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

Answer 1

Answer: (D)

Equation of tangent to circle $x^2+y^2-2x-1=0$ at point $(2,1)$ is
$2x+y-(x+2)-1=0$
$x+y-3=0$
This line is chord of circle $C_2$

So, $OC=\frac{|3-2-3|}{\sqrt{2}}=\sqrt{2}$
Therefore, radius of circle is $\sqrt{C{A}^2+O{C}^2}=\sqrt{4+2}=\sqrt{6}$