This can be generalized a little, to hare step sizes $>2$ and to the tortoise and hare not necessarily starting on the same node. All step sizes $>2$ work if they start on the same node. If they start on different nodes then only a step size of 2 is guaranteed to work no matter what the cycle length.
Let $T =$ the number of nodes before the start of the cycle. Let $C =$ the length of the cycle. Let $S =$ the step size of the hare (2 for the problem as given).
After $n$ iterations, the tortoise has moved $n$ nodes, and the hare $nS$. The hare and tortoise are on the same node whenever $n \geq T$ and $n-T \equiv nS-T$ mod $C$. So we are looking for solutions of $n(S-1) \equiv 0$ mod $C$ with $n \geq T$.
When $S=2$ as in the problem as given, this is simply $n\equiv0$ mod $C$. This happens exactly when $n$ is a multiple of $C$.
When $S>2$ it also happens on every multiple of $C$, but might also happens at other times if $S-1$ and $C$ are not relatively prime.
That covers starting on the same node. For starting on different nodes, let $h =$ the starting node of the hare, and $t=$ the starting node of the tortoise.
Then we need to solve $n(S-1) \equiv t-h$ mod $C$ instead of $n(S-1) \equiv 0$, subject to $t+n\geq T$.
When $S=2$, that is $n\equiv t-h$ mod $C$, which has $n = t-h+kC$ as solutions for all integers $k$. When $S>2$ it is possible that $n(S-1)\equiv t-h$ mod $C$ has no solutions.
Actually, you can even generalize some more, letting the tortoise have a step size $>1$. As long as the tortoise and hare start on the same node, and the hare has a larger step size than the tortoise, it will still work. If $S_h$ is the hare's step size and $S_t$ is the tortoise's step size, then instead of $n(S-1)\equiv 0$ mod $C$ we have to solve $n(S_h-S_t)\equiv 0$ mod $C$.