proving that a curve with constant curvature contained in a sphere its a circle i have a curve $\alpha:I\rightarrow \mathbb{R}^3$ such that his curvature $k$ is constantand $\alpha$ is entirely contained in a sphere, i must prove that this curve is a cirlce.
My try:
I need to prove that $\alpha$ has zero torsion, i supose the sphere with center at origin and radius $r>0$, so i got $|T^{'}(s)|=k$ for all $s$ where $T$ is the tangent line of $\alpha$ and $|\alpha(s)|^2=r^2$ for all $s$, i wanted to prove that the binormal $B$ of $\alpha$ is contants, but i derivatite these two equation a lot and i couln't conclude anything, anyone can help?

thanks to the coments of @willjagy i could conclude my demonstrations, first i find by derivating $|\alpha(s)|^2$ that $<T,\alpha>=0$, derivating again and using Frenet Formulas i get
$k<N,\alpha>+1=0$ and one more time:
$k^{'}<N,\alpha>-k\tau<B,\alpha>=0$, where $\tau$ is the torsion of the curve, so, as $k$ is contant and positive:
$\tau<B,\alpha>=0$ $ \forall s\in I$, 
and then at any cases $\tau=0$ or $<B,\alpha>=0$ i can conclude the $\alpha$ is a circle, for the first i get $\alpha$ in a great circle of the sphere and for the second an arbitrary cirlce with radius less or equal than $r$.
