The center-radius form of the circle equation
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have:
[tex]x^2-4x+y^2-6y=-8[/tex]
Use
[tex](a-b)^2=a^2-2ab+b^2\qquad(*)[/tex]
[tex]x^2-2(x)(2)+y^2-2(y)(3)=-8\qquad\text{add}\ 2^2\ \text{and}\ 3^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(2)+2^2}_{(*)}+\underbrace{y^2-2(y)(3)+3^2}_{(*)}=2^2+3^2-8\\\\(x-2)^2+(y-3)^2=4+9-8\\\\(x-2)^2+(y-3)^2=5\\\\(x-2)^2+(y-3)^2=(\sqrt5)^2\\\\Answer:\\\\\boxed{center:(2,\ 3)}\\\\\boxed{radius:r=\sqrt5}[/tex]
