Answer
At $t=\pi$, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\pi + 4\cos t, - 2 + 4\sin t} \right)$
Work Step by Step
Step 1. Find the radius of the osculating circle
We have ${\bf{r}}\left( t \right) = \left( {t - \sin t,1 - \cos t} \right)$. The first and the second derivatives are ${\bf{r}}'\left( t \right) = \left( {1 - \cos t,\sin t} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {\sin t,\cos t} \right)$, respectively.
We compute the curvature by using Eq. (3) of Theorem 1:
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
To apply Theorem 1, we treat ${\bf{r}}'\left( t \right)$ and ${\bf{r}}{\rm{''}}\left( t \right)$ as vectors in ${\mathbb{R}^3}$ by setting the $z$-components equal to zero. Thus,
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{1 - \cos t}&{\sin t}&0\\
{\sin t}&{\cos t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {\cos t - {{\cos }^2}t - {{\sin }^2}t} \right){\bf{k}}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {\cos t - 1} \right){\bf{k}}$
So, $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)|| = \left| {\cos t - 1} \right|$.
$\kappa \left( t \right) = \frac{{\left| {\cos t - 1} \right|}}{{{{\left( {\sqrt {\left( {1 - \cos t,\sin t} \right)\cdot\left( {1 - \cos t,\sin t} \right)} } \right)}^3}}}$
$\kappa \left( t \right) = \frac{{\left| {\cos t - 1} \right|}}{{{{\left( {\sqrt {{{\left( {1 - \cos t} \right)}^2} + {{\sin }^2}t} } \right)}^3}}} = \frac{{\left| {\cos t - 1} \right|}}{{{{\left( {\sqrt {2 - 2\cos t} } \right)}^3}}}$
At $t=\pi$, $\kappa \left( \pi \right) = \frac{1}{4}$. Since $R = \frac{1}{\kappa }$, at $t=\pi$ the osculating circle has radius $4$.
Step 2. Find the normal vector
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( {1 - \cos t,\sin t} \right)}}{{\sqrt {\left( {1 - \cos t,\sin t} \right)\cdot\left( {1 - \cos t,\sin t} \right)} }} = \frac{{\left( {1 - \cos t,\sin t} \right)}}{{\sqrt {2 - 2\cos t} }}$
${\bf{T}}\left( t \right) = \left( {\frac{{1 - \cos t}}{{\sqrt {2 - 2\cos t} }},\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right)$
To find the derivative of ${\bf{T}}\left( t \right)$:
Evaluate $\frac{d}{{dt}}\left( {\frac{{1 - \cos t}}{{\sqrt {2 - 2\cos t} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{{1 - \cos t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{\left( {\sin t} \right)\sqrt {2 - 2\cos t} - \frac{{\left( {1 - \cos t} \right)\left( {2\sin t} \right)}}{{2\sqrt {2 - 2\cos t} }}}}{{2 - 2\cos t}}$
$\frac{d}{{dt}}\left( {\frac{{1 - \cos t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{\sin t\left( {2 - 2\cos t} \right) - \sin t\left( {1 - \cos t} \right)}}{{{{\left( {2 - 2\cos t} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{{1 - \cos t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{\sin t\left( {1 - \cos t} \right)}}{{{{\left( {2 - 2\cos t} \right)}^{3/2}}}} = \frac{{\sin t}}{{2\sqrt {2 - 2\cos t} }}$
Evaluate $\frac{d}{{dt}}\left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{\left( {\cos t} \right)\sqrt {2 - 2\cos t} - \frac{{2{{\sin }^2}t}}{{2\sqrt {2 - 2\cos t} }}}}{{2 - 2\cos t}}$
$\frac{d}{{dt}}\left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{\cos t\left( {2 - 2\cos t} \right) - {{\sin }^2}t}}{{{{\left( {2 - 2\cos t} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{2\cos t - 2{{\cos }^2}t - {{\sin }^2}t}}{{{{\left( {2 - 2\cos t} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right) = \frac{{2\cos t - {{\cos }^2}t - 1}}{{{{\left( {2 - 2\cos t} \right)}^{3/2}}}} = \frac{{ - {{\left( {1 - \cos t} \right)}^2}}}{{{{\left( {2 - 2\cos t} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}} \right) = - \sqrt {\frac{{1 - \cos t}}{8}} $
Thus, the derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \left( {\frac{{\sin t}}{{2\sqrt {2 - 2\cos t} }}, - \sqrt {\frac{{1 - \cos t}}{8}} } \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\sin }^2}t}}{{8\left( {1 - \cos t} \right)}} + \frac{{1 - \cos t}}{8} = \frac{{8{{\sin }^2}t + 8 - 16\cos t + 8{{\cos }^2}t}}{{64\left( {1 - \cos t} \right)}}$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{16 - 16\cos t}}{{64\left( {1 - \cos t} \right)}} = \frac{{16\left( {1 - \cos t} \right)}}{{64\left( {1 - \cos t} \right)}} = \frac{1}{4}$
$||{\bf{T}}'\left( t \right)|| = \frac{1}{2}$
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = 2\left( {\frac{{\sin t}}{{2\sqrt {2 - 2\cos t} }}, - \sqrt {\frac{{1 - \cos t}}{8}} } \right)$
${\bf{N}}\left( t \right) = \left( {\frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}, - \sqrt {\frac{{1 - \cos t}}{2}} } \right)$
Step 3. Find the the center of the osculating circle
By Eq. (9), the center of the osculating circle is given by
$\overrightarrow {OQ} = {\bf{r}}\left( {{t_0}} \right) + \frac{1}{{{\kappa _P}}}{\bf{N}}$
At $t=\pi$, the center of curvature is
$\overrightarrow {OQ} = {\bf{r}}\left( \pi \right) + \frac{1}{{\kappa \left( \pi \right)}}{\bf{N}}\left( \pi \right)$
$\overrightarrow {OQ} = \left( {\pi ,2} \right) + 4\left( {0, - 1} \right) = \left( {\pi , - 2} \right)$
Step 4. Parametrize the osculating circle
At $t=\pi$, the osculating circle has radius $R=4$ and center $\left( {\pi , - 2} \right)$. So, it can be parametrized by
${\bf{o}}\left( t \right) = \left( {\pi , - 2} \right) + 4\left( {\cos t,\sin t} \right)$
${\bf{o}}\left( t \right) = \left( {\pi + 4\cos t, - 2 + 4\sin t} \right)$
