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JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

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Showing 301-330 of 685 questions

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301 Easy
301 of 685
Application +4 | -1
The length of the tangent from (0,0)(0,0) to the circle 2(x2+y2)+x−y+5=02\left(x^2+y^2\right)+x-y+5=0, is
View answer and explanation
Correct option: D
5/2\sqrt{5 / 2}
Option D (5/2\sqrt{5 / 2}) is correct because it is the length of the tangent calculated using the formula S1\sqrt{S_1}.
302 Medium
302 of 685
Analyzing +4 | -1
If PP and QQ are the points of intersection of the circles x2+y2+3x+7y+2p−5=0x^2+y^2+3 x+7 y+2 p-5=0 and x2+y2+2x+2y−p2=0x^2+y^2+2 x+2 y-p^2=0, then there is a circle passing through P,QP, Q and (1,1)(1,1) for
View answer and explanation
Correct option: A
all values of pp
Option A (all values of pp) is correct because the equation of the circle passing through the intersection points P and Q exists for all values of p.
303 Easy
303 of 685
Analyzing +4 | -1
For the given circles x2+y2−6x−2y+1=0x^2+y^2-6 x-2 y+1=0 and x2+y2+2x−8y+13=0x^2+y^2+2 x-8 y+13=0, which of the following is true?
View answer and explanation
Correct option: D
They touch each other externally
Option D (They touch each other externally) is correct because the distance between the centers equals the sum of the radii.
304 Medium
304 of 685
Analyzing +4 | -1
Tangents OPO P and OQO Q are drawn from the origin OO to the circle x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0. Then, the equation of the circumcircle of the triangle OPQO P Q, is
View answer and explanation
Correct option: B
x2+y2+gx+fy=0x^2+y^2+g x+f y=0
The correct answer is B (x2+y2+gx+fy=0x^2+y^2+g x+f y=0) because the circumcircle of the triangle OPQ passes through the origin and the intersection of the given circle and the chord of contact.
305 Easy
305 of 685
Application +4 | -1
The area of the circle centred at (1,2)\left( {1,2} \right) and passing through (4,6)\left( {4,6} \right), is
View answer and explanation
Correct option: C
25πsq units\mathrm{25\pi sq }\,\mathrm{units}
Option C (25πsq units\mathrm{25\pi sq }\,\mathrm{units}) is correct because the area of the circle is calculated using the distance between the center and a point on the circle as the radius.
306 Easy
306 of 685
Application +4 | -1
An equilateral triangle is inscribed in the circle x2+y2x^2+y^2 =1=1 with one vertex at the point (1,0)(1,0). Length of each side of the triangle is
View answer and explanation
Correct option: D
3\sqrt{3}
Option D (3\sqrt{3}) is correct because the side length of the equilateral triangle inscribed in the circle is calculated using the geometry of the circle and triangle.
307 Easy
307 of 685
Analyzing +4 | -1
The number of common tangents to the circles x2+y2=4{x^2} + {y^2} = 4 and x2+y2−6x−8y+24=0{x^2} + {y^2} - 6x - 8y + 24 = 0 is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the distance between the centers of the circles is greater than the sum of their radii, indicating 4 common tangents.
308 Medium
308 of 685
Application +4 | -1
Let C1C_1 and C2C_2 be two circles of radius r1r_1 and r2r_2 units respectively touching both the axes. If the two circles are orthogonal, then r1r2+r2r1\frac{r_1}{r_2}+\frac{r_2}{r_1} is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because it uses the orthogonality condition for two circles to find the relationship between their radii.
309 Easy
309 of 685
Analyzing +4 | -1
A straight line moves such that the algebraic sum of the perpendicular drawn to it from two fixed points is equal to 2k2 k. Then the straight line always touches a fixed circle of radius
View answer and explanation
Correct option: C
kk
The correct answer is C (kk) because the straight line touches a fixed circle with a radius equal to kk, derived from the given condition.
310 Easy
310 of 685
Application +4 | -1
The number of the tangents that can be drawn from (1,2)\left( {1,2} \right) and x2+y2=5,{x^2} + {y^2} = 5, is
View answer and explanation
Correct option: A
1
Option B (2) is incorrect. Option A (1) is correct because the point (1,2) lies ON the circle x2 + y2 = 5, meaning there is only ONE tangent at that point.
311 Easy
311 of 685
Application +4 | -1
If 2(x2+y2)+4λx+λ2=02\left(x^2+y^2\right)+4 \lambda x+\lambda^2=0 represents a circle of meaningful radius, then the range of real values of λ\lambda, is
View answer and explanation
Correct option: A
RR
Option A (RR) is correct because the condition for the equation to represent a circle of meaningful radius is satisfied for all real values of λ\lambda.
312 Medium
312 of 685
Application +4 | -1
S:x2+y2+6x−14y−6=0S: x^2+y^2+6 x-14 y-6=0 is a circle and L:7x+3y+58=0L: 7 x+3 y+58=0 is a straight line
View answer and explanation
Correct option: B
LL is a chord of SS
Option B (LL is a chord of SS) is correct because the distance from the center of the circle to the line is less than the radius of the circle.
313 Easy
313 of 685
Application +4 | -1
A common tangent to circle x2+y2=16{x^2} + {y^2} = 16 and an ellipse is x249+y24=1\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{4} = 1 is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the calculated common tangent equation does not match any of the given options.
314 Easy
314 of 685
Application +4 | -1
Which of the following equations gives circle?
View answer and explanation
Correct option: A
r=2sin⁡θr = 2\sin \mathrm{\theta }
The correct answer is A (r=2sin⁡θr = 2\sin \mathrm{\theta }) because it represents the polar equation of a circle passing through the origin.
315 Medium
315 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 circles touch each other internally, resulting in only one common tangent.

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316 Easy
316 of 685
Application +4 | -1
The equation of a circle C1C_1 is x2+y2−4x−2y−11=0x^2+y^2-4 x-2 y-11=0. A circle C2C_2 of radius 1 rolls on the outside of the circle C1C_1. The locus of the centre of circle C2C_2 has the equation
View answer and explanation
Correct option: A
x2+y2−4x−2y−20=0x^2+y^2-4 x-2 y-20=0
Option A (x2+y2−4x−2y−20=0x^2+y^2-4 x-2 y-20=0) is correct because the locus of the center of the rolling circle is another circle with the same center as the original circle, but with a radius equal to the sum of the radii of the two circles.
317 Easy
317 of 685
Application +4 | -1
Locus of the point of intersection of perpendicular tangents to the circle x2+y2=16{x^2} + {y^2} = 16 is
View answer and explanation
Correct option: B
x2+y2=32{x^2} + {y^2} = 32
Option B (x2+y2=32{x^2} + {y^2} = 32) is correct because the locus of the intersection of perpendicular tangents to a circle is its director circle, which has a radius of 2\sqrt{2} times the original circle's radius.
318 Medium
318 of 685
Analyzing +4 | -1
The common chord of the circle x2+y2+6x+8y−7=0x^2+y^2+6 x+8 y-7=0 and a circle passing through the origin, and touching the line y=xy=x, always passes through the point
View answer and explanation
Correct option: C
(1/2,1/2)(1 / 2,1 / 2)
Option C ((1/2,1/2)(1 / 2,1 / 2)) is correct because the common chord always passes through the intersection of two lines derived from the given circle equations.
319 Medium
319 of 685
Application +4 | -1
The locus of the centre of the circles which touches both the axes is given by
View answer and explanation
Correct option: A
x2−y2=0x^2-y^2=0
Option A (x2−y2=0x^2-y^2=0) is correct because the locus of the center is given by |x| = |y|, which simplifies to x² - y² = 0.
320 Medium
320 of 685
Application +4 | -1
A line is drawn through the point P(3,11)P\left( {3,11} \right) to cut the circle x2+y2=9{x^2} + {y^2} = 9 at AA and BB. Then, PA.PBPA.PB is equal to
View answer and explanation
Correct option: B
121
Option B (121) is correct because PA.PB is equal to the power of the point P with respect to the circle, which is calculated by substituting the coordinates of P into the circle's equation.
321 Medium
321 of 685
Application +4 | -1
The equation of the circle, which is the mirror image of the circle, x2+y2−2x=0x^2+y^2-2 x=0, in the line, y=3y=3 −x-x is
View answer and explanation
Correct option: A
x2+y2−6x−4y+12=0x^2+y^2-6 x-4 y+12=0
Option A (x2+y2−6x−4y+12=0x^2+y^2-6 x-4 y+12=0) is correct because it represents the circle obtained by finding the mirror image of the center of the original circle with respect to the given line and keeping the same radius.
322 Easy
322 of 685
Application +4 | -1
The lines 3x−4y+4=03x - 4y + 4 = 0 and 6x−8y−7=06x - 8y - 7 = 0 are tangents to the same circle. Then, its radius is
View answer and explanation
Correct option: C
34\frac{3}{4}
The correct answer is 34\frac{3}{4} because the radius is half the distance between the two parallel tangent lines.
323 Easy
323 of 685
Analyzing +4 | -1
Consider the circles x2+(y−1)2=9,(x−1)2+y2=25.{x^2} + {\left( {y - 1} \right)^2} = 9,{\left( {x - 1} \right)^2} + {y^2} = 25. They are such that
View answer and explanation
Correct option: B
One of these circles lies entirely inside the other
Option B (One of these circles lies entirely inside the other) is correct because the distance between the centers is less than the difference of the radii.
324 Medium
324 of 685
Application +4 | -1
Equation of the circle which of the mirror image of the circle x2+y2−2x=0{x^2} + {y^2} - 2x = 0 in the line x+y=2x + y = 2 is
View answer and explanation
Correct option: C
x2+y2−4x−2y+4=0{x^2} + {y^2} - 4x - 2y + 4 = 0
Option C (x2+y2−4x−2y+4=0{x^2} + {y^2} - 4x - 2y + 4 = 0) is correct because it represents the mirror image of the given circle in the line x + y = 2, found by finding the image of the center.
325 Easy
325 of 685
Application +4 | -1
The equation Im⁡(iz−2z−i)+1=0,z∈C,z≠i\operatorname{Im}\left(\frac{i z-2}{z-i}\right)+1=0, z \in \mathrm{C}, z \neq i represents a part of a circle having radius equal to :
View answer and explanation
Correct option: C
34\frac{3}{4}
Option C (34\frac{3}{4}) is correct because it represents the radius of the circle obtained after simplifying the given complex equation.
326 Easy
326 of 685
Analyzing +4 | -1
The two circles x2+y2−2x+6y+6=0{x^2} + {y^2} - 2x + 6y + 6 = 0 and x2+y2−5x+6y+15=0{x^2} + {y^2} - 5x + 6y + 15 = 0 touch each other
View answer and explanation
Correct option: B
Internally
Option B (Internally) is correct because the distance between the centers of the two circles is equal to the difference of their radii, indicating that they touch internally.
327 Medium
327 of 685
Application +4 | -1
A point moves in such a manner that the sum of the squares of its distances from the vertices of a triangle is constant. Then, the locus of the point is
View answer and explanation
Correct option: D
a circle
Option D (a circle) is correct because the locus of the point, satisfying the given condition, results in the equation of a circle.
328 Easy
328 of 685
Application +4 | -1
The tangent at any point to the circle x2+y2x^2+y^2 =r2=r^2 meets the coordinate axes at AA and BB. If lines drawn parallel to the coordinate axes through AA and BB intersect at PP, the locus of PP is
View answer and explanation
Correct option: C
1x2+1y2=1r2\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{r^2}
Option C (1x2+1y2=1r2\frac{{1}}{{x^2}}+\frac{{1}}{{y^2}}=\frac{{1}}{{r^2}}) is correct because it represents the locus of point P based on the tangent to the circle and the parallel lines.
329 Medium
329 of 685
Analyzing +4 | -1
The set of values of k,k, for which the circle C:4x2+4y2−12x+8y+k=0C:4{x^2} + 4{y^2} - 12x + 8y + k = 0 lies inside the fourth quadrant and the point (1,−13)\left( {1, - \frac{1}{3}} \right) lies on or inside the circle C,C, is
View answer and explanation
Correct option: D
(9,929]\left( {9,\frac{{92}}{9}} \right]
Option D (9,929]\left( {9,\frac{{92}}{9}} \right] is correct because it represents the set of k values for which the given circle lies inside the fourth quadrant and the point (1, -1/3) lies on or inside the circle.
330 Medium
330 of 685
Analyzing +4 | -1
The common chord of the circle x2+y2+8x+4y−5=0x^2+y^2+8 x+4 y-5=0 and a circle passing through the origin and touching the line y=xy=x passes through the fixed point
View answer and explanation
Correct option: A
(512,512)\left(\frac{5}{12}, \frac{5}{12}\right)
Option A is correct because the common chord equation is derived from subtracting the two circle equations, and the resulting line passes through the point (5/12, 5/12).
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