Prove that a SRAN curve is part of a circle

Let *alpha*(s): I -> R^3 be smooth, regular, non-singular curve parameterized by arc length with the property that all normal lines to *alpha(*s) pass through the origin. We can assume that *alpha*(s) is non-zero for any s in I.

Prove that *alpha* is part of a circle. (i.e. the trace is contained in a plane and the distance from *alpha* to a fixed point is constant). Hint: explain why *alpha*(s)+f(s)n(s) = 0 for some f:I -> R, and use this together with the Frenet equations to prove the result.

From the hint, I see that since n(s) is the normal vector to *alpha* at s, f can be -*alpha*(s) to give us the zero. But what does this have to do with the Frenet equations? Do we want to prove that curvature is 1/r, thus the trace of *alpha *is a circle at that point?

Re: Prove that a SRAN curve is part of a circle

The normal line to $\displaystyle a(s)$ is given by

$\displaystyle \epsilon(u)=a(s)+u\eta(s),u\in \mathbb{R}$.

Since every normal line passes through the origin, for every $\displaystyle s\in I$ there exists an $\displaystyle u=u(s)$ such that

$\displaystyle 0=a(s)+u(s)\eta(s),s\in I$.

The function $\displaystyle u(s)$ is differentiable (why?). Differentiating the last relation gives

$\displaystyle 0=a'(s)+u'(s)\eta(s)+u(s)\eta'(s) \Rightarrow $

$\displaystyle (1-u(s)k(s))t(s)+u'(s)\eta(s)+u(s)\tau(s)b(s)=0$

where $\displaystyle \{t,\eta,b\}$ is the Serret-Frenet trihedron. Linear independence now implies

$\displaystyle u'(s)=0\Rightarrow u(s)=u_0\neq 0 $

$\displaystyle u_0\tau(s)=0\Rightarrow \tau(s)=0$, so the curve is planar and

$\displaystyle 1-u_0k(s)=0\Rightarrow k(s)=u_0^{-1}$, so the curvature is constant;

Thus $\displaystyle a$ is part of a circle.