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

The solution to x^2+x+y^2+y=0 is Circle with (a,b)=(-1/2 ,-1/2),r=sqrt(1/2)

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

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

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

Solution steps

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

Rewrite

x^{2} + x + y^{2} + 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}

(x - 4)^{2} + (y + 8)^{2} = 4

x^{2} - 2 x + y^{2} - 3 = 0

x^{2} + y^{2} - 12 x + 10 y = - 12

x^{2} + y^{2} - 8 x + 10 y + 25 = 0

(x + 7)^{2} + (y - 1)^{2} = 8

What is x^2+x+y^2+y=0 ?
The solution to x^2+x+y^2+y=0 is Circle with (a,b)=(-1/2 ,-1/2),r=sqrt(1/2)