Let the tangent to the circle C _{1}: x^{2}+y | vedclass

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Question

$OP =\left|\frac{2-3+2}{\sqrt{2}}ight|$

$OP =\frac{3}{\sqrt{2}}$

$AP =\sqrt{ OA ^{2}- OP ^{2}}$

$=\frac{1}{\sqrt{2}}$

$	an 	heta=3$

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

$OP =\left|\frac{2-3+2}{\sqrt{2}}ight|$

$OP =\frac{3}{\sqrt{2}}$

$AP =\sqrt{ OA ^{2}- OP ^{2}}$

$=\frac{1}{\sqrt{2}}$

$	an 	heta=3$

$	herefore \sin 	heta=\frac{3}{\sqrt{10}}=\frac{ AP }{ AN }$

$\Rightarrow AN =\frac{\sqrt{5}}{3}= BN$

Area of $\Delta ANB =\frac{1}{2} \cdot\left( AN ^{2}ight) \sin 2 	heta=\frac{1}{6}$

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

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