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10-1.Circle and System of Circles
hard
Let the tangent to the circle $C _{1}: x^{2}+y^{2}=2$ 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
A
$\frac{1}{2}$
B
$\frac{2}{3}$
C
$\frac{1}{6}$
D
$\frac{5}{3}$
(JEE MAIN-2022)
Solution

$OP =\left|\frac{2-3+2}{\sqrt{2}}\right|$
$OP =\frac{3}{\sqrt{2}}$
$AP =\sqrt{ OA ^{2}- OP ^{2}}$
$=\frac{1}{\sqrt{2}}$
$\tan \theta=3$
$\therefore \sin \theta=\frac{3}{\sqrt{10}}=\frac{ AP }{ AN }$
$\Rightarrow AN =\frac{\sqrt{5}}{3}= BN$
Area of $\Delta ANB =\frac{1}{2} \cdot\left( AN ^{2}\right) \sin 2 \theta=\frac{1}{6}$
Standard 11
Mathematics