# Prove that if the chord length depends only on |s-t|, then it is a line or a part of a circle.

Let $$\alpha : (a,b) \rightarrow \mathbb{R}^2$$ be a smooth map(infinitely differentiable).

Show that if the chord length $$\Vert{\alpha(s)-\alpha(t)}\Vert$$ depends only on $$|s-t|$$, then it is a line or a part of a circle.

It comes from Shifrin's differential geometry notes.

Here is my attempt:

Since the chord length only depends on $$|s-t|$$, for $$s-t=\delta, -\delta$$, the chord length should be same.

$$\Vert \alpha(s+\delta) -\alpha(s) \Vert = \Vert \alpha(s) - \alpha(s-\delta) \Vert$$

Squaring both sides and viewing both sides as an inner product of themselves, respectively.

And expanding Taylor series and discarding terms with higher degree than 2 of $$\delta$$'s.

Then I arrived at $$\alpha(s)' \cdot \alpha(s)'' =0$$.

And I have no idea to do. Could you give me a hint?