Defining curvature via osculating circles I am trying to figure out a geometrically accessible definition for the curvature of a smooth plane curve $c:I \to \mathbb{R}^2$ where $I$ is an interval and $c' \neq 0$ everywhere. My plan is to define the curvature via the osculating circle so that the definition might take the following form: if the osculating circle at a point $c(t_0)$ exists and has radius $r$, define the curvature as $\kappa(t_0):=1/r$. If the circle doesn't exist (for example, if $c$ is a straight line), define $\kappa(t_0):=0$. Moreover, I want to do all this without assuming that $c$ is a unit-speed parametrization.
Question: Does anyone know how such a definition of the curvature could look like in mathematically precise terms?
It is important that the definition is mathematically precise. I also plan to compare the above definition with the more common definition which uses the Frenet frame and show that both definitions yield the same function $\kappa:I \to \mathbb{R}$ up to sign. So, a robust definition is wanted.
My attempts and problems: First of all, we need a precise definition of the osculating circle and a precise and useful criterion when it exists. Here is where I already stuck. I tried to work with something like this: For $h>0$, let $M(t_0,h)$ the center of the circle through the three points $c(t_0-h)$, $c(t_0)$ and $c(t_0+h)$, presupposed they are not located on a straight line. The osculating circle at $c(t_0)$ could then be "defined" as the circle with center $M$ and radius $r$ where
$$M:= \lim_{h \to 0} M(t_0,h),\phantom{aaa}r:=\Vert M-c(t_0)\Vert$$
In order for this definition to work, we must ensure that for sufficiently small $h>0$, the three points $c(t_0 \pm h), c(t_0)$ are not collinear. Otherwise, the points $M(t_0,h)$ do not exist as $h \to 0$ and it makes no sense to talk about their limit. So, we may at least require some condition like this:
$\phantom{aa}$ There is a sequence $(h_n)_{n\in \mathbb{N}}$ with $h_n \to 0$ and all $h_n >0$ such that for each $n$, the points $\phantom{aai}$$c(t_0 \pm h_n), c(t_0)$ are not collinear.
But using only this condition, I have no idea how to prove that the limit $\lim_{h \to 0} M(t_0,h)$ exists which is also necessary in order to show that the definition makes sense.
Needless to say that all this should be done without referring to the curvature since I want to use the osculating circles to define the curvature and avoid circularity.
Remark: During the past days, I already posted some related questions. However, they might have been unclear or to narrowly focused on particular details so that answers didn't really help me. (However, thank you for all the answers.) Some users suggested that I formulate a new question to put all this into a broader context so that the people here get a better idea of what I am trying to achieve and why I have all the questions. I hope this post clarifies the things a bit.
 A: Some hints.  Let $c\colon I\to\mathbb R^2$ be immersive and $\alpha$ such that $\det(c'(\alpha),c''(\alpha))\neq0$. Define $J(x,y):=(-y,x)$.
Show that there is an open interval $ U\subset I$  containing $\alpha$ such that $\det(c'(a),c''(b))\neq0$ for all $a$, $b\in U$.
Let $r$, $s$, and $t\in U$ (with $r<s<t$).
The key point is that there exist $a$ and $b\in U$ such that (invoke the alternating mean value theorem by Schwarz)
$$2\det(c(s)-c(r),c(t)-c(r))=\det(c'(a),c''(b))(s-r)(t-r)(t-s)\neq0.$$
From here $c(s)-c(r)$ and $c(t)-c(r)$ are linearly independent, hence $c(r)$, $c(s)$ and $c(t)$ aren't collinear.  Thus there's exactly one circle $S(r,s,t)$ through these points.  Let $\rho(r,s,t)$ be its radius and $m(r,s,t)$ its center.
Calculate
$$m(r,s,t)=\frac{c(r)+c(t)}{2}+\frac{\langle c(s)-c(r),c(t)-c(s)\rangle}{2\det(c(s)-c(r),c(t)-c(r))}J(c(t)-c(r)).$$
Expand the second term by $1/(s-r)(t-r)(t-s)$.  For $r,s,t\to\alpha$ it will converge to
$$\frac{\|c'(\alpha)\|^2}{\det(c'(\alpha),c''(\alpha))}Jc'(\alpha).$$
Finally $\rho(r,s,t)=\|c(r)-m(r,s,t)\|$.
EDIT The Schwarz Mean Value Theorem
Schwarz stated in the cited paper (in a more "modern" language and a bit generalised) the following:
Let $n$ and $m$ be natural numbers with $n\leq m$, $I$ a (non-empty) interval and $c\colon I\to\mathbb R^m=:V$ (or any other $m$-dimensional real vector space $V$) a $C^n$-path.  Let $\omega\colon V^{n}\to\mathbb R$ be an alternating multilinear form.  Let $t_0,\dots t_n\in I$ satisfying $t_0<\dots <t_n$.
Then there exist $x_i\in[t_0,t_i]$ for $i=0,\dots n$ satisfying $x_0\leq\dots\leq x_n$ such that
$$
\begin{align}
&\frac{\omega\bigl(c(t_1)-c(t_0),\dots, c(t_n)-c(t_0)\bigr)}{\prod_{0\leq i<j\leq n}(t_j-t_i)}=\\
 &\quad\omega\left(\frac{c'(x_0)}{1!},\dots,\frac{c^{(n)}(x_n)}{n!}\right).
\end{align}
$$
(For $m=1$ and $\omega=\det$ this is the "usual" MVT.)
A: One should remember that if a curve $C$ is defined by a regular parametrization $c:I\to \bf R^2$,$c(t)$ and a curve $C'$ is defined by a implicit function $f(x)=0$ with $f'\not (0)$, (here $x$ is a vector of $\bf R^2$) then $C'$ osculate $C$ at the order $n$ iff $f(c(t))=o(t^n)$.
For the calculation, we assume study the problem near the point $t=0$ we assume that $c(0)=0$, and are serching a point $p$ (the center of the circle) so that $\vert c(t)-p\vert ^2-\vert p\vert ^2$=o(t^2)$.
Now $c(t)= c'(0) t+ c"(0) {t^2\over 2}+o(t^2)$
SO $\vert c(t)-p\vert ^2-\vert p\vert ^2= 2 t<p, c'(0)>+t^2<p, c"(0)>+t^2<c'(0),c'(0)>+o(t^2)$
So if $c',c"$ is a frame, namely if the origin is not a flex point, you see that your problem has a unique solution : $<p, c'(0)>=0$ (theismeans that $p$ is on the normal at $c(0$ of the cure , and $<p, c"(0)>=-<c'(0),c'(0)>$
If the parametriztion is normal, $c'(0)=T$ is a unit vector, $c"(0)= k N$ and $p={1\over k} N$
