Showing that a circle is an osculating circle of a unit-speed curve 
Let $\alpha : I\to\mathbb{R}^2$ be a smooth plane curve parametrized by arc length, and assume that $0\in I$. A circle with radius $r$ centred at $p$ is called the osculating circle of $\alpha$ at $0$ if the function $f(s)=\Vert \alpha(s)-p\Vert^2$ satisfies $f(0)=r^2$ and $f'(0)=f''(0)=0$. Prove that if $\kappa(0)\neq 0$, then the circle of radius $\frac{1}{\vert \kappa(0)\vert}$ centred at $p=\alpha(0)+\frac{1}{\kappa(0)}\mathbf{n}(0)$ is the osculating circle of $\alpha$ at $0$.


Here $\kappa(s)$ is the signed curvature of $\alpha$ at $s$, and below $k(s) = |\kappa(s)|$ is the curvature of $\alpha$ at $s$.
I have been able to show both $f(0)=\frac{1}{\vert \kappa(0)\vert^2}=r^2$ and $f'(0)=0$, but I am having trouble with showing that $f''(0)=0$. Here's what I have done so far, letting $\alpha(s)=(x(s), y(s))$:
I've determined an expression for $f''(s)$:
$$
\begin{equation*}
f''(s) = 2x'(s)^2 + 2x(s)x''(s)-2x(0)x''(0)-\frac{2x''(0)x''(s)}{\kappa(0)k(0)} + 2y'(s)^2+2y(s)y''(s)-2y(0)y''(s)-\frac{2y''(0)y''(s)}{\kappa(0)k(0)}
\end{equation*}
$$
From which I can get an expression for $f''(0)$:
$$
\begin{align*}
f''(0) &= 2x'(0)^2-\frac{2x''(0)^2}{\kappa(0)k(0)} + 2y'(0)^2-\frac{2y''(0)^2}{\kappa(0)k(0)} \\
&= 2\left(x'(0)^2+y'(0)^2\right) - 2\left(\frac{x''(0)^2}{\kappa(0)k(0)}+\frac{y''(0)^2}{\kappa(0)k(0)}\right) \\
&= 2\Vert \alpha'(0)\Vert^2 - \frac{2\Vert\alpha''(0)\Vert^2}{\kappa(0)k(0)} \\
&= 2 - \frac{2k(0)}{\kappa(0)}
\end{align*}
$$
Now at this point if I knew that the signed curvature $\kappa(0)$ was
positive, then $k(0)=\kappa(0)$ and $\frac{2k(0)}{\kappa(0)}=2$ and I'd be done. Otherwise, the best that I have is
$$
\begin{equation*}
f''(0) = 2-2\,\textrm{sgn}(\kappa(0))
\end{equation*}
$$
Which is obviously nonzero for $\kappa(0)<0$. What am I missing here?

Since the sign of $\kappa(s)$ depends only on the choice of orientation of $\alpha$ (and in particular has identical magnitude to $k(s)$) I can, without loss of generality, choose $\alpha$ to be oriented such that
$$
\begin{equation*}
\kappa(0) = k(0)
\end{equation*}
$$
And so $\textrm{sgn}({\kappa(0)})=1$, showing the result.
