Solve this following

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 :

 

  1. $\sqrt{53}$

  2. 9

  3. 8

  4. $\sqrt{82}$


Correct Option: , 2

Solution:

$r=\sqrt{6^{2}+(3 \sqrt{5})^{2}}$

$=\sqrt{36+45}=9$

 

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