ANSWER:
[tex](x+4)^{2}+(y-1)^{2}=30[/tex]
Center (4, -1), r = √30
STEP-BY-STEP EXPLANATION:
We have an equation in its general form, we must convert it to its standard form of equation of the circle to determine the center and radius, like this:
[tex]x^2+y^2+8x-2y-13=0[/tex]
In the following way, we solve correctly
[tex]\begin{gathered} \text{ The equation of the circle:} \\ \\ \left(x−h\right)^2+\left(y−k\right)^2=r^2\: \\ \\ (h,k)\text{ is the center and r is the radius } \\ \\ \text{ Therefore:} \\ \\ x^2+y^2+8x-2y-13=0 \\ \\ x^2+y^2+8x-2y=13 \\ \\ \left(x^2+8x\right)+\left(y^2-2y\right)=13 \\ \\ \left(x^2+8x+16\right)+\left(y^2-2y+1\right)=13+16+1 \\ \\ \left(x+4\right)^2+\left(y-1\right)^2=30 \\ \\ \text{ Thus} \\ \\ \text{ The center is \lparen-4,1\rparen} \\ \\ \text{ The radius is }\sqrt{30} \end{gathered}[/tex]
Center (4, -1), r = √30
