Answer:
[tex]\huge\boxed{(-5,\ -6)}[/tex]
Step-by-step explanation:
[tex]\text{The standard form of an equation of a circle:}\\\\(x-h)^2+(y-k)^2=r^2\\\\(h,\ k)-\text{center}\\r-\text{radius}[/tex]
[tex]\text{We have}\\\\x^2+y^2+10x+12y+25=0\\\\\text{You can use}\\\\(x-h)^2+(y-k)^2=r^2\\x^2-2hx+h^2+y^2-2ky+k^2=r^2\qquad\text{subtract}\ r^2\ \text{from both sides}\\x^2-2hx+h^2+y^2-2ky+k^2-r^2=0\\x^2+y^2-2xh-2yk+h^2+k^2-r^2=0\\\\\text{Used}\ (a-b)^2=a^2-2ab+b^2[/tex]
[tex]\begin{array}{cccccc}x^2&y^2&10x&12y&25\\\downarrow&\downarrow&\downarrow&\downarrow&\downarrow\\x^2&y^2&-2xh&-2yk&h^2+k^2-r^2\end{array}\\\\\text{Therefore}\\\\\begin{array}{ccc}-2xh=10x&-2yk=12y&h^2+k^2-r^2=25\\\boxed{h=-5}&\boxed{k=-6}\end{array}[/tex]
[tex]\text{If you want calculate the radius:}\\\text{Substitute}\ h=-5,\ k=-6\ \text{to}\ h^2+k^2-r^2=25\\\\(-5)^2+(-6)^2-r^2=25\\25+36-r^2=25\\61-r^2=25\qquad\text{subtract 61 from both sides}\\-r^2=-36\qquad\text{change the signs}\\r^2=36\to r=\sqrt{36}\to \boxed{r=6}[/tex]
