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

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

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

center

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

(2, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(2, 0)

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

9 x^{2} + 9 y^{2} + 18 x - 18 y + 14 = 0

(x + 3)^{2} - 49 = - y^{2}

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

(x + \frac{5}{2})^{2} + (y - 14)^{2} = 20