Assume that $α$ is a regular curve in $\mathbb{R}^2$ and all the normal lines of the curve pass though the origin. Prove that $α$ is contained in a circle around the origin. (Recall the normal line at $α(t)$ is the line through $α(t)$ pointing in the direction of the normal vector $N(t)$.)
If I can prove that the curvature is constant, then it would be solved, but I don't know how to do that. Any hint or suggestion?
Note: There is an old answer similar to this one, but it wasn't solved using the curvature. Also, it is not well explained from my point of view.
Best Answer
Let $\mathbf r(t)$ be the curve. The tangent vector to the curve at $\mathbf r(t)$ is $\dot {\mathbf r}(t)$.
The normal to the curve at $\mathbf r(t)$ is a line that passes through $\mathbf r(t)$. If it also passes through the origin, then it has direction vector $\mathbf r(t) - \mathbf 0 = \mathbf r(t)$.
But it is also perpendicular to the tangent vector $\dot {\mathbf r}(t)$. That is, $$\dot {\mathbf r} \cdot \mathbf r = 0$$
Now, integrate with respect to $t$.