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Comprehension Based Questions: Circle - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Comprehension Based Questions: Circle (8 Questions)

You can prepare effectively for JEE Crack JEE with 35 Years of Previous Year Solved Papers with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Comprehension Based Questions: Circle". These 8 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 8

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Test: Comprehension Based Questions: Circle - Question 1

PASSAGE-1
ABCD is a square of side length 2 units. C1 is the circle touching all the sides of the square ABCD and C2 is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point.

Q. If P is any point of C1 and Q is another point on C2, then

is equal to (2006 - 5M, –2)  

Detailed Solution: Question 1

Without loss of generality we can assume the square ABCD with its vertices A (1, 1), B  (–1, 1), C (–1, –1), D (1, –1)
P to be the point (0, 1) and Q as (,0 ).

 

Then, 

Test: Comprehension Based Questions: Circle - Question 2

PASSAGE-1
ABCD is a square of side length 2 units. C1 is the circle touching all the sides of the square ABCD and C2 is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point.

Q. If a circle is such that it touches the line L and the circle C1 externally, such that both the circles are on the same side of the line, then the locus of centre of the circle is (2006 - 5M, –2) 

Detailed Solution: Question 2

Let C' be the said circle touching C1 and L, so that C1 and C' are on the same side of L. Let us draw a line T parallel to L at a distance equal to the radius of circle C1, on opposite side of L.
Then the centre of C' is equidistant from the centre of C1 and from line T.
⇒ locus of centre of C' is a parabola.

Test: Comprehension Based Questions: Circle - Question 3

PASSAGE-1
ABCD is a square of side length 2 units. C1 is the circle touching all the sides of the square ABCD and C2 is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point.

Q. A line L' through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts L' at T2 and T3 and AC at T1, then area of ΔT1T2T3 is (2006 - 5M, –2) 

Detailed Solution: Question 3

Since S is equidistant form A and line BD, it traces a parabola. Clearly, AC is the axis, A (1, 1) is the focus

and  is the vertex of parabola.

T2 T3 = latus rectum of parabola

∴ Area (ΔT1T2T3) =

Test: Comprehension Based Questions: Circle - Question 4

PASSAGE-2
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation
and the point D is

 Further, it is given that the origin and the centre of C are on the same side of the line PQ.

Q. The equation of circle C is (2008) 

Detailed Solution: Question 4

Slope of CD = 

∴ Parametric equation of CD is

∴ Two possible coordinates of C are

 or 

As (0, 0) and C lie on the same side of PQ

  should be the coordinates of C.
NOTE THIS STEP :  Remember (x1, y1) and (x2, y2) lie on the same or opposite side of a line ax + by + c = 0 according as 

∴ Equation of the circle is

Test: Comprehension Based Questions: Circle - Question 5

PASSAGE-2
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation
and the point D is

 Further, it is given that the origin and the centre of C are on the same side of the line PQ.

Q. Points E and F are given by (2008) 

Detailed Solution: Question 5

ΔPQR is an equilateral triangle, the incentre  C must coincide with centriod of ΔPQR and D, E, F must concide with the mid points of sides PQ, QR and RP respectively.

Also  

Writing the equation of side PQ in symmetric form we

get, 

∴ Coordinates of P = 

 and 

coordinates of  

Let coordinates of R be (α,β) , then using the formula for centriod of Δ we get

⇒ α = 0 and β = 0

∴ Coordinates of R = (0, 0)

Now coordinates of E = mid point of QR  =

and coordinates of F = mid point of PR = 

Test: Comprehension Based Questions: Circle - Question 6

PASSAGE-2
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation
and the point D is

 Further, it is given that the origin and the centre of C are on the same side of the line PQ.

Q. Equations of the sides QR, RP are (2008) 

Detailed Solution: Question 6

Equation of side QR is y = and equation of side RP is  y = 0

Paragraph 3 Given the implicit function y3 – 3y + x = 0

For x ∈(–∞, –2) ∪ (2,∞) it is y = f (x) real valued differentiable function and for x ∈ (–2, 2) it is y = g(x) real valued differentiable function.

Test: Comprehension Based Questions: Circle - Question 7

PASSAGE-3
A tangent PT is drawn to the circle x2 + y2 = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1.    (2012)

Q. A possible equation of L is 

Detailed Solution: Question 7

Equation of tangent PT to the circle x2 + y2 = 4 at the point  

Let the line L, perpendicular to tangent PT be

As it is tangent to the circle (x – 3)2 + y2 = 1

∴ length of perpendicular from centre of circle to the tangent = radius of circle.

= – 1 or  – 5

∴ Equation of L can be 

Test: Comprehension Based Questions: Circle - Question 8

PASSAGE-3
A tangent PT is drawn to the circle x2 + y2 = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1.    (2012)

Q. A common tangent of the two circles is 

Detailed Solution: Question 8

From the figure it is clear that the intersection point of two direct common tangents lies on x-axis.
Also ΔPT1C1 ~ ΔPT2C2 ⇒ PC1 : PC2 = 2 : 1
or P divides C1C2 in the ratio 2 : 1 externally
∴ Coordinates of P are (6, 0) Let the equation of tangent through P be y = m (x – 6)
As it touches x2 + y2 = 4

36 m2 = 4(m2 + 1)

∴ Equations of common tangents are

Also x = 2 is the common tangent to the two circles.

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