I need to show this result:
Let $\alpha :I\rightarrow \mathbb{R}^2$ a smooth curve, where $I$ is a compact interval of the real line. If $\lVert \alpha (s) - \alpha (t) \rVert$ depends only on $|s-t|$, for all $s$, $t$ in $I$, then $\alpha$ must be a subset of a line or a circle.
I have tried calling $\lVert \alpha (s) - \alpha (t) \rVert=f(|s-t|)$, where $f$ is smooth but this led me nowhere. Also, I had a suggestion to fixing $t$ as $0$ and $\alpha(0)=0$ through reparametrization and a rigid movement, so I would have $\lVert\alpha(s) \rVert=f(|s|)$, $f$ is smooth; this lead to nothing also. I strongly believe I must show that the curvature of this function if constant, either $0$ (then it is a line) or it is a constant $\neq 0$ from where it is in a circle.
Can someone please give me a hint?