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? 

$\newcommand{\e}{\epsilon}
\renewcommand{\d}{\pm\Delta s}
\renewcommand{\n}[1][]{\,n\left(s_{0}#1\right)}
\newcommand{\gs}{\gamma\left(s_{0}\right)}
\newcommand{\gd}{\gamma\left(s_{0}\d\right)}
\newcommand{\dg}{\gamma\,'\left(s_{0}\right)}
\newcommand{\ddg}{\gamma\,''\left(s_{0}\right)}
\newcommand{\O}[1][]{\mathcal{O}\big(\Delta^{#1}\big)}$
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}
