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

The solution to 2x^2+2y^2+2x-y=0 is Circle with (a,b)=(-1/2 , 1/4),r=(sqrt(5))/4

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

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

(a, b) = (- \frac{1}{2}, \frac{1}{4}), r = \frac{5}{4}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - \frac{1}{4})^{2} = (\frac{5}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, \frac{1}{4}), r = \frac{5}{4}

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

144 x^{2} + 144 y^{2} + 72 x + 192 y + 37 = 0

x^{2} - 4 x + y^{2} = 21

x^{2} + (y - 1)^{2} = 36

x^{2} + y^{2} - 5 x + 4 y - 46 = 0

What is 2x^2+2y^2+2x-y=0 ?
The solution to 2x^2+2y^2+2x-y=0 is Circle with (a,b)=(-1/2 , 1/4),r=(sqrt(5))/4