Answer:
Center has coordinate [tex](-4; 1)[/tex] and radius [tex]\sqrt{30}[/tex]
Step-by-step explanation:
Let's start with rewriting terms as it was done:
[tex](x^2+8x) +(y^2-2y)-13=0[/tex]
Now let's focus on the two brackets: each one is one term from the square of a binomial, namely [tex](x+4)^2[/tex] and [tex](y-1)^2[/tex]. By expanding them we get:
[tex]x^2+8x+16[/tex] and [tex]y^2-2y+1[/tex]. Let's make the third term appear by adding and subtracting it (so we're not changing it) to our original equation:
[tex](x^2+8x+16)+(y^2-2y+1)-13-16-1=0[/tex] at this point it's just crunching numbers and wrapping the squares back up
[tex](x+4)^2 +(y-1)^2 = 30[/tex]
At this point we can use the center-radius form of the equation of a circle ([tex](x-\alpha)^2 +(y-\beta)^2 = r^2[/tex] to determine what we need by confronting the two
In particular, center has coordinate [tex](-4; 1)[/tex] and radius [tex]\sqrt{30}[/tex]