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

The center x^2+y^2-2x-4y-4=0 is (1,2)

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

center

(1, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 2), r = 3

And the center is:

(1, 2)