I am interested in the following result:
"If all the normal planes of a curve pass through a particular point, then the curve is contained in a sphere".
My approach:
Let $\alpha: I \to \mathbb{R}^3$ be the curve (assume it is arc length parametrized). Let $P$ be the common point of all the normal planes. It is a well-known fact that, if $\tau(s) \ne 0 \quad \forall s$ and $k'(s) \ne 0 \quad \forall s$, then $\alpha$ lies in a sphere if and only if $\dfrac{\tau(s)}{k(s)} = \left(\dfrac{k'(s)}{\tau(s)k^2(s)}\right)'$.
Consider a point $\alpha(s)$. Since $P$ belongs to the normal plane of $\alpha$ at $\alpha(s)$, then
$$ \alpha(s)-P = x(s)\overrightarrow{n}(s) + y(s)\overrightarrow{b}(s) $$
Deriving and applying Frenet formulas, we get $$ 0 = -\bigl(1+x(s)k(s)\bigl)\overrightarrow{t} + \bigl(x'(s)-y(s)\tau(s)\bigl)\overrightarrow{n} + \bigl(x(s)\tau(s)+y'(s)\bigl)\overrightarrow{b} $$
so
$$ \left\{ \begin{array}{l} 1+x(s)k(s) = 0 \\ x'(s) -y(s)\tau(s) = 0\\ x(s)\tau(s) + y'(s) = 0 \end{array} \right. $$
Playing with these equations and assuming $\tau(s) \ne 0$, I get $\dfrac{\tau(s)}{k(s)} = \left(\dfrac{k'(s)}{\tau(s)k^2(s)}\right)'$ as desired. However, I don't see how I could prove that $\tau(s) \ne 0$ and $k'(s) \ne 0$.
I am aware that, as Shifrin notes suggests, using the fact that $\lVert f(t) \rVert = \text{constant} \iff f \cdot f' = 0$ helps deriving the result without using the characterization of spherical curves in terms of their curvature, torsion and their derivatives. However, I'd like to complete (if possible) the proof I have been working on.
