The standard equation of a circle with center (h,k) and radius rr is given by:
\(\displaystyle{\left({x}−{h}\right)}^{{2}}+{\left({y}−{k}\right)}^{{2}}={r}^{{2}}\)

So, we write the given in this form.

Isolate the constant and group the x’s and y’s:

\(\displaystyle{\left({x}^{{2}}−{x}\right)}+{\left({y}^{{2}}+{y}\right)}={4}\)

Complete the square:

\(\displaystyle{\left({\left({x}^{{2}}\right)}-{x}+{\left(\frac{{1}}{{2}}\right)}^{{2}}\right)}+{\left({\left({y}^{{2}}+{y}+{1}{\left(\frac{{1}}{{2}}\right)}^{{2}}\right)}={4}+{\left(\frac{{1}}{{2}}\right)}^{{2}}+{\left(\frac{{1}}{{2}}\right)}^{{2}}\right.}\)

\(\displaystyle{\left({x}-{\left(\frac{{1}}{{2}}\right)}\right)}+{\left({y}+{\left(\frac{{1}}{{2}}\right)}\right)}^{{2}}={4}+\frac{{1}}{{4}}+\frac{{1}}{{4}}\)

\(\displaystyle{\left({x}-{\left(\frac{{1}}{{2}}\right)}\right)}^{{2}}+{\left({y}+\frac{{1}}{{2}}\right)}^{{2}}=\frac{{18}}{{4}}\)

Hence, we can solve for the radius rr by writing: \(\displaystyle{r}^{{2}}=\frac{{18}}{{4}}\)

\(\displaystyle{r}=\frac{\sqrt{{18}}}{{2}}\)

\(\displaystyle{r}={3}\frac{\sqrt{{2}}}{{2}}\)

So, we write the given in this form.

Isolate the constant and group the x’s and y’s:

\(\displaystyle{\left({x}^{{2}}−{x}\right)}+{\left({y}^{{2}}+{y}\right)}={4}\)

Complete the square:

\(\displaystyle{\left({\left({x}^{{2}}\right)}-{x}+{\left(\frac{{1}}{{2}}\right)}^{{2}}\right)}+{\left({\left({y}^{{2}}+{y}+{1}{\left(\frac{{1}}{{2}}\right)}^{{2}}\right)}={4}+{\left(\frac{{1}}{{2}}\right)}^{{2}}+{\left(\frac{{1}}{{2}}\right)}^{{2}}\right.}\)

\(\displaystyle{\left({x}-{\left(\frac{{1}}{{2}}\right)}\right)}+{\left({y}+{\left(\frac{{1}}{{2}}\right)}\right)}^{{2}}={4}+\frac{{1}}{{4}}+\frac{{1}}{{4}}\)

\(\displaystyle{\left({x}-{\left(\frac{{1}}{{2}}\right)}\right)}^{{2}}+{\left({y}+\frac{{1}}{{2}}\right)}^{{2}}=\frac{{18}}{{4}}\)

Hence, we can solve for the radius rr by writing: \(\displaystyle{r}^{{2}}=\frac{{18}}{{4}}\)

\(\displaystyle{r}=\frac{\sqrt{{18}}}{{2}}\)

\(\displaystyle{r}={3}\frac{\sqrt{{2}}}{{2}}\)