Answer
The osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\frac{{17}}{2}\sqrt {17} \cos t + \frac{{25}}{2},\frac{{17}}{2}\sqrt {17} \sin t - 32} \right)$
Work Step by Step
Write $f\left( x \right) = y = \sqrt x $. Thus, we can use the parametrization:
${\bf{r}}\left( x \right) = \left( {x,f\left( x \right)} \right) = \left( {x,\sqrt x } \right)$
Step 1. Find the radius of the osculating circle
By Eq. (5) of Theorem 2 (Section 14.4), the curvature is
$\kappa \left( x \right) = \frac{{\left| {f{\rm{''}}\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}} = \frac{{\left| { - \frac{1}{4}{x^{ - 3/2}}} \right|}}{{{{\left( {1 + {{\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( 4 \right) = \frac{{1/32}}{{{{\left( {17/16} \right)}^{3/2}}}} = \frac{2}{{17\sqrt {17} }}$
Thus, the osculating circle has radius $R = 1/\kappa \left( 1 \right) = \frac{{17}}{2}\sqrt {17} $.
Step 2. Find the unit normal vector ${\bf{N}}$ at $x=4$
We have ${\bf{r}}'\left( x \right) = \left( {1,\frac{1}{{2\sqrt x }}} \right)$. The unit tangent vector at $x=4$ is
${\bf{T}}\left( 4 \right) = \frac{{{\bf{r}}'\left( 4 \right)}}{{||{\bf{r}}'\left( 4 \right)||}} = \frac{{\left( {1,\frac{1}{4}} \right)}}{{||\left( {1,\frac{1}{4}} \right)||}}$
${\bf{T}}\left( 4 \right) = \frac{1}{{\sqrt {17/16} }}\left( {1,\frac{1}{4}} \right)$
${\bf{T}}\left( 4 \right) = \frac{4}{{\sqrt {17} }}\left( {1,\frac{1}{4}} \right)$
Since ${\bf{N}}\left( 4 \right)$ must be orthogonal to ${\bf{T}}\left( 4 \right)$, we can have ${\bf{N}}\left( 4 \right) = \frac{4}{{\sqrt {17} }}\left( { - \frac{1}{4},1} \right)$ or ${\bf{N}}\left( 4 \right) = \frac{4}{{\sqrt {17} }}\left( {\frac{1}{4}, - 1} \right)$ as the candidates for ${\bf{N}}\left( 4 \right)$ because they satisfy ${\bf{T}}\left( 4 \right)\cdot{\bf{N}}\left( 4 \right) = 0$. However, we need to choose ${\bf{N}}\left( 4 \right)$ such that it points inward of the curve at $x=4$. Thus,
${\bf{N}}\left( 4 \right) = \frac{4}{{\sqrt {17} }}\left( {\frac{1}{4}, - 1} \right)$
Step 3. Find the center of the osculating circle $Q$
By Eq. (9) of Section 14.4, the center of the osculating circle at $x=4$
$\overrightarrow {OQ} = {\bf{r}}\left( 4 \right) + R{\bf{N}}\left( 4 \right) = \left( {4,2} \right) + \frac{{17}}{2}\sqrt {17} \cdot\frac{4}{{\sqrt {17} }}\left( {\frac{1}{4}, - 1} \right)$
$\overrightarrow {OQ} = \left( {4,2} \right) + 34\left( {\frac{1}{4}, - 1} \right) = \left( {\frac{{25}}{2}, - 32} \right)$
Step 4. Parametrize the osculating circle
At $x=4$, the osculating circle has radius $R = \frac{{17}}{2}\sqrt {17} $ and center $\left( {\frac{{25}}{2}, - 32} \right)$. So, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\frac{{25}}{2}, - 32} \right) + \frac{{17}}{2}\sqrt {17} \left( {\cos t,\sin t} \right)$
${\bf{o}}\left( t \right) = \left( {\frac{{17}}{2}\sqrt {17} \cos t + \frac{{25}}{2},\frac{{17}}{2}\sqrt {17} \sin t - 32} \right)$
