Centre of the circle

Question

Another approach to the curvature of a unit-speed plane curve $\gamma$ at a point $\gamma (s_0)$ is to look for the ‘best approximating circle’ at this point. We can then define the curvature of $\gamma$ to be the reciprocal of the radius of this circle. Carry out this programme by showing that the centre of the circle which passes through three nearby points $\gamma (s_0)$ and $\gamma (s_0 \pm \delta_s)$ on $\gamma$ approaches the point $$\epsilon (s_0) = \gamma (s_0) + \frac{1}{\kappa_s (s_0)}n_s(s_0)$$ as $\delta_s$ tends to zero. The circle $C$ with centre $\epsilon (s_0)$ passing through $\gamma (s_0)$ is called the osculating circle to $\gamma$ at the point $\gamma (s_0)$, and $\epsilon (s_0)$ is called the centre of curvature of $\gamma$ at $\gamma (s_0)$. The radius of $C$ is $\frac{1}{|\kappa_s (s_0)|} = \frac{1}{\kappa (s_0)}$, where $\kappa$ is the curvature of $\gamma$– this is called the radius of curvature of $\gamma$ at $\gamma (s_0)$.

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I have done the following: The three points $\gamma (s_0), \gamma (s_0 + \delta_s), \gamma (s_0 - \delta_s)$ are on the circle with radius $r$ and centre $\epsilon$. So $$r^2=\|\gamma (s_0)-\epsilon\|^2=\|\gamma (s_0 + \delta_s)-\epsilon\|^2=\|\gamma (s_0 - \delta_s)-\epsilon\|^2$$

Since we want show that the centre of the circle tends to $\epsilon (s_0)$ we do the following:

\begin{align} |\epsilon (s_0)-\epsilon| &=|\gamma (s_0) + \frac{1}{\kappa_s (s_0)}n_s(s_0)-\epsilon| \\ &\leq | \gamma (s_0) -\epsilon|+\frac{1}{|\kappa_s (s_0)|}|n_s(s_0)| \\ &=r+\frac{1}{|\kappa_s (s_0)|}|n_s(s_0)|. \end{align}

Is this correct so far? How could we continue?

EDIT:

We have that the radius of $C$ is $\frac{1}{|\kappa_s (s_0)|}$ so we get \begin{align} |\epsilon (s_0)-\epsilon| &=|\gamma (s_0) + \frac{1}{\kappa_s (s_0)}n_s(s_0)-\epsilon| \\ &\leq | \gamma (s_0) -\epsilon|+\frac{1}{|\kappa_s (s_0)|}|n_s(s_0)| \\ &=r+\frac{1}{|\kappa_s (s_0)|}|n_s(s_0)| \\ &=\frac{1}{|\kappa_s (s_0)|}+\frac{1}{|\kappa_s (s_0)|}|n_s(s_0)| \\ &=\frac{1}{|\kappa_s (s_0)|}(1+|n_s(s_0)|) \\ &=\frac{1}{|\kappa (s_0)|}(1+|n_s(s_0)|). \end{align}

What do we get from that?

Answer 1

$\newcommand{\e}{\epsilon} \renewcommand{\d}{\pm\Delta s} \renewcommand{\n}[1][]{\,n\left(s_{0}#1\right)} \newcommand{\gs}{\gamma\left(s_{0}\right)} \newcommand{\gd}{\gamma\left(s_{0}\d\right)} \newcommand{\dg}{\gamma\,'\left(s_{0}\right)} \newcommand{\ddg}{\gamma\,''\left(s_{0}\right)} \newcommand{\O}[1][]{\mathcal{O}\big(\Delta^{#1}\big)}$ Assume circle with radius $ R $ passes through points $\,\gs,\,\gd$. Then vectors connecting these points with center $\e$ have the same length $R$. \begin{align} \left\|\gs-\e\right\| = \left\|\gd-\e\right\| = R \end{align} Observe that these vectors are, in fact, unit normals $\,\n,\,\n[\pm\Delta],$ at points $\,\gs,\,\gd$ multiplied by radius $ R $:

\begin{align} \gs-\e &= R\n \\ \gd-\e &= R\n[\d] \end{align}

These vectors point to the same place, which is center of circle, i.e. $$\gs-R\n=\gd-R\n[\d]=\e$$

Therefore \begin{align} \gs - \gd = R\,\Big(\n-\n[\d]\Big) \implies R = \dfrac{\gd - \gs}{\n[\d]-\n} \end{align} Consider Taylor expansion $\, \gd = \gs + \Delta\,\dg + \dfrac{\Delta^{2}}{2}\,\ddg + \O[3]$.

Using Taylor expansion $\,\gd\,$ and $\,\n[\d] \,$ we get \begin{align} R = \dfrac{\dg + \O[2]}{n\,'\left(s_0\right) + \O[2]} \approx \dfrac{\dg}{n\,'\left(s_0\right)} = \dfrac{\dg}{-\dg\cdot\kappa} = -\dfrac{1}{\kappa} \end{align} where $\, \dfrac{dn}{ds} = -\kappa\left(s_{0}\right)\,\dg\,$ by Frenet formula.

Therefore the center of the circle can be expressed as \begin{align} \boxed{\;\e = \gs - R\n = \gs + \dfrac{\n}{\kappa\left(s_{0}\right)\,}\;} \end{align}

Answer 2

Too large for the comments:

$\epsilon = \gamma (s_0) + r\ n(s_0)$

$\epsilon$ is the vector sum of $\gamma (s_0)$ and the unit normal from it times the radius, $r\ n(s_0)$.

The proof would involve showing that as $\delta_s$ approaches zero the unit normal of the circle and the curve $\gamma(s)$ converge and by definition $\frac{1}{\kappa_s (s_0)} = r$.

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Edit

An approach may be to show that the tangent to the circle at $\gamma (s_0)$ becomes equal to the tangent to the curve $\gamma (s)$ as $\delta_s \rightarrow 0$. Since there are three common points $\gamma (s_0)$, $\gamma (s_0 \pm \delta_s)$ to the circle and the curve then both tangents are parallel to $\displaystyle \lim_{ \delta_s \rightarrow 0} \frac {\gamma (s_0 + \delta_s) - \gamma (s_0 - \delta_s)}{2\delta_s}$. The unit normals are perpendicular to the same tangent. The direction of the normal $(\pm)$ is the same because its the same three points. Its the same unit normal to the circle and the curve $\gamma (s)$ at $s_0$.

Still need an $\frac{1}{\kappa_s (s_0)} = r$ argument.

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Edit 2

Introduction to Vector Analysis -- fifth edition -- Davis & Snider , pg. 80

"define the curvature k as the rate at which the unit tangent vector turns with respect to an arc length along the curve:"

$$ k = \left|\frac{dT}{ds} \right| = \frac{|dT/dt|}{ds/dt}$$

o.k. Since there are three common points we can create two tangents by limits.

$$ k = \left|\frac{dT}{ds} \right| = \displaystyle \lim_{ \delta_s \rightarrow 0} \frac{\left|\frac{ \frac {\gamma (s_0 + \delta_s) - \gamma (s_0)}{\delta_s}}{\left|\frac {\gamma (s_0 + \delta_s) - \gamma (s_0)}{\delta_s}\right|} - \frac{\frac {\gamma (s_0) - \gamma (s_0 - \delta_s)}{\delta_s}}{\left|\frac {\gamma (s_0) - \gamma (s_0 - \delta_s)}{\delta_s}\right|}\right|}{\left|\delta_s\right|}$$

Since the three points are common to the circle and the curve $\gamma(s)$ they have the same limit curvature equation. i.e. The same curvature at $s_0$. Given the curvature of the circle is $\frac{1}{r}$ and the vector definition of $\epsilon = \gamma (s_0) + r\ n(s_0)$ then $\epsilon (s_0)$ is the same as $\epsilon$.

$$\lim_{ \delta_s \rightarrow 0} \epsilon = \epsilon (s_0) = \gamma (s_0) + \frac{1}{\kappa_s (s_0)}n_s(s_0)$$

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Summary

Because there are three common points to the circle and the curve $\gamma (s)$:

The unit tangent to the circle and the curve are the same at $s_0$.

The unit normal to the circle and the curve are perpendicular to the same tangent and are the same at $s_0$.

The curvature of the circle and the curve are the same at $s_0$ because they have the same limit equation for curvature.

The concluding equation is a vector sum.

Answer 3

Carry out this programme by showing that the centre of the circle which passes through three nearby points $\gamma (s_0)$ and $\gamma (s_0 \pm \delta_s)$ on $\gamma$ approaches the point $\epsilon (s_0)$ ...

This doesn't "carry out the programme", it only verifies a special case. To justify the idea of a circle of curvature requires convergence of the circumcircle of any inscribed triangle whose vertices approach $\gamma(s_0)$. Even with one vertex pinned at $\gamma(s_0)$, the triangles with the other two points at $\gamma(s_0 \pm \delta)$ are not generic.