Answer
Center of circle is $\left( 4,-1 \right)$ and radius is $r=2$.
Work Step by Step
Standard equation of the circle is:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (equation - 1)
And equation of circle is ${{x}^{2}}+{{y}^{2}}-8x+2y=-13$ (equation - 2)
Now add $16$ and $1$ on both the sides of (equation - 2) to complete the square twice.
$\begin{align}
& {{x}^{2}}+{{y}^{2}}-8x+2y+16+1=-13+16+1 \\
& \left( {{x}^{2}}-8x+16 \right)+\left( {{y}^{2}}+2y+1 \right)=4 \\
& {{\left( x-4 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( 2 \right)}^{2}}
\end{align}$
Now compare the (equation – 1) with the equation${{\left( x-4 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( 2 \right)}^{2}}$.
Center coordinate of circle is $\left( h=4,k=-1 \right)$.
And radius of circle is $r=2$.
To graph, we plot the points $\left( 4,0.414 \right)$, $\left( 4,-2.414 \right)$, $\left( 2.585,-1 \right)$, and $\left( 5.414,-1 \right)$ which are, respectively, $\sqrt2$ units above, below, left and right of $\left( 4,-1 \right)$.
