Let the tangent to the circle ${\mathrm{C}}_1:{\mathrm{x}}^2+{\mathrm{y}}^2=2$ at the point M(-1,1) intersect the circle $C_2:(x-3{)}^2+(y-2{)}^2=5$, at two distinct points A and B. If the tangents to $C_2$ at the points A and B intersect at N, then the area of the triangle ANB is equal to :

Equation of tangent to circle ${\mathrm{x}}^2+{\mathrm{y}}^2=2$ at point P(–1, 1) is
$\begin{array}{l}\mathrm{T}=0\\ \mathrm{x}(-1)+\mathrm{y}\left(1\right)=2\\ -\mathrm{x}+\mathrm{y}=2\end{array}$
x – y + 2 = 0 ………(1)
Let point N(h, k)
Equation of chord of contact of circle (2) drawn from point N(h, k) is
T = 0
$\begin{array}{l}\mathrm{hx}+\mathrm{ky}-3(\mathrm{x}+\mathrm{h})-2(\mathrm{y}+\mathrm{k})+8=0\\ (\mathrm{h}-3)\mathrm{x}+(\mathrm{k}-2)\mathrm{y}-3\mathrm{h}-2\mathrm{k}+8=0 .......\left(2\right)\end{array}$
by comparing (1) and (2)
$\begin{array}{l}\frac{\mathrm{h}-3}{1}=\frac{\mathrm{k}-2}{-1}=\frac{3\mathrm{h}+2\mathrm{k}-8}{-2}\\ -\mathrm{h}+3=\mathrm{k}-2\text{ and }-2\mathrm{h}+6=3\mathrm{h}+2\mathrm{k}-8\\ \mathrm{h}+\mathrm{k}=5 5\mathrm{h}+2\mathrm{k}=14\end{array}$
So, $\mathrm{N}\left(\frac{4}{3},\frac{11}{3}\right)$
Point of intersection of chord AB and circle (2)
$\begin{array}{l}(\mathrm{x}-3{)}^2+{\mathrm{x}}^2=5\\ 2{\mathrm{x}}^2-6\mathrm{x}+4=0\\ {\mathrm{x}}^2-3\mathrm{x}+2=0\\ \Rightarrow \mathrm{x}=1,2\end{array}$
So, A(1,3) and B(2,4)
Now area of $\mathrm{\Delta ANB}=\frac{1}{2}\left|\begin{array}{ccc}1& 3& 1\\ 2& 4& 1\\ \frac{4}{3}& \frac{11}{3}& 1\end{array}\right|=\frac{1}{2}\left[1\left(4-\frac{11}{3}\right)-3\left(2-\frac{4}{3}\right)+1\left(\frac{22}{3}-\frac{16}{3}\right)\right]$
$=\frac{1}{2}\left[\frac{1}{3}-\frac{6}{3}+\frac{6}{3}\right]=\frac{1}{6}$ square unit.