#### Answer

See below:

#### Work Step by Step

We know that for a circle, the standard form is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$,
where the center is $\left( h,k \right)$ and the radius is $r$.
So, the equation of the circle can be written as:
$\begin{align}
& {{x}^{2}}-2x+{{y}^{2}}-15=0 \\
& \left( {{x}^{2}}-2x \right)+{{y}^{2}}=15 \\
& \left( {{x}^{2}}-2x+1 \right)+{{y}^{2}}=15+1 \\
& {{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=16
\end{align}$
So, the equation of the circle in the standard form is given as:
${{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{4}^{2}}$
Now, compare this equation to the standard form, to get the value of $h=1,k=0,\,\,\text{ and }\,\,r=4$.
Therefore, the center is $\left( 1,0 \right)$ and the radius is $4$ units.