What is the radius x^2+y^2+x+y-1/2 =0 ?

The radius x^2+y^2+x+y-1/2 =0 is r=1

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radius x^2+y^2+x+y-1/2 =0

radius

x^{2} + y^{2} + x + y - \frac{1}{2} = 0

r = 1

Solution steps

x^{2} + y^{2} + x + y - \frac{1}{2} = 0

Rewrite

x^{2} + y^{2} + x + y - \frac{1}{2} = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 1

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