Skip to main content
Back to CIRCLES

Your Progress

0 / 685 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • CIRCLES

Circles practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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.

Take a chapter quiz →
Showing 1-30 of 685 questions
1 Hard
1 of 685
Application +4 | -1
The sum of the coefficients in the expansion of (α2x2−2αx+1)51,{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}}, as a polynomial in xx, vanishes. Position of the point (α,2α2)\left( {\alpha ,2{\alpha ^2}} \right) with respect to the circle x2+y2=4,{x^2} + {y^2} = 4, is
View answer and explanation
Correct option: A
Outside
Option A (Outside) is correct because substituting α=1\alpha = 1 into the circle's equation shows that the point (1, 2) lies outside the circle.
2 Easy
2 of 685
Application +4 | -1
One of the diameter of the circle x2+y2−2x+4y−4=0{x^2} + {y^2} - 2x + 4y - 4 = 0 is
View answer and explanation
Correct option: A
x−y−3=0x - y - 3 = 0
Option A (x−y−3=0x - y - 3 = 0) is correct because the diameter of a circle passes through its center, and this line equation satisfies the center's coordinates.
3 Medium
3 of 685
Application +4 | -1
The locus of the middle point of the chords of the circle x2+y2=a2{x^2} + {y^2} = {a^2} such that the chords pass through a given point (x1,y1)\left( {{x_1},{y_1}} \right), is
View answer and explanation
Correct option: A
x2+y2−xx1−yy1=0{x^2} + {y^2} - x{x_1} - y{y_1} = 0
Option A (x2+y2−xx1−yy1=0{x^2} + {y^2} - x{x_1} - y{y_1} = 0) is correct because it represents the locus of the midpoint of the chords passing through a given point.
4 Easy
4 of 685
Analyzing +4 | -1
α,β\quad \alpha, \beta and γ\gamma are the parametric angles of three points P,QP, Q, and R respectively, on the circle x2+y2=1x^2+y^2=1, and AA is the point (−1,0)(-1,0). If the lengths of the chords AP,AQA P, A Q and ARA R are in GPG P, then cos⁡α/2,cos⁡β/2\cos \alpha / 2, \cos \beta / 2 and cos⁡γ/2\cos \gamma / 2 are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the lengths of the chords AP, AQ, and AR being in GP implies that cos(alpha/2), cos(beta/2), and cos(gamma/2) are also in GP.
5 Easy
5 of 685
Application +4 | -1
The radius of the circle passing through the point (6,2), two of whose diameters are x+y=6x+y=6 and x+2y=4x+2 y=4 is
View answer and explanation
Correct option: B
252 \sqrt{5}
Option B (252 \sqrt{5}) is correct because it is the distance between the center of the circle (intersection of diameters) and a point on the circle (6,2).
6 Easy
6 of 685
Application +4 | -1
If a circle has two of its diameters along the lines x+y=5x+y=5 and x−y=1x-y=1 and has area 9π9 \pi, then the equation of the circle is
View answer and explanation
Correct option: A
x2+y2−6x−4y+4=0x^2+y^2-6 x-4 y+4=0
Option A (x2+y2−6x−4y+4=0x^2+y^2-6 x-4 y+4=0) is correct because the center is the intersection of the diameters, and the radius is derived from the area.
7 Easy
7 of 685
Analyzing +4 | -1
Four distinct points (2k,3k),(1,0),(0,1)\left( {2k,3k} \right),\left( {1,0} \right),\left( {0,1} \right) and (0,0)\left( {0,0} \right) lie on a circle for
View answer and explanation
Correct option: D
For two values of kk
Option D (For two values of kk) is correct because there are two non-trivial values of kk for which the four points lie on a circle.
8 Easy
8 of 685
Application +4 | -1
The radius of the circle whose arc of length 15 π  cm15\,\pi \,\,\mathrm{cm} makes an angle of 3 π4\frac{{3\,\pi }}{4} radian at the centre is
View answer and explanation
Correct option: B
20  cm20\,\,\mathrm{cm}
Option B (20 cm) is correct because the radius is calculated using the formula relating arc length, angle, and radius: arc length = radius * angle.
9 Easy
9 of 685
Application +4 | -1
Equation of the tangent to the circle at the point (1,−1)(1,-1) whose centre is the point of intersection of the straight lines x−y=1x-y=1 and 2x+y−3=02 x+y-3=0, is
View answer and explanation
Correct option: B
x+4y+3=0x+4 y+3=0
The correct answer is x+4y+3=0x+4 y+3=0 because it is derived by substituting the point and center into the tangent equation.
10 Hard
10 of 685
Analyzing +4 | -1
The values of θ\theta in [0,2π]\left[ {0,2\pi } \right] so that circles x2+y2+2(sin⁡α)x+2(cos⁡α)y+sin⁡2θ=0{x^2} + {y^2} + 2\left( {\sin \alpha } \right)x + 2\left( {\cos \alpha } \right)y + {\sin ^2}\theta = 0 always lies inside the square of unit side length, is /are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct range for θ\theta is [π/3,2π/3]∪[4π/3,5π/3]\left[ {\pi /3,2\pi /3} \right] \cup \left[ {4\pi /3,5\pi /3} \right], which is not listed in the options.
11 Easy
11 of 685
Application +4 | -1
The equation of the image of the circle x2+y2+16x−24y+183=0{x^2} + {y^2} + 16x - 24y + 183 = 0 by the line mirror 4x+7y+13=04x + 7y + 13 = 0 is
View answer and explanation
Correct option: D
x2+y2+32x+4y+235=0{x^2} + {y^2} + 32x + 4y + 235 = 0
Option D (x2+y2+32x+4y+235=0{x^2} + {y^2} + 32x + 4y + 235 = 0) is correct because it represents the image of the given circle after reflection across the line, found by finding the image of the center.
12 Medium
12 of 685
Application +4 | -1
If a chord of contact of tangents drawn from a point PP with respect to the circle x2+y2=9x^2+y^2=9 is x=2x=2, then area, in square units, of triangle formed by tangents drawn from PP to the circle and their chord of contact is equal to
View answer and explanation
Correct option: C
552\frac{5 \sqrt{5}}{2}
Option C (552\frac{{5 \sqrt 5 }}{{2}}) is correct because it is the area of the triangle formed by the tangents and chord of contact, calculated using the formula with the given circle and point P.
13 Easy
13 of 685
Application +4 | -1
If the image of the point (−4,5)(-4,5) in the line x+2y=2x+2 y=2 lies on the circle (x+4)2+(y−3)2=r2,(x+4)^2+(y-3)^2=r^2, then rr is equal to:
View answer and explanation
Correct option: B
2
Option B (2) is correct because by finding the image of the point (-4,5) in the line x+2y=2 and substituting it into the circle's equation, we can solve for r and find that r=2.
14 Medium
14 of 685
Application +4 | -1
If the angle between tangents drawn to x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 from (0,0)(0,0) is π2\frac{\pi}{2}, then
View answer and explanation
Correct option: B
g2+f2=2cg^2+f^2=2 c
Option B (g2+f2=2cg^2+f^2=2c) is correct because it represents the condition for the angle between tangents from (0,0) to be π2\frac{{\pi}}{{2}}.
15 Easy
15 of 685
Application +4 | -1
The radius of the circle touching the straight lines x−2y−1=0x-2 y-1=0 and 3x−6y+7=03 x-6 y+7=0, is
View answer and explanation
Correct option: B
53\frac{\sqrt{5}}{3} unit
Option A (53\frac{\sqrt{5}}{3} unit) is correct because the radius is half the distance between the two parallel lines.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Medium
16 of 685
Analyzing +4 | -1
If each of the lines 5x+8y=135 x+8 y=13 and 4x−y=34 x-y=3 contains a diameter of the circle x2+y2−2(a2−7a+11)x−2(a2−6a+6)y+b3+1=0x^{2}+y^{2}-2\left(a^{2}-7 a+11\right) x-2\left(a^{2}-6 a+6\right) y+b^{3}+1=0, then
View answer and explanation
Correct option: D
a=5a=5 and b∈(−∞,1)b \in(-\infty, 1)
Option D (a=5a=5 and b∈(−∞,1)b \in(-\infty, 1)) is correct because the center of the circle must be the intersection of the diameters, and the radius must be real.
17 Easy
17 of 685
Application +4 | -1
If the lines 2x+3y+1=02 x+3 y+1=0 and 3x−y−4=03 x-y-4=0 lie along diameters of a circle of circumference 10π10 \pi, then the equation of the circle is
View answer and explanation
Correct option: D
x2+y2−2x+2y−23=0x^2+y^2-2 x+2 y-23=0
Option D (x2+y2−2x+2y−23=0x^2+y^2-2 x+2 y-23=0) is correct because the center is the intersection of the diameters, and the radius is derived from the circumference.
18 Easy
18 of 685
Application +4 | -1
The locus of the centre of a circle touching the lines x+2y=0x+2 y=0 and x−2y=0x-2 y=0, is
View answer and explanation
Correct option: A
xy=0x y=0
Option A (xy=0) is correct because the locus of the center is found to be xy=0, representing the x and y axes.
19 Easy
19 of 685
Application +4 | -1
The number of points (x,y)(x, y) having integral coordinates satisfying the condition x2+y2<25x^2+y^2<25 is
View answer and explanation
Correct option: A
69
Option A (69) is correct because after considering the integral coordinates within the circle x2+y2<25x^2 + y^2 < 25, and excluding the invalid cases, the total number of points is 69.
20 Medium
20 of 685
Application +4 | -1
If a circles x2+y2+2g1x+2f1y=0x^2+y^2+2 g_1 x+2 f_1 y=0 and x2+y2+2g2x+2f2y=0x^2+y^2+2 g_2 x+2 f_2 y=0 touch each other, then
View answer and explanation
Correct option: B
f1g1=f2g2\frac{f_1}{g_1}=\frac{f_2}{g_2}
Option B (f1g1=f2g2\frac{f_1}{g_1}=\frac{f_2}{g_2}) is correct because when two circles touch each other, the line joining their centers passes through the point of contact, implying the ratio of their y-coordinates to x-coordinates are equal.
21 Easy
21 of 685
Application +4 | -1
An equation of the normal at the point (2,3)(2,3) to the circle x2+y2−2x−2y−3=0x^2+y^2-2 x-2 y-3=0 is
View answer and explanation
Correct option: C
2x−y−1=02 x-y-1=0
Option C (2x−y−1=02 x-y-1=0) is correct because the normal to a circle at a point passes through the center of the circle.
22 Easy
22 of 685
Application +4 | -1
The coordinates of the point on the circle x2+y2−2x−4y−11=0x^2+y^2-2 x-4 y-11=0 farthest from the origin are
View answer and explanation
Correct option: B
(1+4/5,2+8/5)(1+4 / \sqrt{5}, 2+8 / \sqrt{5})
Option B ((1+4/5,2+8/5)(1+4 / \sqrt{5}, 2+8 / \sqrt{5})) is correct because the point farthest from the origin lies on the line connecting the origin and the center of the circle, and it is at a distance of radius from the center.
23 Easy
23 of 685
Application +4 | -1
The equations of the tangents to circle x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0, which are parallel to line 4x+3y+154x + 3y + 15 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=04x + 3y + 19 = 0 and 4x+3y−31=04x + 3y - 31 = 0) is correct because these are the equations of the tangents to the circle that are parallel to the given line.
24 Easy
24 of 685
Application +4 | -1
The locus of the middle points of chords of the circle x2+y2=25x^2+y^2=25 which are parallel to the line x−2y+3=0x-2 y+3=0, is
View answer and explanation
Correct option: B
2x+y=02 x+y=0
Option B (2x+y=02 x+y=0) is correct because it represents the locus of the midpoints of the chords parallel to the given line, derived from the properties of circles and lines.
25 Medium
25 of 685
Analyzing +4 | -1
Let ϕ(x,y)=0\phi(x, y)=0 be the equation of a circle. If ϕ(0,λ)=0\phi(0, \lambda)=0 has equal roots λ=2,2\lambda=2,2 and ϕ(λ,0)=0\phi(\lambda, 0)=0 has roots λ=45,5\lambda=\frac{4}{5}, 5, then the centre of the circle is
View answer and explanation
Correct option: B
(2910,2)\left(\frac{29}{10}, 2\right)
Option B ((2910\frac{{29}}{{10}}, 2)) is correct because it represents the center of the circle that satisfies the given conditions for ϕ(0,λ)=0\phi(0, \lambda)=0 and ϕ(λ,0)=0\phi(\lambda, 0)=0.
26 Easy
26 of 685
Analyzing +4 | -1
For the given circles x2+y2−6x−2y+1=0{x^2} + {y^2} - 6x - 2y + 1 = 0 and x2+y2+2x−8y+13=0{x^2} + {y^2} + 2x - 8y + 13 = 0, which of the following is true?
View answer and explanation
Correct option: D
They touch each other externally
The correct answer is D. By calculating the distance between the centers and comparing it to the sum of the radii, it's found that the circles touch each other externally.
27 Medium
27 of 685
Analyzing +4 | -1
Let L1L_1 be a straight line passing through the origin and L2L_2 be the straight line x+y=1x+y=1. If the intercepts made by the circle x2+y2−x+3y=0x^2+y^2-x+3 y=0 on L1L_1 and L2L_2 are equal, then which of the following equations can represent L1L_1 ?
View answer and explanation
Correct option: B
x−y=0,x+7y=0x-y=0, x+7 y=0
Option B (x−y=0,x+7y=0x-y=0, x+7 y=0) is correct because these lines satisfy the condition that the intercepts made by the circle on them are equal to the intercept made on the line x+y=1x+y=1.
28 Easy
28 of 685
Application +4 | -1
The number of integral values of λ\lambda for which x2+y2+λx+(1−λ)y+5=0x^2+y^2+\lambda x+(1-\lambda) y+5=0 is the equation of a circle whose radius cannot exceed 5 , is
View answer and explanation
Correct option: C
16
Option C (16) is correct because the number of integral values of λ\lambda that satisfy the condition for the radius of the circle not exceeding 5 is 16.
29 Easy
29 of 685
Application +4 | -1
Area of the triangle formed by tangent of the curve x=t2+t+2,y=t2−t+2,t≥0\mathrm{x}=\mathrm{t}^2+\mathrm{t}+2, \mathrm{y}=\mathrm{t}^2-\mathrm{t}+2, \mathrm{t} \geq 0 which passes through (2,2)(2,2) and co-ordinate axes is equal to
View answer and explanation
Correct option: D
8
Option D (8) is correct because the area of the triangle formed by the tangent to the curve at (2,2) and the coordinate axes is calculated to be 8.
30 Easy
30 of 685
Application +4 | -1
If the radii of the circles (x−1)2+(y−2)2=1(x-1)^2+(y-2)^2=1 and (x−7)2+(y−10)2=4(x-7)^2+(y-10)^2=4 are increasing uniformly w.r.t. time as 0.3 unit/s0.3 \mathrm{\,unit} / \mathrm{s} and 0.4 unit/s0.4 \mathrm{\,unit} / \mathrm{s} respectively, then they will touch each other at tt equals to
View answer and explanation
Correct option: B
90 s90 \mathrm{~s}
Option B (90 s) is correct because it represents the time at which the distance between the centers equals the sum of the radii.
Revision notes and key takeaways

Review CIRCLES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 685 single-correct-answer MCQs on CIRCLES.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 685 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - CIRCLES MCQs (Easy) 480 MCQs → ENGINEERING MATHEMATICS - CIRCLES MCQs (Hard) 16 MCQs → ENGINEERING MATHEMATICS - CIRCLES MCQs (Medium) 189 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs
📚 Switch Set (ALL)
Chapter Sets
ALL 685 MCQs EASY 480 MCQs HARD 16 MCQs MEDIUM 189 MCQs
0 /685
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →