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

The center x^2+y^2+x+y=0 is (-1/2 ,-1/2)

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center x^2+y^2+x+y=0

center

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

(- \frac{1}{2}, - \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 center is:

(- \frac{1}{2}, - \frac{1}{2})

x^{2} + y^{2} + 2 x + 18 y - 143 = 0

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x^{2} + y^{2} - 2 x - 15 = 0

4 x^{2} + 4 y^{2} - 16 x + 24 y - 12 = 0

x^{2} + 10 x + y^{2} - 16 y - 11 = 0