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JEE • MATHEMATICS • CIRCLES

Circles practice · Easy

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Showing 211-240 of 480 questions

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211 Easy
211 of 480
Application +4 | -1
The equation of the circle which touches both the axes and the straight line 4x+3y=64x + 3y = 6 in the first quadrant and lies below it is
View answer and explanation
Correct option: A
4x2+4y2−4x−4y+1=04{x^2} + 4{y^2} - 4x - 4y + 1 = 0
Option A (4x2+4y2−4x−4y+1=04{x^2} + 4{y^2} - 4x - 4y + 1 = 0) is correct because it represents a circle that touches both axes and the given line in the first quadrant, lying below the line.
212 Easy
212 of 480
Application +4 | -1
Tangents are drawn to x2+y2=1x^2+y^2=1 from any arbitrary point PP on the line 2x+y−4=02 x+y-4=0. The corresponding chord of contact passes through a fixed point whose coordinates are
View answer and explanation
Correct option: C
(12,14)\left(\frac{1}{2}, \frac{1}{4}\right)
Option C ((12,14)\left(\frac{{1}}{{2}}, \frac{{1}}{{4}}\right)) is correct because the chord of contact passes through this fixed point, determined by substituting the point P on the line into the chord of contact equation.
213 Easy
213 of 480
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}.
214 Easy
214 of 480
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.
215 Easy
215 of 480
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.
216 Easy
216 of 480
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.
217 Easy
217 of 480
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.
218 Easy
218 of 480
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.
219 Easy
219 of 480
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.
220 Easy
220 of 480
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.
221 Easy
221 of 480
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.
222 Easy
222 of 480
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.
223 Easy
223 of 480
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.
224 Easy
224 of 480
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.
225 Easy
225 of 480
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.

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226 Easy
226 of 480
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.
227 Easy
227 of 480
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.
228 Easy
228 of 480
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.
229 Easy
229 of 480
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.
230 Easy
230 of 480
Application +4 | -1
If the area of the circle 4x2+4y2−8x+16y+k=04 x^2+4 y^2-8 x+16 y+k=0 is 9π9 \pi square units, then the value of kk is
View answer and explanation
Correct option: C
−16-16
Option C (-16) is correct because when the area of the circle is 9π9\pi, the value of k is -16.
231 Easy
231 of 480
Application +4 | -1
The pole of the line lx+my+n=0l x+m y+n=0 with respect to the circle x2+y2=a2,x^2+y^2=a^2, is
View answer and explanation
Correct option: B
(−a2ln,−a2mn)\left(\frac{-a^2 l}{n}, \frac{-a^2 m}{n}\right)
Option B ((−a2ln,−a2mn)\left(\frac{{-a^2 l}}{{n}}, \frac{{-a^2 m}}{{n}}\right)) is correct because it represents the coordinates of the pole of the given line with respect to the circle, derived from comparing the equation of the polar.
232 Easy
232 of 480
Analyzing +4 | -1
The number of rational point (s) (a point (a,b)\left( {a,b} \right) is rational, if a and b both are rational numbers) on the circumference of a circle having centre (π,e)\left( {\pi ,e} \right) is
View answer and explanation
Correct option: A
At most one
Option A (At most one) is correct because if there were two rational points on the circle, the center would also have to be rational, contradicting the given irrational center.
233 Easy
233 of 480
Application +4 | -1
The locus of the point (l,m)\left( {l,m} \right) so that lx+my=1lx + my = 1 touches the circle x2+y2=a2{x^2} + {y^2} = {a^2}, is
View answer and explanation
Correct option: B
x2+y2=1a2{x^2} + {y^2} = \frac{1}{{{a^2}}}
The correct answer is x2+y2=1a2{x^2} + {y^2} = \frac{1}{{{{a^2}}}} because it represents the condition for the line to be tangent to the circle.
234 Easy
234 of 480
Application +4 | -1
Equation of the circle with centre (a2,b2)\left(\frac{a}{2}, \frac{b}{2}\right) and radius a2+b24\sqrt{\frac{a^2+b^2}{4}} is
View answer and explanation
Correct option: D
x2+y2−ax−by=0x^2+y^2-a x-b y=0
Option D (x2+y2−ax−by=0x^2+y^2-a x-b y=0) is correct because it represents the equation of a circle with the given center and radius.
235 Easy
235 of 480
Application +4 | -1
Area of an equilateral triangle inscribed in a circle of radius aa is
View answer and explanation
Correct option: C
33a2/43 \sqrt{3} a^2 / 4
Option C (33a2/43 \sqrt{3} a^2 / 4) is correct because it is the area of an equilateral triangle inscribed in a circle of radius a.
236 Easy
236 of 480
Application +4 | -1
The normal to the circle x2+y2−3x−6y−10=0x^2+y^2-3 x-6 y-10=0 at the point (−3,4)(-3,4) is
View answer and explanation
Correct option: A
2x+9y−30=02 x+9 y-30=0
Option A (2x+9y−30=02 x+9 y-30=0) is correct because it represents the equation of the normal to the given circle at the point (-3,4), derived using the normal equation formula.
237 Easy
237 of 480
Application +4 | -1
The angle between two tangents drawn from the origin to the circle (x−7)2+(y+1)2=25(x-7)^2+(y+1)^2=25, is
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{\pi}{2}) is correct because the origin lies on the director circle of the given circle, implying the tangents from the origin are perpendicular.
238 Easy
238 of 480
Application +4 | -1
If the circle x2+y2+4x+22y+I=0x^2+y^2+4 x+22 y+I=0 bisects the circumference of the circle x2+y2−2x+8y−m=0x^2+y^2-2 x+8 y-m=0, then l+ml+m is equal to
View answer and explanation
Correct option: B
50
Option B (50) is correct because the condition for one circle bisecting the circumference of another leads to the equation l+m=50l+m=50.
239 Easy
239 of 480
Application +4 | -1
The square of the length of the tangent from (3,−4)\left( {3, - 4} \right) to the circle x2+y2−4x−6y+3=0{x^2} + {y^2} - 4x - 6y + 3 = 0, is
View answer and explanation
Correct option: C
40
The correct answer is 40 because it is the square of the length of the tangent from the point (3, -4) to the given circle.
240 Easy
240 of 480
Application +4 | -1
If the point (a,−a)(a,-a) lies inside the circle x2+y2−4x+2y−8=0x^2+y^2-4 x+2 y-8=0, then a lies in the interval
View answer and explanation
Correct option: A
(−1,4)(-1,4)
Option A ((−1,4))((-1,4)) is correct because it represents the interval in which 'a' must lie for the point (a, -a) to be inside the given circle.
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