# Curve with constant torsion and curvature is a circular helix.

I am trying to find a proof for the 9th question of section 2.4, from the book Elementary Differential Geometry by Barrett O'Neill. I want to show that a curve $\alpha$ with curvature $\kappa$ and torsion $\tau$ both constant is a circular helix.

My thoughts: I know $\alpha$ must be a cylindrical helix since $\tau/\kappa$ is a constant. Remains to show that it is circular. I'm not sure how to do this. I found this question here : Is the helix the unique path with constant curvature and constant torsion? where the answer gives a hint to prove N′′(s)=−(κ2+τ2)N(s). I can do this, but where does it head me toward? I cannot see where this will lead me.

Random thought: It occurs to me that if I can project the curve $\alpha$ on the xz plane, it must be a circle. But I don't know how I would go about showing this.

Any help is appreciated. Thank you for your time.

The intent of my hint was to then deduce that $\vec N(s)=\cos(ks)\vec c_1+\sin(cs)\vec c_2$ with $k^2=\kappa^2+\tau^2$ and $\vec c_1,\vec c_2$ fixed vectors. Show then that $\vec c_1$ and $\vec c_2$ must be orthogonal and show that you get circular motion in the plane they span. Can you finish from there?