Geometry Question Involving Side Lengths I can't figure out how to find the minimum value of $y$ here. I know why its maximum value is 8, and the textbook says the minimum value is $12-\sqrt{5}$. Thank you!
Edit: I forgot to add that $x=4$ and I believe $y$ is minimized when segment $CD$ is a diameter of the circle.
 A: Let $t=CD$ and $\angle BAF = \phi$. By the PoP with respect to $E$ we have $$EF\cdot EA = ED\cdot EC\implies x(x+12) = y(y+t)$$
By the PoP with respect to $B$ we have $$BA^2 = BD\cdot BC\implies \boxed{64 = y(y+t)}$$ 
so we have $x^2+12x-64 = 0\implies x=4$. By the law of cosine we have:
$$ (2y+t)^2 = 16^2+8^2-2\cdot 16\cdot 8\cos \phi$$ so
using boxed equation we have:
$$ 4\cdot 64 +t^2 = 4y^2+4yt+t^2 = 16^2+8^2-2\cdot 16\cdot 8\cos \phi$$ 
so $$t^2 = 8^2-2\cdot 16\cdot 8\cos \phi <8^2\cdot 5$$
When $\phi \to -\pi $ we get $ t= 8\sqrt{5}$. From boxed equation we also get $$ y ={128\over \sqrt{t^2+256}+t}> {128\over 24+8\sqrt{5}}=12-4\sqrt{5}$$ 
A: Let $O$ be the center of the circle, $t$ its distance from $AF$. We have:
$$
\cos(\angle EAB)=-\sin(\angle EAO)=-{t\over\sqrt{36+t^2}}.
$$
Hence, by the cosine rule:
$$
EB^2=64\left(5+4{t\over\sqrt{36+t^2}}\right).
$$
On the other hand, from $BC\cdot BD=64$ one obtains:
$$
y={1\over2}\left(EB-\sqrt{EB^2-256}\right)={128\over EB+\sqrt{EB^2-256}}.
$$
Hence $y$ is minimum when $EB$ gets its supremum value, which is $24$ (for $t\to+\infty$). We thus find that the infimum of $EB$ is $4(3-\sqrt5)$.
