JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Circle

Let the tangent to the circle

C_{1} : x^{2} + y^{2} = 2

at the point

M (- 1, 1)

intersect the circle

C_{2}

(x - 3)^{2} + (y - 2)^{2} = 5

, at two distinct points A and B . If the tangents to

C_{2}

at the points A and B intersect at N , then the area of the triangle ANB is equal to :

Answer: (C)

Solution

The tangent to

C_{1} : x^{2} + y^{2} = 2

M (- 1, 1)

is:

- x + y = 2

x - y + 2 = 0

.

The perpendicular distance from center

P (3, 2)

C_{2}

to this line is:

O P = \frac{∣3 - 2 + 2∣}{2} = \frac{3}{2}

Radius of

C_{2}

5

. The distance from

P

to chord

A B

is:

A P = 5 - \frac{9}{2} = \frac{1}{2}

Since tangents from

N

C_{2}

touch at

A

and

B

, and

tan θ = \frac{A P}{O P} = \frac{1}{3}

, we have

sin θ = \frac{1}{10}

and

cos θ = \frac{3}{10}

.

From geometry:

A N = \frac{5}{s i n θ} \cdot cos θ = \frac{5}{3}

.

Area of

△ A N B = \frac{1}{2} A N^{2} sin 2 θ = \frac{1}{2} \cdot \frac{5}{9} \cdot \frac{3}{5} = \frac{1}{6}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option