What is the center (x-4)^2+(y-1)^2=4 ?

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

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center (x-4)^2+(y-1)^2=4

center

(x - 4)^{2} + (y - 1)^{2} = 4

(4, 1)

Solution steps

(x - 4)^{2} + (y - 1)^{2} = 4

Rewrite

(x - 4)^{2} + (y - 1)^{2} = 4

in the form of the standard circle equation

(x - 4)^{2} + (y - 1)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (4, 1), r = 2

And the center is:

(4, 1)

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

4 x^{2} + 4 y^{2} + 12 x - 4 y - 90 = 0

x^{2} + y^{2} - 6 x + 4 y + 9 = 0

x^{2} + y^{2} + 4 x - 6 y = 12

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