Let C be the centre of the circle x 2 + y 2 - x + 2 y = 11 4 and P be a point on the circle. A line passes through the point C , makes an angle of π 4 with the line C P and intersects the circle at the points Q and R . Then the area of the triangle P Q R (in unit 2 ) is

Step-by-step solution

Given, x 2 + y 2 - x + 2 y = 11 4 On rearranging terms we get, ⇒ x - 1 2 2 + y + 1 2 = 2 2 Now given in ∆ P Q R , ∠ P C R = π 4 so ∠ P Q R = 22 1 2 by circle property, radius is 2 so R C = Q C = 2 Now, P R = Q R sin 22 1 2 = 4 sin π 8 And P Q = Q R cos 22 1 2 = 4 cos 22 1 2 Now area of Δ P Q R = 1 2 P R × P Q = 1 2 4 sin π 8 4 cos π 8 = 4 × 2 sin π 8 cos π 8 = 4 sin π 4 = 4 2 = 2 2