1. Tardigrade
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  4. Let the tangent to the circle C 1: x2+y2=2 at the point M(-1,1) intersect the circle C2: ( 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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Q. 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 :

JEE MainJEE Main 2022Conic Sections

Solution:

$OP =\left|\frac{2-3+2}{\sqrt{2}}\right|$
image
$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 $\triangle ANB =\frac{1}{2} \cdot\left( AN ^{2}\right) \sin 2 \theta=\frac{1}{6}$