What is the center x^2+y^2-4x-4y=0 ?

The center x^2+y^2-4x-4y=0 is (2,2)

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center x^2+y^2-4x-4y=0

center

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

(2, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - 2)^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (2, 2), r = 22

And the center is:

(2, 2)

x^{2} + y^{2} - 12 x + 20 y + 15 = 0

x^{2} + y^{2} = \frac{1}{4}

x^{2} + y^{2} - 2 x + 6 y - 15 = 0

4 x^{2} - 4 x + 4 y^{2} - 63 = 0

(x - 3)^{2} + (y + 4)^{2} = 25