Let C _ 1 and C _ 2 be the centres of the circles x ^ 2 + y ^ 2 -...

Equation of given circles are

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 4$ and ${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 4$.

Hence, ${C}_{1}\left(1, 1\right)$ and ${r}_{1} = 2$; ${C}_{2}\left(3, 3\right)$ and ${r}_{2} = 2$

$⇒ P{C}_{1} = P{C}_{2} = 2$

Now, by distance formula,

${C}_{1}{C}_{2} = \sqrt{{\left(3 - 1\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{{2}^{2} + {2}^{2}} = \sqrt{8}$

$⇒ P{C}_{1}^{2} + P{C}_{2}^{2} = {C}_{1}{C}_{2}^{2}$

$⇒ ∠{C}_{1}P{C}_{2} = \dfrac{\pi}{2}$ (by converse of pythagoras theorem in $△ P{C}_{1}{C}_{2}$)

Hence, area of quadrilateral $P{C}_{1}Q{C}_{2} = 2 \times$area of $△ P{C}_{1}{C}_{2} = 2 \times$area of $△ Q{C}_{1}{C}_{2}$

$= 2 \times \dfrac{1}{2} \times 2 \times 2 = 4$