The base is a chord of the circle, so its perpendicular bisector passes through the centre. We know the distance from the chord to the other side is $h$, so the distance from the chord to the centre is $h - r$. We now have a right-angled triangle with hypotenuse $r$, and sides $h-r$ and $b/2$ (because we bisected the length $b$). Hence, \begin{align*} r^2 &= (h-r)^2 + \left(\frac{b}{2}\right)^2\\ &= h^2 - 2hr + r^2 + \frac{b^2}{4}\\ 2hr &= h^2 + \frac{b^2}{4}\\ r &= \frac{h^2 + b^2/4}{2h}\\ &= \frac{h}{2} + \frac{b^2}{8h}. \end{align*}