What is the center (x-3)^2+(y-4)^2=4 ?

The center (x-3)^2+(y-4)^2=4 is (3,4)

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center (x-3)^2+(y-4)^2=4

center

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

(3, 4)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, 4), r = 2

And the center is:

(3, 4)

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

x^{2} + y^{2} + 4 x + 4 y + 7 = 0

x^{2} + y^{2} - 10 x - 16 y + 53 = 0

x^{2} + y^{2} - 12 x + 8 y - 5 = 0

(x - 5)^{2} + (y - 3)^{2} = 49