The osculating circle of a curve $\alpha$ at the point $p \in \alpha$ is the circle $\mathbb{S}^1$ which is tangent to $\alpha$ at $p$ and has radius $\frac{1}{k(p)}$. Show that, if $k'(p) \neq 0$, then the osculating circle at $p$ intersects $\alpha$.
I'm trying to do this by using the canonical forms at $p$ of both $\alpha$ and the osculating circle. Obviously I know the canonical form of $\alpha$, but I'm having a hard time computing the canonical form of the osculating circle, even though I know it's equation is:
$$S = c(t) + \frac{1}{k(t)}(\cos(t),\sin(t)) =\alpha(t) +\frac{1}{k(t)} N(t)+ \frac{1}{k(t)}(\cos(t),\sin(t)) \\= \alpha(t)+\frac{1}{k(t)}(N(t)+(\cos(t),\sin(t)))$$