Answer
Center: $(-4,1)$ $;$ Radius: $5$
Work Step by Step
$x^{2}+y^{2}+8x-2y-8=0$
Take $8$ to the right side:
$x^{2}+y^{2}+8x-2y=8$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}+8x)+(y^{2}-2y)=8$
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, $b=8$ and for the second group, $b=-2$
$\Big[x^{2}+8x+\Big(\dfrac{8}{2}\Big)^{2}\Big]+\Big[y^{2}-2y+\Big(-\dfrac{2}{2}\Big)^{2}\Big]=8+\Big(\dfrac{8}{2}\Big)^{2}+\Big(-\dfrac{2}{2}\Big)^{2}$
$(x^{2}+8x+16)+(y^{2}-2y+1)=8+16+1$
$(x^{2}+8x+16)+(y^{2}-2y+1)=25$
Factor the expressions inside each parentheses, which are perfect square trinomials:
$(x+4)^{2}+(y-1)^{2}=25$
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 $(h,k)=(-4,1)$, so the center of the circle is the point $(-4,1)$
The radius is:
$r^{2}=25$
$\sqrt{r^{2}}=\sqrt{25}$
$r=5$
The graph is:
