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

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