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

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

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

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

And the center is:

(0, 0)

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

- 20 - x^{2} = - 16 x - 20 y + y^{2}

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

x^{2} + y^{2} - 8 x - 4 y - 16 = 0

x^{2} + y^{2} = 72