Where I live there are bike racks that look like circles of metal embedded in the ground. If the rack has a height of $h$ and a base length of $b$, find the radius, $r$, of the circle.
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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*}