Answer
Center of circle is $\left( 0,0 \right)$ and radius is $r=\frac{1}{2}$
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 $4{{x}^{2}}+4{{y}^{2}}=1$ (equation - 2)
Multiply $\frac{1}{4}$on both the sides of equation $4{{x}^{2}}+4{{y}^{2}}=1$.
$\begin{align}
& 4\times \frac{1}{4}{{x}^{2}}+4\times \frac{1}{4}{{y}^{2}}=\frac{1}{4} \\
& {{x}^{2}}+{{y}^{2}}=\frac{1}{4}
\end{align}$
Now compare the standard equation with equation ${{x}^{2}}+{{y}^{2}}=\frac{1}{4}$.
$\begin{align}
& {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=\frac{1}{4} \\
& {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}} \\
\end{align}$
Center coordinate of circle is $\left( h=0,k=0 \right)$.
And radius of circle is $r=\frac{1}{2}$.
Graph:
To graph, we plot the points $\left( 0,0.5 \right)$, $\left( 0,-0.5 \right)$, $\left( -0.5,0 \right)$, and $\left( 0.5,0 \right)$ which are, respectively, $\frac{1}{2}$ units above, below, left and right of $\left( 0,0 \right)$.
