Comprehension Based Questions of Circle, Past year Questions JEE Advance, Class 11, Maths

# Comprehension Based Questions of Circle, Past year Questions JEE Advance, Class 11, Maths - Notes - Class 11

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

is equal to (2006 - 5M, –2)
(a) 0.75
(b) 1.25
(c)1
(d) 0.5

Ans. Sol. (a) 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,

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. 2. 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)
(a) ellipse
(b) hyper bola
(c) parabola
(d) pair of straight line

Ans. Sol. 2. (b) 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.

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.3. 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)
(a) sq. units
(b) sq. units
(c) 1 sq. units
(d) 2 sq. units

Ans. Sol. (c) 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) =

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. 4. The equation of circle C is (2008)

(a)

(b)

(c)

(d)

Ans. Sol. (d) 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

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

(a)

(b)

(c)

(d)

Ans. Sol. (a) 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 =

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

(a)

(b)

(c)

(d)

Ans. Sol. (d) 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.

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. 7. A possible equation of L is

(a)

(b)

(c)

(d)

Ans. Sol. (a) 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

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. 8. A common tangent of the two circles is

(a) x = 4
(b) y = 2
(c)
(d)

Ans. d

Sol.

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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