This $O(n)$ solution is not intuitional. I proved using modulo arithmetic.
Let $x$ be the index of the node where fast pointer and slow pointer meets.
Let $y$ be the index of the tail.
Let $z$ be the index of the node that tail's next pointer is connected to.
From the definition of $x$:
$2x-x =x= 0 \mod N$
From the definition of $y$ and$z$:
Therefore, $x=N=y-z+1 \mod N$