# Centre of the circle

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 $$\gamma (s_0)$$.

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?

• $|\epsilon (s_0)-\epsilon| \lt\lt r$ and $\frac{1}{\kappa_s (s_0)} \approx r$ The error of the two centers should be much smaller than the radius. The inverse of the curvature is approximately equal to the radius. Roughly the math says |-r + r| < |r| + |r| which wont converge to zero. – arthur Oct 20 '15 at 21:21
• $|n_s(s_0)|$ = 1. The unit normal magnitude is 1. – arthur Oct 21 '15 at 0:46
• – Han de Bruijn Oct 21 '15 at 13:35
• So, we have that $$|\epsilon (s_0 )-\epsilon | \leq \frac{1}{|\kappa (s_0)|}(1+|n_s (s_0)|) \Rightarrow |\epsilon (s_0 )-\epsilon | \leq \frac{1}{|\kappa (s_0)|}(1+1) \Rightarrow |\epsilon (s_0 )-\epsilon | \leq \frac{2}{|\kappa (s_0)|}$$ Does this give us the desired result? @arthur – Mary Star Oct 21 '15 at 15:29
• @Mary Star - No. We want $|\epsilon (s_0 )-\epsilon | => 0$ as $\delta_s => 0$. – arthur Oct 21 '15 at 20:53


\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}

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

$\epsilon$ 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 $\frac{1}{\kappa_s (s_0)} = r$.

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.

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 $\epsilon$.

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

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.

• corrected unit tangents in $k = \left|\frac{dT}{ds} \right|$ equation. – arthur Oct 22 '15 at 1:18
• Why is the direction of $n(s_0)$ towards the center of the circle? – Mary Star Nov 11 '15 at 1:02
• The circle and the curve share three points. The common tangent is defined using the two points $\gamma(s_0 \pm \delta_s)$. The mid point $\gamma(s_0)$ is common and determines the direction of the normal. More rigorously $N = \frac{\frac{dT}{ds}}{\left| \frac{dT}{ds} \right|}$ which uses the same three common points used in the $k = \left|\frac{dT}{ds} \right|$ equation. The normal to the circle and curve use the same three common points in the limit equation so they are equal. The radius (normal to the circle) points to the center of the circle, the normal to the curve is equal to it. – arthur Nov 11 '15 at 1:29
• Also the normal points to the concave side, for a circle that is inward toward the center. – arthur Nov 11 '15 at 1:49

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.