JEE Main 2022 (Online) 29th June Morning ShiftQuestion 1
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 :
Circle
Solution
Concept. $N$ is the pole of $AB$ for $C_2$: it is on the perpendicular from the centre to $AB$ with $CN=\frac{r^2}{d}$. Formulas used. (1) Tangent to $x^2+y^2=a^2$ at $(x_1,y_1)$: $xx_1+yy_1=a^2$(2) Chord $=2\sqrt{r^2-d^2}$; pole distance $CN=\frac{r^2}{d}$Step 1 - Line $AB$, (1). At $M(-1,1)$: $-x+y=2$, i.e. $x-y+2=0$. Step 2 - $C_2$. Centre $C(3,2)$, $r^2=5$; $d=\frac{|3-2+2|}{\sqrt2}=\frac{3}{\sqrt2}$. Step 3 - Chord, (2). $AB=2\sqrt{5-\frac92}=\sqrt2$. Step 4 - Height. $CN=\frac{5}{3/\sqrt2}=\frac{5\sqrt2}{3}$; $N$ is on the same side as $AB$, so its distance from $AB$ is $$\frac{5\sqrt2}{3}-\frac{3}{\sqrt2}=\frac{10-9}{3\sqrt2}=\frac{\sqrt2}{6}.$$ Step 5 - Area. $\frac12\cdot\sqrt2\cdot\frac{\sqrt2}{6}=\frac16$. Answer: C