Marvin Mccormick
2021-08-06
Answered

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

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doplovif

Answered 2021-08-07
Author has **71** answers

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

So, we write the given in this form.

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

$({x}^{2}-x)+({y}^{2}+y)=4$

Complete the square:

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

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

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

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

$r=\frac{\sqrt{18}}{2}$

$r=3\frac{\sqrt{2}}{2}$

So, we write the given in this form.

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

Complete the square:

Hence, we can solve for the radius rr by writing:

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