# Radius of curvature for a strictly convex curve

Suppose I have a strictly convex curve $\gamma$ parametrised by arc length $s$ in the plane. I take a point $s_{1}$ on $\gamma$ and issue a straight line from $s_{1}$ to another point $s_{2}$ on $\gamma$. Assume $s_{2} > s_{1}$. Let's denote the length of the line by $L$. Now $L = L(s_{1}, s_{2})$ as it depends on the points $s_{1}$ and $s_{2}$.

I position a circle $C$ of radius $R$ on the midpoint of $L$ and choose $R$ such that the circle $C$ is $\textit{tangent}$ to $\gamma$ at some point $s_{0} \in (s_{1}, s_{2})$.

How do I obtain a expression for the radius $R$ in terms of $s$ and $L$?

Would appreciate any suggestions.

HINT: This will have to be the osculating circle of $\gamma$ at the point $\gamma(s_0)$, so the radius will be the reciprocal of the curvature, and the midpoint of $L$ will have to lie on the normal line at that point. It follows that that (at least usually) $\gamma(s_0)$ will be the point on $\gamma$ (between $\gamma(s_1)$ and $\gamma(s_2)$) closest to the midpoint of $L$.