Answer
Center: $(2,6)$ $;$ Radius: $7$
Work Step by Step
$x^{2}+y^{2}-4x-12y-9=0$
Take $9$ to the right side:
$x^{2}+y^{2}-4x-12y=9$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}-4x)+(y^{2}-12y)=9$
Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside each parentheses and also on the right side of the equation.
For the first group, $b=-4$ and for the second group, $b=-12$
$\Big[x^{2}-4x+\Big(-\dfrac{4}{2}\Big)^{2}\Big]+\Big[y^{2}-12y+\Big(-\dfrac{12}{2}\Big)^{2}\Big]=9+\Big(-\dfrac{4}{2}\Big)^{2}+\Big(-\dfrac{12}{2}\Big)^{2}$
$(x^{2}-4x+4)+(y^{2}-12y+36)=9+4+36$
$(x^{2}-4x+4)+(y^{2}-12y+36)=49$
Factor the expressions inside each parentheses, which are perfect square trinomials:
$(x-2)^{2}+(y-6)^{2}=49$
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 $(h,k)=(2,6)$, so the center of the circle is the point $(2,6)$
The radius is:
$r^{2}=49$
$\sqrt{r^{2}}=\sqrt{49}$
$r=7$
The graph is:
