Skip to main content
Back to CIRCLES

Your Progress

0 / 16 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • CIRCLES

Circles practice · Hard

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 →
1 Hard
1 of 16
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 Hard
2 of 16
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.
3 Hard
3 of 16
Analyzing +4 | -1
The set of values of 'cc' so that the equation y=∣x∣+cy = \left| x \right| + c and x2+y2−8∣x∣−9=0{x^2} + {y^2} - 8\left| x \right| - 9 = 0 have no solution, is
View answer and explanation
Correct option: D
(52−4,  ∞)\left( {5\sqrt 2 - 4,\;\infty } \right)
Option C is correct because eliminating the parameter 't' results in the equation of a parabola.
4 Hard
4 of 16
Analyzing +4 | -1
If the circles x2+y2+2ax+cy+a=0{x^2} + {y^2} + 2ax + cy + a = 0 and x2+y2−3ax+dy−1=0{x^2} + {y^2} - 3ax + dy - 1 = 0 intersect in two distinct points PP and QQ, then the line 5x+by−a=05x + by - a = 0 passes through PP and QQ for
View answer and explanation
Correct option: C
No value of aa
Option C (No value of aa) is correct because the condition for the line to pass through the intersection points leads to a quadratic equation with no real solutions for aa.
5 Hard
5 of 16
Application +4 | -1
If the tangent at the point PP on the circle x2+y2+6x+6y=2{x^2} + {y^2} + 6x + 6y = 2 meets the straight line. 5x−2y+6=05x - 2y + 6 = 0 at a point QQ on the yy-axis, then the length of PQPQ is
View answer and explanation
Correct option: C
5
Option C (5) is correct because it represents the length of the tangent segment from the point Q on the y-axis to the circle.
6 Hard
6 of 16
Application +4 | -1
One of the diameters of the circle circumscribing the rectangle ABCDABCD is 4y=x+7.4y = x + 7. If AA and BB are the points (−3,4)\left( { - 3,4} \right) and (5,4)\left( {5,4} \right) respectively, then the area of the rectangle is
View answer and explanation
Correct option: C
32 sq. units
Option C (32 sq. units) is correct because it is derived from finding the center of the circle, the length of the rectangle, and the perpendicular distance to determine the width.
7 Hard
7 of 16
Application +4 | -1
The equation to the circle having y=mxy = mx as a diameter where y=mxy = mx is a chord of the circle, through the origin, of radius a and having the x-axis as diameter is
View answer and explanation
Correct option: A
(1+m2)(x2+y2)−2a(x+my)=0\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) - 2a\left( {x + my} \right) = 0
The correct answer is (1+m2)(x2+y2)−2a(x+my)=0\left( {1 + {m^2}} \right)\left( {{x^2} + {y^2}} \right) - 2a\left( {x + my} \right) = 0 because it represents the equation of the circle with the given conditions.
8 Hard
8 of 16
Application +4 | -1
If P(−1,−3)P\left( { - 1, - 3} \right) is a centre of similitude for the circles x2+y2=1{x^2} + {y^2} = 1 and x2+y2−2x−6y+6=0,{x^2} + {y^2} - 2x - 6y + 6 = 0, then the length of the common tangent through PP to the circles is
View answer and explanation
Correct option: B
3
The correct answer is 3 because it represents the length of the common tangent through P to the circles, calculated using the lengths of tangents from P to each circle.
9 Hard
9 of 16
Application +4 | -1
If a point PP has co-ordinates (0,−2)(0,-2) and QQ is any point on the circle, x2+y2−5x−y+5=0x^{2}+y^{2}-5 x-y+5=0, then the maximum value of (PQ)2(P Q)^{2} is :
View answer and explanation
Correct option: B
14+5314+5 \sqrt{3}
Option B (14+5314+5 \sqrt{3}) is correct because it represents the maximum value of (PQ)² calculated by parameterizing the circle and maximizing the distance.
10 Hard
10 of 16
Analyzing +4 | -1
Three sides of a triangle have the equations Lr≡y−mrx−cr=0,r=1,2,3.{L_r} \equiv y - {m_r}x - {c_r} = 0,r = 1,2,3. If λ,μ,v\lambda ,\mu ,v are non-zero real numbers such that λ  L2L3+μL3L1+v  L1L2=0\lambda \;{L_2}{L_3} + \mu {L_3}{L_1} + v\;{L_1}{L_2} = 0 represents the circumcircle of the triangle, then
View answer and explanation
Correct option: C
Both (1) and (2) hold together
The correct answer is C. For the given equation to represent the circumcircle, it must satisfy the conditions for a circle, which leads to both equations in options A and B holding true.
11 Hard
11 of 16
Application +4 | -1
The equation of the circle radius 222\sqrt 2 whose centre lies on the line x−y=0x - y = 0 and which touches the line x+y=4,x + y = 4, and whose centre is coordinate satisfy x+y>4,x + y > 4, is
View answer and explanation
Correct option: A
x2+y2−8x−8y+24=0{x^2} + {y^2} - 8x - 8y + 24 = 0
Option A (x2+y2−8x−8y+24=0{x^2} + {y^2} - 8x - 8y + 24 = 0) is correct because the center of the circle lies on the line x−y=0x - y = 0, touches the line x+y=4x + y = 4, and the coordinates of the center satisfy x+y>4x + y > 4.
12 Hard
12 of 16
Analyzing +4 | -1
If a variable tangent of the circle x2+y2=1{x^2} + {y^2} = 1 intersects the ellipse x2+2y2=4{x^2} + 2{y^2} = 4 at points PP and QQ, then the locus of the point of intersection of tangent at PP and QQ is
View answer and explanation
Correct option: C
An ellipse with eccentricity 32\frac{{\sqrt 3 }}{2}
The correct answer is C. The locus of the intersection point is an ellipse, and its eccentricity is calculated to be 32\frac{{\sqrt 3 }}{{2}}.
13 Hard
13 of 16
Analyzing +4 | -1
The two lines through (2,3)\left( {2,3} \right) from which the circle x2+y2=25{x^2} + {y^2} = 25 intercepts chords of length 8 units have equations
View answer and explanation
Correct option: B
y=3,12x+5y=39y = 3,12x + 5y = 39
Option B (y=3,12x+5y=39y = 3,12x + 5y = 39) is correct because these lines satisfy the condition of intercepting chords of length 8 units from the given circle when passing through (2,3).
14 Hard
14 of 16
Application +4 | -1
Consider a family of circles, which are passing through the point (−1,  1)\left( { - 1,\;1} \right) and are tangent to xx-axis. If (h,  k)\left( {h,\;k} \right) are the coordinates of the centre of the circles, then the set of values of kk is given by the interval
View answer and explanation
Correct option: B
k≥12k \ge \frac{1}{2}
Option B (k≥12k \ge \frac{1}{2}) is correct because it represents the range of possible y-coordinates for the center of the circle, derived from the condition that the circle is tangent to the x-axis and passes through (-1, 1).
15 Hard
15 of 16
Analyzing +4 | -1
Two concentric circles of which smallest is x2+y2=4x^2+y^2=4, have the difference in radii as dd, if line y=x+1y=x+1 cuts the circles in real points, then dd lies in the interval
View answer and explanation
Correct option: A
(−∞,−2−12)∪(−2+12,∞)\left(-\infty,-2-\frac{1}{\sqrt{2}}\right) \cup\left(-2+\frac{1}{\sqrt{2}}, \infty\right)
The correct answer is A (−∞,−2−12)∪(−2+12,∞)\left(-\infty,-2-\frac{{1}}{{\sqrt{{2}}}}\right) \cup\left(-2+\frac{{1}}{{\sqrt{{2}}}}, \infty\right) because it represents the interval for the difference in radii such that the line intersects both circles in real points.
16 Hard
16 of 16
Application +4 | -1
The equation of the circle which passes through the intersection of x2+y2+13x−3y=0{x^2} + {y^2} + 13x - 3y = 0 and 2x2+2y2+4x−7y−25=02{x^2} + 2{y^2} + 4x - 7y - 25 = 0 and whose centre lies on 13x+30y=013x + 30y = 0, is
View answer and explanation
Correct option: B
4x2+4y2+30x−13y−25=04{x^2} + 4{y^2} + 30x - 13y - 25 = 0
Option C (4x2+4y2+30x−13y−25=04{x^2} + 4{y^2} + 30x - 13y - 25 = 0) is correct because it represents the equation of a circle passing through the intersection of two given circles and whose center lies on the given line.
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 16 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.
0 /16
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

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.