$\textbf{preliminaries}$
I am in the course of geometry of curves and surfaces, clearly frenet's trihedron is $\{T(s),N(s),B(s)\}$ where $T(s)=\alpha'(s)$ , $N(s)=\frac{T'(s)}{\|T'(s)\|}$ and $B(s)=T(s) \times N(s)$ The Serret-Frenet equations of the curve $\alpha$ are:
$$T'(s)=k(s)N(s)$$ $$N'(s)=-k(s)T(s)-\tau(s)B(s)$$ $$B'(s)=\tau(s)N(s)$$
$\textbf{Question}$ Let $\alpha: I \to \mathbb{R}^3$ be a curve p.b.a.l. With positive curvature such that $k'(s)\not = 0$ and $\tau(s)\not = 0$ for each $s\in I$. Prove that the trace of $\alpha$ is contained in a sphere of radius $r>0$ if and only if $$\frac{1}{k(s)^2}+\frac{k'(s)^2}{k(s)^4\tau(s)^2}=r^2$$
$\textbf{My attempt:}$
If the trace of $\alpha$ lies in a sphere of radius $r>0$, then $\|\alpha\|^2=r^2$ Differentiating this expression three times, we have: $$\langle \alpha,T \rangle=0$$ $$1+k\langle \alpha, N \rangle=0$$ and $$k'\langle \alpha, N\rangle -k\tau \langle \alpha, B \rangle=0$$ therefore: $r^2=\|\alpha\|^2=\langle \alpha , T \rangle^2+\langle \alpha, N \rangle^2+ \langle \alpha, B\rangle^2=(\frac{1}{k})^2+(\frac{k'}{\tau k^2})^2$.
Here I assumed the following equality $\|\alpha\|^2=\langle \alpha , T \rangle^2+\langle \alpha, N \rangle^2+ \langle \alpha, B\rangle^2$ but in reality I don't know how to justify it, can you tell me how to justify this equality?
conversely this turns out
$\textbf{imagge attached}$
