What is the radius x^2+y^2+x+y=0 ?

The radius x^2+y^2+x+y=0 is r=sqrt(1/2)

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radius x^2+y^2+x+y=0

radius

x^{2} + y^{2} + x + y = 0

r = \frac{1}{2}

Solution steps

x^{2} + y^{2} + x + y = 0

Rewrite

x^{2} + y^{2} + x + y = 0

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, - \frac{1}{2}), r = \frac{1}{2}

And the radius is:

r = \frac{1}{2}

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