What is x^2+x+y^2+y=4 ?

The solution to x^2+x+y^2+y=4 is Circle with (a,b)=(-1/2 ,-1/2),r= 3/(sqrt(2))

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x^2+x+y^2+y=4

x^{2} + x + y^{2} + y = 4

(a, b) = (- \frac{1}{2}, - \frac{1}{2}), r = \frac{3}{2}

Solution steps

x^{2} + x + y^{2} + y = 4

Rewrite

x^{2} + x + y^{2} + y = 4

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, - \frac{1}{2}), r = \frac{3}{2}

(x - 4)^{2} + (y - 3)^{2} = 2 [(x + 2)^{2} + (y - 1)^{2}]

x^{2} + y^{2} + 2 x - 4 y = 11

(x - 37.5)^{2} + (y + 2.5)^{2} = 1^{2}

(y - 2)^{2} + (x - 8)^{2} = 64

x^{2} + y^{2} - 18 x + 4 y + 60 = 0

What is x^2+x+y^2+y=4 ?
The solution to x^2+x+y^2+y=4 is Circle with (a,b)=(-1/2 ,-1/2),r= 3/(sqrt(2))