The tangent to the circle $C_1: x^2+y^2-2 x-1=0$ at the point $(2,1)$ cuts off a chord of length 4 from a…
The tangent to the circle $C_1: x^2+y^2-2 x-1=0$ 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
$\sqrt{6}$
2
$\sqrt{2}$
3
Solution
Here, equation of tangent on $C_1$ at $(2,1)$ is: $2 x+y-(x+2)-1=0$
Or $x+y=3$
If it cuts off the chord of the circle $C_2$ then
the equation of the chord is:
$
x+y=3
$
$\therefore \quad$ distance of the chord from $(3,-2)$ is :
$
d=\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}
$
Also, length of the chord is $l=4$
$
\begin{aligned}
\therefore \text { radius of } C_2=r &=\sqrt{\left(\frac{l}{2}\right)^2+d^2} \\
&=\sqrt{(2)^2+(\sqrt{2})^2}=\sqrt{6}
\end{aligned}
$