Question: Consider a circle $C$ which touches the $y$-axis at $(0,6)$ and cuts off an intercept $6 \sqrt{5}$ on the x-axis. Then the radius of the circle $\mathrm{C}$ is equal to :
Correct Option: , 2
Solution:
$r=\sqrt{6^{2}+(3 \sqrt{5})^{2}}$
$=\sqrt{36+45}=9$
