Answer
$\left( 5,-1 \right)$ and $6$
Work Step by Step
${{x}^{2}}-10x+{{y}^{2}}+2y=10$
Now grouping x-terms and y-terms;
$\left( {{x}^{2}}-10x \right)+\left( {{y}^{2}}+2y \right)=10$
Adding 25 and 1 on both sides:
$\begin{align}
& \left( {{x}^{2}}-10x+25 \right)+\left( {{y}^{2}}+2y+1 \right)=10+25+1 \\
& \left( {{x}^{2}}-10x+25 \right)+\left( {{y}^{2}}+2y+1 \right)=36 \\
\end{align}$
The equation of the circle is:
${{\left( x-5 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( 6 \right)}^{2}}$
Now compare it with formula:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$
Thus, the center and radius of the equation, ${{x}^{2}}-10x+{{y}^{2}}+2y=10$, are $\left( 5,-1 \right)$ and $6$.
