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

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

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

center

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

(1, 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 1), r = 2

And the center is:

(1, 1)

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

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

2 x^{2} + 2 y^{2} = 18

x^{2} + y^{2} + 5 x + 4 y - 34 = 0

(x - 4)^{2} + (y - 2)^{2} = 49