Question

A circle having centre $(0,k)$ with $k>6$ is tangent to the lines pair $x^2-y^2=0$ and $y=6$. The radius of the circle is/are:

• A. $6(\sqrt{2}-1)$
• B. $12$
• C. $6\sqrt{2}$
• D. $6\sqrt{2}+6$

Answer 1

The correct option is A $6(\sqrt{2}-1)$

Let R be the radius of the circle.

Draw the two radii that meet the points of tangency to the lines $y=\pm x$.

We can see that a square is formed by the origin, two tangency points, and the center of the circle.

The side lengths of this square are R and the diagonal is $k=R+6$.

The diagonal of a square is $\sqrt{2}$ times the side length.

Therefore, $R+6=R\sqrt{2}\Rightarrow R=\frac{6}{\sqrt{2}-1}=6+6\sqrt{2}\Rightarrow E$.