# 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?

## 1 Answer

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.