Q2Q3=12
R2R3=4√6
Area of the triangle OR2R3 is 6√2
Area of the triangle PQ2Q3 is 4√2
x² + y² = 3, x² = 2y ⇒ y² + 2y - 3 = 0 ⇒ y = 1, x = √2
Tangent at (√2, 1) to C1 is √2x + y = 3
Let centre of circle touching this tangent be (0, y)
Hence, |√2(0) + y - 3|/√(√2² + 1²) = 2√3
|y - 3| = 6 ⇒ y = 9, -3
Hence, Q2(0, 9), Q3(0, -3), Q2Q3 = 12
Now C2: x² + (y - 9)² = (2√3)²; C3: x² + (y + 3)² = (2√3)²
Now foot of perpendicular from Q2 and Q3 on tangent √2x + y = 3 is
√2x - y = -9 ⇒ x = -√2, y = 7, R2(-√2, 7)
√2x - y = 3 ⇒ x = 2√2, y = 1, R3(2√2, 1)
Hence, R2R3 = √((4√2)² + (8)²) = 4√6
Area of ΔOR2R3 = 1/2|0(7 - 1) - √2(1 - 1) + 2√2(1 - 7)| = |√2(6)| = 6√2
Area of ΔPQ2Q3 = 1/2|√2(9 - 3) + 0(7 - 1) + 0(1 - 9)| = 3√2
Area of ΔPQ2Q3 = 1/2 |√2(1)(-3 - 9) + 0 + 0| = 6√2