Mathematics - Circle Question with Solution | TestHub

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  $OP=\left|\frac{3-2+2}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$

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

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

Now $\tan \theta =\frac{OP}{AP}=3$

$\therefore \sin \theta =\frac{3}{\sqrt{10}}=\frac{AP}{AN} \& \cos \theta =\frac{1}{\sqrt{10}}$

$\Rightarrow AN=\frac{\sqrt{10}}{3}AP=\frac{\sqrt{10}}{3}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{5}}{3}=BN$

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