The circle C 1 : x 2 + y 2 = 3 , with centre at O, intersects the parabola x 2 = 2 y at the point P in the first quadrant. Let the tangent to the circle C 1 at P touches other two circles C 2 and C 3 at R 2 and R 3 , respectively. Suppose C 2 and C 3 have equal radii 2 3 and centres Q 2 and Q 3 , respectively. If Q 2 and Q 3 lie on the y - axis, then

Step-by-step solution

On solving x 2 + y 2 = 3 and x 2 = 2 y we get point P 2 , 1 Equation of tangent at P 2 x + y = 3 Let Q 2 be (0, k) and radius is 2 3 ∴ 2 0 + k - 3 2 + 1 = 2 3 ∴ k = 9 , - 3 Q 3 Q 2 0 , 9 and Q 3 ( 0 , - 3 ) C 2 : ( x − 0 ) 2 + ( y − 9 ) 2 = 12 C 3 : ( x − 0 ) 2 + ( y + 3 ) 2 = 12 Hence Q 2 Q 3 = 12 ⇒ a 2 + 12 = 36 ⇒ a = 2 6 R 2 R 3 = 2 a = 4 6 Perpendicular distance of origin O from R 2 R 3 is equal to distance of O from tangent 2 x + y = 3 which is same as radius of circle C 1 = 3 Hence area of Δ O R 2 R 3 = 1 2 ×