Question

The tangent to the circle $C_1:x^2+y^2-2x-1=0$ at the point $(2,1)$ cuts off a chord of length $4$ from a circle $C_2$ whose centre is $(3,-2)$. The radius of $C_2$ is :

• A. $\sqrt{6}$
• B. $\sqrt{2}$
• C. $2$
• D. $3$

Answer 1

The correct option is A $\sqrt{6}$

The tangent to the circle $x^2+y^2-2x-1=0$ at the point $(2,1)$ is
$2x+y-x-2-1=0$
$\Rightarrow x+y-3=0$
$AB$ is the tangent to circle $C_1$ and chord of cirle $C_2$
Perpendicular length $C_{2}M=\frac{3\left(1\right)-2\left(1\right)-3}{\sqrt{1^2+1^2}}=\sqrt{2}$
From right $△C_{2}MA$
$A{C}_2=\sqrt{2^2+(\sqrt{2}{)}^2}=\sqrt{6}$
Hence, radius of circle $C_2=\sqrt{6}$ unit