Originally Posted by

**Deveno** Let's say $(a,b)$ is the point on the upper semi-circle where the tangent line hits. Note that $b = \sqrt{64 - a^2}$. Now the tangent line has formula:

$y = \dfrac{\sqrt{64 - a^2}}{a - 12}(x - 12)$.

The radial line has slope $\dfrac{b}{a} = \dfrac{\sqrt{64 - a^2}}{a}$, and by perpendicularity, we know that:

$\dfrac{\sqrt{64-a^2}}{a-12}\cdot\dfrac{\sqrt{64-a^2}}{a} = -1$ that is:

$64 - a^2 = 12a - a^2 \implies 64 = 12a \implies a = \dfrac{16}{3}$, so $b = \dfrac{8\sqrt{5}}{3}$.

Now, what we WANT to know is the value of $y$ on the tangent line when $x = 0$, doubling this gives the answer.

And we have everything we need to calculate this:

$y = \dfrac{\frac{8\sqrt{5}}{3}}{\frac{-20}{3}}\cdot(-12) = \dfrac{24\sqrt{5}}{5}$ so the desired answer is $\dfrac{48\sqrt{5}}{5}$.

At least, I hope that's right...my old addled brain does make mistakes.