If the tangents drawn at the point O 0 , 0 and P 1 + 5 , 2 on the circle x 2 + y 2 - 2 x - 4 y = 0 intersect at the point Q , then the area of the triangle O P Q is equal to

Step-by-step solution

Tangent at R x 1 , y 1 when point R is on the circle x 2 + y 2 + 2 g x + 2 f y + c = 0 can be written as x x 1 + y y 1 + g x + x 1 + f y + y 1 + c = 0 . Tangent at O will be - x + 0 - 2 y + 0 = 0 ⇒ x + 2 y = 0 Tangent at P will be x 1 + 5 + 2 y - x + 1 + 5 - 2 y + 2 = 0 ⇒ 5 x = 5 + 5 ⇒ x = 5 + 1 and y = - 5 + 1 2 So Q 5 + 1 , - 5 + 1 2 Length of tangent L = O Q = 6 + 2 5 + 3 + 5 2 = 5 2 + 5 2 2 = 5 + 5 2 We know area of triangle O P Q = R L 3 R 2 + L 2 Here, radius of circle R = 5 So area = 5 &#21