# What is the radius length of the circle with equation

What is the radius length of the circle with equation ${x}^{2}+{y}^{2}-x+y-4=0$?
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doplovif
The standard equation of a circle with center (h,k) and radius rr is given by: ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
So, we write the given in this form.
Isolate the constant and group the x’s and y’s:
$\left({x}^{2}-x\right)+\left({y}^{2}+y\right)=4$
Complete the square:
$\left(\left({x}^{2}\right)-x+{\left(\frac{1}{2}\right)}^{2}\right)+\left(\left({y}^{2}+y+1{\left(\frac{1}{2}\right)}^{2}\right)=4+{\left(\frac{1}{2}\right)}^{2}+{\left(\frac{1}{2}\right)}^{2}$
$\left(x-\left(\frac{1}{2}\right)\right)+{\left(y+\left(\frac{1}{2}\right)\right)}^{2}=4+\frac{1}{4}+\frac{1}{4}$
${\left(x-\left(\frac{1}{2}\right)\right)}^{2}+{\left(y+\frac{1}{2}\right)}^{2}=\frac{18}{4}$
Hence, we can solve for the radius rr by writing: ${r}^{2}=\frac{18}{4}$
$r=\frac{\sqrt{18}}{2}$
$r=3\frac{\sqrt{2}}{2}$