Limit definition of the osculating circle

Question

Let $I$ be an interval and $c:I \to \mathbb{R}^2$ a smooth curve with $c' \neq 0$ everywhere. I am currently trying to figure out how to define the osculating circle at a point $c(t_0)$, $t_0 \in I$, without assuming that the curvature has previously been defined (which could for example be done via the Frenet frame).

One way to define it might be the following way using a limit: Assume that locally around the point $c(t_0)$ the curve is not a straight line. Consider a value $h>0$ with $t_0 \pm h \in I$ such that the three points $c(t_0-h)$, $c(t_0)$ and $c(t_0+h)$ are not located on a straight line. Let $M(t_0,h)$ be the center of the circle defined by those three points. The osculating circle at $c(t_0)$ might then be defined as being the circle with center $M$ and radius $r$ where these values are given by $$M:=\lim_{h \to 0} M(t_0,h)\phantom{aaa}\text{and}\phantom{aaa}r:=\Vert M-c(t_0) \Vert$$ The limit runs only over those values of $h$ for which the three points $c(t_0-h)$, $c(t_0)$ and $c(t_0+h)$ are not located on a straight line. (This is not really a restriction. Since $c$ is not a straight line around $c(t_0)$, just choose $h$ small enough.)

My question: How to show that the limit from the above definition exists? Or is there something wrong with the definition?

Some background: A similar definition for the osculating circle is indicated in the book "Modern Differential Geometry of Curves and Surfaces with Mathematica" by Gray et al. in section 4.4. However, they do not explain why the limit should exist or least I do not understand this.

Addendum: Finally, I learned that there must be a "mistake" in the definition. Since this is not mentioned in any of the answers, I decided to explain it here: The condition that $c$ is not a straight line locally around $c(t_0)$ is not sufficient for the osculating circle to exist. Under this condition, it may still happen that for every sufficiently small $h>0$, you can find a straight line through the points $c(t_0-h)$, $c(t_0)$ and $c(t_0+h)$. Take for example $c(t)=(t,t^3)$, $-1/2 < t < 1/2$ and $t_0=0$. Therefore: No osculating circle at $c(t_0)$ in that case. The points $M(t_0,h)$ are not well-defined because they don't exist. So, currently I am wondering what the "correct" condition on $c$ might be guaranteeing that the points $M(t_0,h)$ exist for sufficiently small $h$...

Answer 1

HINTS: Without loss of generality, assume $c$ is arclength-parametrized, $t_0=0$, $c(0)=0$, and $c'(0)=(1,0)$.

Take the equation of the circle through $c(-h), 0, c(h)$ to be $f_h(x,y)=(x-a_h)^2 + (y-b_h)^2 - (a_h^2+b_h^2) = 0$. Let $g_h(t) = f_h(c(t))$. Apply Rolle's Theorem a few times to get $-h<\theta_h<0<\tau_h<h$ with $g'_h(\theta_h)=g'_h(\tau_h)=0$, and again to get $\xi_h$ with $g''_h(\xi_h)=0$. You should be able to argue now that $g(t)=\lim\limits_{h\to 0^+} g_h(t)$ exists and $g(0)=g'(0)=g''(0)=0$. This will give you what you want. (You can use a Taylor expansion of $c(t)$ or give an implicit function theorem argument that $(a_h,b_h)$ is a smooth function of $h$.)

Answer 2

The equation of a circle is $x^2+y^2-ax-by-c=0$ say $F(x,y ; a,b,c)=0$ $1/2(a,b)$ is the center $a^2+b^2+c$ the square of the radius.

So the circle through 3 points is obtained by solving a $(3,3)$ linear system (the unknown are $a,b,c$) with coefficient which are polynomial in the coordinates if these point, so that this circle (the parameter $(a,b,c)$ is unique and continuously depends on the three point. Of course, one must be careful and check that the determinant is $\not 0$, which means that the three points are not one the same line.

Here, one can argue further and make the computation in the Frenet frame, so that the equation of the circle through $M(-s), M(s), 0$ are

$c=0$

$x^2(s)+y^2(s)-ax(s)-by(s)=0$

$x^2(-s)+y^2(-s)-ax(-s)-by(-s)=0$

But $(x(s),y(s))= s(1,0)+ {s^2\over 2 R}(0,1)+o(s^2)$ (Frenet)

so the system is :

$a( s+o(s))+b( {s^2\over 2 R}+o(s^2))= s^2+o(s^2)$

$a( -s+o(s))+b( {s^2\over 2 R}+o(s^2))= s^2+o(s^2)$

And a simple computation yields $a=o(s^2), b=R+o(s)$

Answer 3

Responding to the addendum:

A test for collinearity of the three distinct points $(x_1,y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ uses determinants. These three distinct points are collinear if and only if $$ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \text{.} $$

You may also be hoping for too much with symmetric choices in the parameter space. You may be happier using $c(t_0 + k)$, $c(t_0)$, and $c(t_0 + h)$ in the limit as $h \rightarrow 0$ and $k \rightarrow 0$, independently. This won't solve the problem you found with the map $t \mapsto (t, t^3)$, where the "$y$ as a function of $x$" graph has an inflection point at $x = 0$, since (finite radius) circles do not approximate the graph of $y = x^3$ at its inflection point.