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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 $h$...

russoo
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  • Concerning the addendum - What you can do is broaden your concept of the "limits existing" to include the case of $\lim_{h\to 0+} r = \infty$ and $\exists \alpha, \lim_{h\to 0+} M\cdot \alpha/|M| = 0$. This is what happens when the osculating circle is a straight line (where $\cdot$ is the Euclidean inner product). – Paul Sinclair Dec 30 '23 at 12:46

3 Answers3

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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 $h$.)

Ted Shifrin
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  • Thanks. So far, my only idea is to solve the equations $g_h'(\theta_h)=0$ and $g_h''(\xi_h)=0$ for $a_h$ and $b_h$ and then conclude convergence. But this won't make use of the limit $g_h(t)$ for $h \to 0$ and also doesn't use a Taylor expansion or implicit functions. Not sure if that is way you have in mind... – russoo Apr 26 '21 at 12:50
  • No, I wouldn't do that. Argue that $g'(0)=g''(0)=0$ (so some sort of uniform convergence on a closed interval around $0$). I wasn't committing to Taylor or implicit function theorem, but they are possible ingredients in different arguments. – Ted Shifrin Apr 26 '21 at 16:56
  • And how to show that $g'(0)=0$? – russoo Apr 27 '21 at 16:40
  • And by $g'(0)=0$, do you mean $\lim g_h'(0)=0$ for $h \to 0$? – russoo Apr 27 '21 at 16:51
  • I guess I'm suggesting that the function $G(t,h) = g_h(t)$ (with $g_0 = g$) is smooth. – Ted Shifrin Apr 27 '21 at 16:53
  • Sorry. I really don't see how this should help with the problem. – russoo Apr 27 '21 at 18:02
  • Have you actually tried to do the computation? What I suggested gives you $a=a_0=0$ and $b=b_0=1/\kappa(0)$, as it should. – Ted Shifrin Apr 27 '21 at 18:04
  • What I've done is computing $g_h,g_h', g_h''$ and then tried various ways to find expressions for $a_h,b_h$ and also tried to figure out how these expressions behave as $h$ tends to $0$ but I am not sure if this is what you meant. Up to now, this didn't give me the final result. Moreover, I don't want to use the curvature $\kappa$. – russoo Apr 27 '21 at 18:25
  • To summarize my problem: It is clear for me that $g_h'(\theta), g_h'(\tau_h), g_h''(\xi_h) \to 0$ as $h \to 0$ (trivial). But how does it follow from that that $a_h,b_h \to 0$ as $h \to 0$? – russoo Apr 28 '21 at 19:42
  • just wanted to let you know that I understand all parts of you answer and was able to solve the problem I stated above. Therefore accpeted your answer. What was necessary is to assume the curvature is not zero, or equivalently $\det(c'(0),c''(0))\neq 0$ and that confused me quite a bit. Many thanks! Finally, after a few years... ;) – russoo Oct 01 '23 at 14:51
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    Yes, I just read your edit (because you finally responded). Indeed, nonzero curvature at the point is necessary. – Ted Shifrin Oct 01 '23 at 19:18
  • @TedShifrin How to show that $g_h$ converges uniformly to $g$? – Lorenzo Vanni Dec 28 '23 at 20:18
  • @LorenzoVanni On a compact neighborhood of the origin, sure. That should be immediate. – Ted Shifrin Dec 29 '23 at 00:01
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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.

Eric Towers
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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 $a=o(s^2), b=R+o(s)$

Thomas
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  • Cannot really follow you. What I see is that you get a $(2 \times 2)$ linear system for the center of the circle and I see that the determinant is $\neq 0$. However, the determinant tends to $0$ for $h \to 0$ which is a problem. But then, I don't understand what you are doing by introducing the Frenet Frame. – russoo Apr 28 '21 at 19:30
  • Certainly the detreminant goes to 0, but the solution not (compute it). I introduced the Frenet frame to have a nice farme in which the computation is easy. – Thomas Apr 29 '21 at 05:24
  • I tried to compute it but I don't get anything from which I can see that we have convergence. – russoo Apr 29 '21 at 16:09
  • If I just try to compute the solution in a straight forward way, I get a complicated term in which the denominator tends to $0$ as $h$ does and I do not see a way to simplify or manipulate the term so that this problem disappears. – russoo Apr 29 '21 at 16:51
  • come one : it is (2,2) linear system ! – Thomas Apr 29 '21 at 17:10
  • As I said: I can compute the solution but I cannot show convergence for it. – russoo Apr 29 '21 at 17:14