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 of the circle can be written as:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}-x+2y+1=0 \\
& \left( {{x}^{2}}-x \right)+\left( {{y}^{2}}+2y \right)=-1 \\
& \left( {{x}^{2}}-x+\frac{1}{4} \right)+\left( {{y}^{2}}+2y+1 \right)=-1+\frac{1}{4}+1 \\
& {{\left( x-\frac{1}{2} \right)}^{2}}+{{\left( y+1 \right)}^{2}}=\frac{1}{4}
\end{align}$
So, the equation of the circle in the standard form is given as:
${{\left( x-\frac{1}{2} \right)}^{2}}+{{\left( y-\left( -1 \right) \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}}$
Compare this equation to the standard form, to get the value of $h=\frac{1}{2},k=-1\,,\,\text{ and }\,\,r=\frac{1}{2}$.
Therefore, the center is $\left( \frac{1}{2},-1 \right)$ and the radius is $\frac{1}{2}$ units.
