, 2) and its radius is 2.
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