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+y^2-10x-16y+53=0[/tex]
Use [tex](a-b)^2=a^2-2ab+b^2\qquad(*)[/tex]
[tex]x^2-10x+y^2-16y+53=0\qquad\text{subtract 53 from both sides}\\\\x^2-2(x)(5)+y^2-2(y)(8)=-53\qquad\text{add}\ 5^2\ \text{and}\ 8^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(5)+5^2}_{(*)}+\underbrace{y^2-2(y)(8)+8^2}_{(*)}=5^2+8^2-53\\\\(x-5)^2+(y-8)^2=25+64-53\\\\(x-5)^2+(y-8)^2=36\\\\(x-5)^2+(y-8)^2=6^2\\\\Answer:\\\\\boxed{center:(5,\ 8)}\\\\\boxed{radius:r=6}[/tex]
