Question

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

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{6}$
• D. $\frac{5}{3}$

Answer 1

Answer: (C)

$\frac{1}{6}$

Answer 2

The correct option is(C): $=\frac{1}{6}$

Tangent to C 1 at M : – x + y = 2 ≡ T

Intersection of T with C 2⇒ (x – 3)2 + x 2 = 5

⇒ x = 1, 2

A(1, 3) and B(2, 4)

Let N ≡ (α, β)

Then – x + y = 2 shall be chord of contact for x2 + y2 – 6x – 4y + 8 = 0

$∴αx+βy-3x-3α-2y-2β+8=0$

– x + y = 2

After simplification

$=\frac{1}{6}$ Units.