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

The solution to 2x^2+2y^2+x+y=0 is Circle with (a,b)=(-1/4 ,-1/4),r= 1/(2sqrt(2))

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

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

(a, b) = (- \frac{1}{4}, - \frac{1}{4}), r = \frac{1}{2 2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{4}))^{2} + (y - (- \frac{1}{4}))^{2} = (\frac{1}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{4}, - \frac{1}{4}), r = \frac{1}{2 2}

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

9 x^{2} + 9 y^{2} = 16

x^{2} + y^{2} + 4 y - 117 = 0

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} + \frac{5}{2} x + 10 y + 5 = 0

(x - 7)^{2} + (y + 2)^{2} = 100

What is 2x^2+2y^2+x+y=0 ?
The solution to 2x^2+2y^2+x+y=0 is Circle with (a,b)=(-1/4 ,-1/4),r= 1/(2sqrt(2))