Change in function's chord length depends on $|s-t|$ I am stuck on this question and any help would go a long way.

Show that if $\alpha : [a, b] \rightarrow \mathbb{R}$ is a regular smooth curve and $||α(s) − α(t)||$ depends only on $|s − t|$, then $\alpha$ must be a subset of a circle or a line.

I have shown that the speed of such a curve is constant, but I don't know where further to go. Also, the answer given here did not seem to help.
Any help will be extremely appreciated.
Thank you!

EDIT
As per the comment, I will elaborate on my answer and what I understood from the answer attached.
I have understood that the speed of such a curve must be constant.
With the speed being constant, and knowing the relation $\langle \alpha'(t)-\alpha'(s),\alpha(t)-\alpha(s)\rangle=0,$ which one can derive as the solution given does, we get that the angles formed by $\alpha'(t)$ and $\alpha'(s)$ with $\alpha(t)-\alpha(s)$ are equal.
Now, the solution says that the directions of $\alpha'(t),\alpha'(s)$ are different... Why?
Suppose this was true; then the solution says that one can see that the curve alpha must satisfy the equation
$$r\frac{d\theta}{dr}=\tan \theta.$$
Why does this follow?

Help with these doubts will be much appreciated. A different approach altogether also is be fantastic. Thank you.
 A: I agree with you that the equation tells you only that the vectors $\alpha(s)$ and $\alpha(t)$ make the same angle with the chord $\alpha(t)-\alpha(s)$. One must consider those two cases separately.
To derive the differential equation, assume arclength parametrization and set $s=0$, with $\alpha(0)=0$ and $\alpha'(0)=(1,0)$. Then $\alpha(t)=r(t)(\cos\theta(t),\sin\theta(t))$. So
\begin{align*}
\alpha'(t)\cdot\alpha(t)&=r(t)r'(t)=r(t)\cos\theta(t) \quad\text{and either} \\
\alpha'(t)\cdot (-\sin\theta(t),\cos\theta(t))&=r\theta'(t)=\cos(\pi/2-\theta(t)) \quad\text{or} \\
\alpha'(t)\cdot (-\sin\theta(t),\cos\theta(t))&=r\theta'(t)=\cos(\pi/2+\theta(t)).
\end{align*}
Thus, you end up with either
$$ r'(t)=\cos\theta(t) \quad\text{and}\quad r\theta'(t) = \sin\theta(t) $$
or
$$r'(t)=\cos\theta(t) \quad\text{and}\quad r\theta'(t) = -\sin\theta(t).$$
I leave you to finish the analysis.
By the way, the problem can be done with just continuity, not assuming differentiability. This is a rather nice solution, though.
