What is the center 2x^2-2x+2y^2+4y=-1 ?

The center 2x^2-2x+2y^2+4y=-1 is (1/2 ,-1)

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center 2x^2-2x+2y^2+4y=-1

center

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

(\frac{1}{2}, - 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(\frac{1}{2}, - 1)

4 x^{2} + 4 y^{2} - 4 x + 2 y + 1 = 0

x^{2} + 2 x + y^{2} - 3 = 6

x^{2} + y^{2} + 14 x - 16 y = 87

4 x^{2} + 4 x + 4 y^{2} - 16 y - 19 = 0

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

What is the center 2x^2-2x+2y^2+4y=-1 ?
The center 2x^2-2x+2y^2+4y=-1 is (1/2 ,-1)