Spherical Curve

Question

That seems logic :

"The osculating circle of a spherical curve is the intersection of the sphere and the osculating plane".

I don't how we can prove it geometrically instead of "the osculating circle is the best circle that approach the curve and the curve lies on the sphere".

How do you give the osculating circle of any space curve? It lies in the osculating plane and what are its center and radius? (By the way, have you checked the claim for the case where the curve is a latitude circle on the sphere?) — Ted Shifrin, Oct 18 '16 at 22:20
The center of the osculating circle is in the direction of N (Frenet). And the radius of the circle is 1/k with k the curvature. But what more ? — doeup, Oct 19 '16 at 10:52

score 1 · Answer 1 · edited Apr 13 '17 at 12:19

HINT: One standard way (see, for example, this MSE question) to work with spherical curves $\alpha$ is to assume the sphere is centered at the origin and write $\alpha(s)$ as a linear combination $$\alpha(s)=\lambda(s)T(s)+\mu(s)N(s)+\nu(s)B(s)$$ and solve for $\lambda(s),\mu(s),\nu(s)$ in terms of $\kappa(s)$ and $\tau(s)$. First you should see that $\lambda(s)=0$ for all $s$. Thinking about the geometry of $\alpha(s)=\mu(s)N(s)+\nu(s)B(s)$, you should be able to see that the osculating plane at $\alpha(s)$ intersects the sphere in a circle of radius $|\mu(s)|$.

Spherical Curve

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