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 ______.

पर्याय

• `1/2`

• `2/3`

• `1/6`

• `5/3`

Answer 1

Let the tangent to the circle C₁: x² + y² = 2 at the point M(–1, 1) intersect the circle C₂: (x – 3) + (y – 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 `underlinebb(1/6)`.

Explanation:

Circles C₁: x² + y² = 2 and C₂: (x – 3)² + (y – 2)² = 5

Now, equation of tangent to the circle C₁ and M(–1, 1) is given by x(–1) + y(1) = 2

⇒ x – y + 2 = 0

Let ∠ANB = 2α

∴ ∠CAD = α

Now, CD = `|(3 - 2 + 2)/sqrt(1^2 + 1^2)| = 3/sqrt(2)`

Apply Pythagoras' theorem in ΔACD, we get

(CD)² + (AD)² = (AC)² 

⇒ AD = `sqrt(5 - 9/2)`

⇒ AD = `1/sqrt(2)`

⇒ tan α = `(CD)/(AD)` = 3

⇒ sin α = `(CD)/(AC) = 3/sqrt(10)`

Now, in ΔADN,

sin α = `(AD)/(AN) = 3/sqrt(10)`

⇒ AN = `sqrt(10)/3 . 1/sqrt(2) = sqrt(5)/3`

Now, area of ΔANB = `1/2(AN)^2 sin 2α`

= `1/2(5/9)(2 sin α cos α)`

= `5/9 xx 3/sqrt(10) xx 1/sqrt(10)`

= `1/6`square unit