What is 2x^2+2y^2-x+y=1 ?

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

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2x^2+2y^2-x+y=1

2 x^{2} + 2 y^{2} - x + y = 1

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

Solution steps

2 x^{2} + 2 y^{2} - x + y = 1

Rewrite

2 x^{2} + 2 y^{2} - x + y = 1

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} + y^{2} - 6 x + 6 y - 7 = 0

x^{2} + y^{2} - 18 x + 20 y + 177 = 0

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

x^{2} + y^{2} + 2 x + 4 y - 44 = 0

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