Answer
Center: $(5,3)$ $;$ Radius: $8$
Work Step by Step
$x^{2}+y^{2}-10x-6y-30=0$
Take $16$ to the right side:
$x^{2}+y^{2}-10x-6y=30$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}-10x)+(y^{2}-6y)=30$
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, $b=-10$ and for the second group, $b=-6$
$\Big[x^{2}-10x+\Big(\dfrac{-10}{2}\Big)^{2}\Big]+\Big[y^{2}-6y+\Big(\dfrac{-6}{2}\Big)^{2}\Big]=30+\Big(\dfrac{-10}{2}\Big)^{2}+\Big(\dfrac{-6}{2}\Big)^{2}$
$(x^{2}-10x+25)+(y^{2}-6y+9)=30+25+9$
$(x^{2}-10x+25)+(y^{2}-6y+9)=64$
Factor the expressions inside each parentheses, which are perfect square trinomials:
$(x-5)^{2}+(y-3)^{2}=4$
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 $(h,k)=(5,3)$, so the center of the circle is the point $(5,3)$
The radius is:
$r^{2}=64$
$\sqrt{r^{2}}=\sqrt{64}$
$r=8$
The graph is:
