We will operate as follows:
[tex]x^2-8x+y^2+2y+1=0\Rightarrow(x^2-8x)+(y^2+2y+1)=0[/tex][tex]\Rightarrow x(x-8)+(y^{}+1)^2=0\Rightarrow(x-4)^2+(y+1)^2=16[/tex]Now, from this, we can see that the center of the circle is located at (4, -1).
The graph of the circle is the following:
