Answer
Center: $(0,3)$ $;$ Radius: $4$
Work Step by Step
$x^{2}+y^{2}-6y-7=0$
Take $7$ to the right side of the equation:
$x^{2}+y^{2}-6y=7$
Group the terms with $y$ together:
$x^{2}+(y^{2}-6y)=7$
Complete the square for the expression inside parentheses by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside and also on the right side of the equation.
In this case, $b=-6$
$x^{2}+\Big[y^{2}-6y+\Big(-\dfrac{6}{2}\Big)^{2}\Big]=7+\Big(-\dfrac{6}{2}\Big)^{2}$
$x^{2}+(y^{2}-6y+9)=7+9$
$x^{2}+(y^{2}-6y+9)=16$
Factor the expression inside parentheses, which is a perfect square trinomial:
$x^{2}+(y-3)^{2}=16$
This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$
It can be seen from the obtained equation that $(h,k)=(0,3)$, so the center of the circle is the point $(0,3)$
The radius is:
$r^{2}=16$
$\sqrt{r^{2}}=\sqrt{16}$
$r=4$
The graph is:
