Question

Let the tangent to the circle C₁: x² + y² = 2 at the point M(–1, 1) intersect the circle C₂: (x – 3)² + (y – 2)² = 5, at two distinct points A and B. If the tangents to C₂ 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

The Correct Option is C

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

Tangent toC₁ at M: –x + y = 2 ≡T

Intersection of T with C₂⇒ (x – 3)² + x² = 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.