What is the center 2x^2+2y^2-2x-2y-3=0 ?

The center 2x^2+2y^2-2x-2y-3=0 is (1/2 , 1/2)

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center 2x^2+2y^2-2x-2y-3=0

center

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

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

Solution steps

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

Rewrite

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

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

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

x^{2} + y^{2} + 8 y = 33

x^{2} + y^{2} - 4 x + 2 y - 11 = 0

x^{2} + y^{2} + 6 x - 2 y + 1 = 0

(x - 3)^{2} + (y + 2)^{2} = 10

What is the center 2x^2+2y^2-2x-2y-3=0 ?
The center 2x^2+2y^2-2x-2y-3=0 is (1/2 , 1/2)