Let C1and C2 be the centers of the circles x2 + y2-2x - 2y- 2 = 0 and x2 + y2- 6x - 6y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2is

Let $S_{1} \equiv x^{2} + y^{2} - 2 x - 2 y - 2 = 0$

$\begin{aligned} S_{2} \equiv x^{2} + y^{2} - 6 x - 6 y + 14 = 0 \\ ∴ C_{1} \equiv (1, 1) and C_{2} \equiv (3, 3) \end{aligned}$

Radius of $S_{1}, P C_{1} = \sqrt{1 + 1 + 2} = 2$

Radius of $S_{2}, P C_{2} = \sqrt{9 + 9 - 14} = 2$

$∴ P C_{1} = Q C_{1} = P C_{2} = Q C_{2} = 2$

Now, $2 g_{1} g_{2} + 2 f_{1} f_{2} = 2 \times 3 + 2 \times 3 = 6 + 6 = 12$ and

$c_{1} + c_{2} = 14 - 2 = 12$

Here, $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

$∴$ Both circles are orthogonal. So, $P C_{1} Q C_{2}$ is a square.

$∴$ Area of $P C_{1} Q C_{2} = 2 \times 2 = 4 sq. units$ .

Let $C_{1} a n d C_{2}$ be the centers of the circles $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ and $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ respectively. If $P a n d Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P C_{1} Q C_{2}$ is