A triangle is formed by the tangents at the point ( 2 , 2 ) on the curves y 2 = 2 x and x 2 + y 2 = 4 x , and the line x + y + 2 = 0 . If r is the radius of its circumcircle, then r 2 is equal to

Step-by-step solution

Let S 1 : y 2 = 2 x S 2 : x 2 + y 2 = 4 x P ( 2 , 2 ) is common point on S 1 &   S 2 Let T 1 is tangent to S 1 at P 2 , 2 , then T 1 : 2 y = x + 2 ⇒ T 1 : x - 2 y + 2 = 0 And, T 2 is tangent to S 2 at P 2 , 2 , then T 2 : 2 x + 2 y = 2 x + 2 ⇒ T 2 : y = 2 And, L 3 : x + y + 2 = 0 P Q = a = 20 Q R = b = 8 R P = c = 6 Also, ar ∆ P Q R = 1 2 2 2 1 - 2 0 1 - 4 2 1 ⇒ ar ∆ P Q R = 1 2 2 2 1 - 4 - 2 0 - 6 0 0 ⇒ ar ∆ P Q R = 1 2 - 6 0 + 2 = 6 ∴ r = a b c 4 Δ = 1