# If $\beta(s)\cdot\beta'(s) = 0$, then the curve $\beta$ is (part of) a circle.

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 $$\beta(s)\cdot\beta'(s) = 0$$ 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: $$\kappa = T' \cdot N$$. So to be constant, it needs to be: $$\kappa' = 0 = T''\cdot N+ T'\cdot N'$$ I was thinking about using the formulas $$T = \beta'$$, $$T' = \kappa N$$ and $$N' = -\kappa T$$. But together with the given condition: $$\beta(s)\cdot\beta'(s) = 0$$, I got stuck.

• How about just differentiating $\beta(s)\cdot\beta(s)$? – Angina Seng Jun 22 '19 at 15:31

Note that a curve lies on a circle if $$\beta(s)\cdot \beta(s) = c = r^2$$ for some (necessarily positive) constant $$c$$. Since $$I$$ is connected, it suffices to check that the derivative of $$\beta(s)\cdot \beta(s)$$ is $$0$$. However the derivative of $$\beta(s)\cdot \beta(s)$$ is $$2\beta(s)\cdot \beta'(s)$$, and we already know this is $$0$$ by the given information.
In general, if the curve lies in $$\Bbb{R}^n$$, this shows that the curve lies on a sphere.
• why lies a curve on a circle if $\beta(s)\beta(s)$ =c – questmath Aug 20 '20 at 19:14
• @mathmath Because that says that the length of $\beta$ is constant, so all points of the curve lie on a circle of radius $\sqrt{c}$. – jgon Aug 20 '20 at 20:47
Hint: Since $$||\beta(s)||^2=\beta(s)\cdot\beta(s)$$ you can use the product rule to show $$\frac{\partial}{\partial s}||\beta(s)||^2=0$$.