Answer
See below:
Work Step by Step
We know that for a circle the standard form is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$,
Where the center is $\left( h,k \right)$ and the radius is $r$.
So, the equation can be written as:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}+x+y-\frac{1}{2}=0=0 \\
& \left( {{x}^{2}}+x \right)+\left( {{y}^{2}}+y \right)=\frac{1}{2} \\
& \left( {{x}^{2}}+x+\frac{1}{4} \right)+\left( {{y}^{2}}+y+\frac{1}{4} \right)=\frac{1}{2}+\frac{1}{4}+\frac{1}{4} \\
& {{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( y+\frac{1}{2} \right)}^{2}}=1
\end{align}$
So, the equation of the circle in the standard form is given as:
${{\left( x-\left( -\frac{1}{2} \right) \right)}^{2}}+{{\left( y-\left( -\frac{1}{2} \right) \right)}^{2}}={{1}^{2}}$
Now compare this equation to the standard form, to get the value of $h=-\frac{1}{2},k=-\frac{1}{2},\,\,\text{ and }\,\,r=1$.
Therefore, the center is $\left( -\frac{1}{2},-\frac{1}{2} \right)$ and the radius is $1$ unit.
