Assume we have an arc-length parameterized curve $\beta :I\to {\mathbb{E}}^{2}$ with I a random interval. I want to show that if $\beta (s)\cdot {\beta}^{\prime}(s)=0$ for arc-length parameter $s$, then $\beta $ is (part of) a circle.

I was thinking about using the Frenet-formulas for proving this. If we want to show something is a circle, one needs to show that the curvature is constant. In the Frenet-frame $(T,N)$, the curvature is: $\kappa ={T}^{\prime}\cdot N$. So to be constant, it needs to be:

${\kappa}^{\prime}=0={T}^{\u2033}\cdot N+{T}^{\prime}\cdot {N}^{\prime}$

I was thinking about using the formulas $T={\beta}^{\prime}$, ${T}^{\prime}=\kappa N$ and ${N}^{\prime}=-\kappa T$. But together with the given condition: $\beta (s)\cdot {\beta}^{\prime}(s)=0$, I got stuck.

I was thinking about using the Frenet-formulas for proving this. If we want to show something is a circle, one needs to show that the curvature is constant. In the Frenet-frame $(T,N)$, the curvature is: $\kappa ={T}^{\prime}\cdot N$. So to be constant, it needs to be:

${\kappa}^{\prime}=0={T}^{\u2033}\cdot N+{T}^{\prime}\cdot {N}^{\prime}$

I was thinking about using the formulas $T={\beta}^{\prime}$, ${T}^{\prime}=\kappa N$ and ${N}^{\prime}=-\kappa T$. But together with the given condition: $\beta (s)\cdot {\beta}^{\prime}(s)=0$, I got stuck.