Answer
Center of circle is $\left( 0,0 \right)$ and radius is $r=2\sqrt{2}$
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 the equation of the circle is ${{x}^{2}}+{{y}^{2}}=8$ (equation - 2)
Now compare both the equations.
$\begin{align}
& {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=8 \\
& {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 2\sqrt{2} \right)}^{2}} \\
\end{align}$
Center coordinate of the circle is $\left( h=0,k=0 \right)$.
And radius of the circle is $r=2\sqrt{2}$.
To graph, we plot the points $\left( 0,2.828 \right)$, $\left( 0,-2.828 \right)$, $\left( -2.828,0 \right)$, and $\left( 2.828,0 \right)$ which are, respectively, $2\sqrt{2}$ units above, below, left and right of $\left( 0,0 \right)$.
