Let $\alpha: I \mapsto \mathbb{R^3}$ be a regular unit speed curve such that $k(s) > 0$ and $\tau(s) \neq 0, \forall s \in I$. Prove that:
a) If $\alpha(I)$ is contained in a sphere $S$ centered at $c$ with radius $R$, then $$\alpha(s) - c = -\frac{1}{k(s)}N(s) - \frac{k'(s)}{k^2(s)\tau(s)}B(s)$$ and thus $$r^2 = \frac{1}{k^2(s)} + \left( \frac{k'(s)}{k^2(s) \tau(s)} \right)^2$$
b) Conversely, if $r^2 = \frac{1}{k^2(s)} + \left( \frac{k'(s)}{k^2(s) \tau(s)} \right)^2$ is a constant, with $k'(s) \neq 0$, then $\alpha$ is contained in a sphere of radius $r$.
Warning: the book I'm using defines $B'(s) = \tau(s) N(s)$, so that $$N'(s) = -k(s)T(s) - \tau(s)B(s)$$
I managed to do a) just fine, it's b) I'm having trouble with. My strategy is to prove $W'(s) = 0$, where $$W(s) = \alpha(s) + pN(s) - p' \sigma B(s)$$
$$p = \frac{1}{k(s)}, \sigma = \frac{1}{\tau(s)}$$
Differentiating, we have:
$$W'(s) = B(s)\left(- \frac{p}{\sigma} - (p' \sigma)' \right) $$
And although I've tried a lot, I can't see how to go futher. If I prove $W'(s) = 0$, then $$||\alpha(s) - W(s)|| = r^2$$ and my problem is solved, but I'm having a hard time with that. I'd appreciate any help.
Actually, I've solved it now:
$$r^2 = p^2 + (p' \sigma)^2 \Rightarrow2pp' + 2(p' \sigma)(p'\sigma)' = 0$$ Since $k(s), k'(s)$ and $\tau(s)$ are all non zero, then, dividing by $2p'\sigma$: $$\frac{p}{\sigma} + (p'\sigma)'=0$$ as desired.