Question

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. $\frac{5}{3}$

Answer 1

Answer: (C)

$OP=\left|\frac{2-3+2}{\sqrt{2}}\right|$

$OP=\frac{3}{\sqrt{2}}$
$AP=\sqrt{O{A}^2-O{P}^2}$
$=\frac{1}{\sqrt{2}}$
$\tan \theta =3$
$\therefore \sin \theta =\frac{3}{\sqrt{10}}=\frac{AP}{AN}$
$\Rightarrow AN=\frac{\sqrt{5}}{3}=BN$
Area of $△ANB=\frac{1}{2}\cdot \left(A{N}^2\right)\sin 2\theta =\frac{1}{6}$