Question

What is the radius length of the circle with equation x^2 + y^2 - x + y - 4 = 0?

Circles
What is the radius length of the circle with equation $$\displaystyle{x}^{{2}}+{y}^{{2}}-{x}+{y}-{4}={0}$$?

2021-07-01

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: $$\displaystyle{r}^{{2}}=\frac{{18}}{{4}}$$
$$\displaystyle{r}=\frac{\sqrt{{18}}}{{2}}$$
$$\displaystyle{r}={3}\frac{\sqrt{{2}}}{{2}}$$