Differential Geometry Question

[mXCSNT]

Homework Statement

Assume that \tau(s) \neq 0 and k'(x) \neq 0 for all s \in I. Show that a necessary and sufficient condition for \alpha(I) to lie on a sphere is that R^2 + (R')^2T^2 = const where R = 1/k, T = 1/\tau, and R' = \frac{dr}{ds}

Homework Equations

\alpha(s) is a curve in R3 parametrized by arc length
k = curvature = |\alpha''|
\tau = torsion = -\frac{\alpha' \times \alpha'' \cdot \alpha'''}{k^2} (note sign; this is opposite of some conventions)

The Attempt at a Solution

I've approached this from 2 directions, but I haven't gotten them to meet. First, a necessary and sufficient condition is that |\alpha - P| is constant, where P is the center of the circle. Alternatively, (\alpha - P) \cdot \alpha' = 0.
And I've expanded out R^2 + (R')^2T^2 = const to get
\frac{(\alpha' \times \alpha'' \cdot \alpha''')^2 + (\alpha'' \cdot \alpha''')^2}{(\alpha'' \cdot \alpha'')(\alpha' \times \alpha'' \cdot \alpha''')^2} = const
Also, I'm going to guess that the const on the right hand side is some function of the radius of the sphere, maybe the square of it (which would be (\alpha - P) \cdot (\alpha - P)), because what else is constant in a sphere?

But I don't know where to go from here. I'm just looking for a hint at an avenue of approach, please nothing specific.

 

[mXCSNT]

So if I assume that the "const" in question is the square of the radius, that could make R and (R')T the lengths of respective legs of a right triangle whose hypotenuse is the radius. I know that R is the radius of curvature, so drawing a diagram I'm guessing that the center P is
P = Rn - (R')Tb
(where n is the normal vector \alpha''/|\alpha''| and b = t \times n is the binormal vector).
So then the sphere radius I want to test, (\alpha-P) \cdot (\alpha-P), becomes (\alpha - Rn + (R')Tb) \cdot (\alpha - Rn + (R')Tb)
= \alpha \cdot \alpha - 2 R \alpha \cdot n + 2 R' T \alpha \cdot b + R^2 + (R')^2T^2
Now if I assume that R^2 + (R')^2T^2 is constant I need to show that \alpha \cdot \alpha - 2 R \alpha \cdot n + 2 R' T \alpha \cdot b is constant to show the sphere radius is constant. Am I on the right track?

 

[mXCSNT]

Revised guess for center P:
P = \alpha + Rn - (R')Tb
so now the constant radius follows immediately and I simply have to show that P itself is constant.