Hello Mustard

Welcome to Math Help Forum!

I have an answer, based on the Tangent-Secant Theorem - and a whole lot of algebra. It is:$\displaystyle EF = \sqrt{\frac{2bh^3}{b^2+h^2}}$

This checks out for two special cases:When $\displaystyle b = h, EF = h$, which is so, because EF is then the diameter of the circle, which is of length $\displaystyle h$.

As $\displaystyle b \to \infty, EF \to 0$, which is also correct.

Perhaps you'd like to check my working.

To keep the algebra simpler, I let:$\displaystyle h = 2r$

$\displaystyle b = r+s$

and, in the diagram you have drawn:$\displaystyle DE = p$

$\displaystyle BF = q$

$\displaystyle BD = t$

$\displaystyle EF = x$

Then the lengths of the tangents from D and B to the circle, are $\displaystyle r$ and $\displaystyle s$ respectively. So, using the Tangent-Secant Theorem:$\displaystyle p(p+x) = r^2$ ...(1)

$\displaystyle q(q+x) = s^2$ ...(2)

So, from (1):$\displaystyle p^2+px-r^2 = 0$

$\displaystyle \Rightarrow p = \frac{-x+\sqrt{x^2+4r^2}}{2}$, taking the positive root

Similarly$\displaystyle \Rightarrow q = \frac{-x+\sqrt{x^2+4s^2}}{2}$

But$\displaystyle p+x+q = t$

Therefore$\displaystyle \frac{-x+\sqrt{x^2+4r^2}}{2}+x+\frac{-x+\sqrt{x^2+4s^2}}{2}=t$

$\displaystyle \Rightarrow\sqrt{x^2+4r^2}+\sqrt{x^2+4s^2}=2t$

$\displaystyle \Rightarrow x^2+4r^2+x^2+4s^2+2\sqrt{(x^2+4r^2)(x^2+4s^2)}=4t^ 2$

and, then if we re-arrange, square both sides, and re-arrange again, this simplifies to:$\displaystyle 4t^2x^2=4(t^2-r^2-s^2)^2-16r^2s^2$ ...(3)

So we now have an expression for x that we can write in terms of $\displaystyle b$ and $\displaystyle h$.

Now:$\displaystyle t^2 = b^2+h^2$ (Pythagoras' Theorem on $\displaystyle \triangle DBC$)

$\displaystyle = (r+s)^2+(2r)^2$

$\displaystyle = 5r^2 + 2rs+s^2$

So (3) becomes:$\displaystyle 4(b^2+h^2)x^2 = 4(4r^2+2rs)^2 - 16r^2s^2$$\displaystyle =4(16r^4+16r^3s+4r^2s^2)-16r^2s^2$

$\displaystyle (b^2+h^2)x^2=16r^4+16r^3s$

$\displaystyle \Rightarrow x^2= \frac{16r^3(r+s)}{b^2+h^2}$

$\displaystyle \Rightarrow x =\sqrt{\frac{2bh^3}{b^2+h^2}}$

Grandad