Let the tangent to the circle C _ 1 : x ^ 2 + y ^ 2 = 2 at the...

Let $O\left(3, 2\right)$ be the centre of the circle ${C}_{2}$.

The tangent to the circle ${C}_{1}: {x}^{2} + {y}^{2} - 2 = 0$ at $M\left(- 1, 1\right)$ is $x - y + 2 = 0$

Now the perpendicular distance from $O$ to the tangent is $OP = \left|\dfrac{3 - 2 + 2}{\sqrt{2}}\right| = \dfrac{3}{\sqrt{2}}$

Also $AP = \sqrt{O{A}^{2} - O{P}^{2}} = \sqrt{5 - \dfrac{9}{2}}$

$= \dfrac{1}{\sqrt{2}}$

Now $tan\theta = \dfrac{OP}{AP} = 3$

$∴ sin\theta = \dfrac{3}{\sqrt{10}} = \dfrac{AP}{AN} \text{ and } cos\theta = \dfrac{1}{\sqrt{10}}$

$⇒ AN = \dfrac{\sqrt{10}}{3}AP = \dfrac{\sqrt{10}}{3} \times \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{5}}{3} = BN$

Area of $\delta ANB = \dfrac{1}{2} \cdot \left(A{N}^{2}\right)\sin 2\theta = \dfrac{1}{2} \times \dfrac{10}{9} \times \dfrac{3}{5} = \dfrac{1}{6}$