Answer
$\left( 2,-1 \right)$ and $\sqrt{11}$
Work Step by Step
${{x}^{2}}+{{y}^{2}}-4x+2y=6$
Now grouping x-terms and y-terms;
$\left( {{x}^{2}}-4x \right)+\left( {{y}^{2}}+2y \right)=6$
Adding 4 and 1 to both sides of the equation to get standard form;
$\begin{align}
& \left( {{x}^{2}}-4x+4 \right)+\left( {{y}^{2}}+2y+1 \right)=6+4+1 \\
& \left( {{x}^{2}}-4x+4 \right)+\left( {{y}^{2}}+2y+1 \right)=11 \\
\end{align}$
This gives:
${{\left( x-2 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( \sqrt{11} \right)}^{2}}$
Now compare it with the formula:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$
Thus, the center and radius of the function, ${{x}^{2}}+{{y}^{2}}-4x+2y=6$, are $\left( 2,-1 \right)$ and $\sqrt{11}$.
