#### 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 can be written as:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}-6y-7=0 \\
& {{x}^{2}}+\left( {{y}^{2}}-6y \right)=7 \\
& {{x}^{2}}+\left( {{y}^{2}}-6y+9 \right)=7+9 \\
& {{\left( x-0 \right)}^{2}}+{{\left( y-3 \right)}^{2}}=16
\end{align}$
So, the equation of the circle in the standard form is:
${{\left( x-0 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{4}^{2}}$
Now, compare this equation to the standard form, to get the value of $h=0,k=1\,,\,\text{ and }\,\,r=4$.
Therefore, the center is $\left( 0,3 \right)$ and the radius is $4$ units.