Let B be the centre of the circle x 2 + y 2 - 2 x + 4 y + 1 = 0 . Let the tangents at two points P and Q on the circle intersect at the point A ( 3 , 1 ) . Then 8 area Δ APQ area Δ BPQ is equal to .

Step-by-step solution

Given, centre of circle x 2 + y 2 - 2 x + 4 y + 1 = 0   . . . . . i Points B   1 , - 2   &   A 3 , 1 For point P ,   y = - 2 Putting value of y in equation i x 2 + - 2 2 - 2 x + 4 × - 2 + 1 = 0 x 2 - 2 x - 3 = 0 x - 3 x + 1 = 0 x = - 1 (not possible) x = 3 Point P 3 , - 2 A P = 3 - 3 2 + - 2 - 1 2 A P = 3 = A Q r = 1 2 + - 2 2 - 1 r = 1 + 4 - 1 r = 2 tan θ = 3 2 Area ( Δ A P Q ) Area ( Δ B P Q ) = A R R B = 3 sin θ 2 cos θ = 9 4 8 Area ( Δ A P Q ) Area