What is the center (x-5)^2+(y+4)^2=4 ?

The center (x-5)^2+(y+4)^2=4 is (5,-4)

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center (x-5)^2+(y+4)^2=4

center

(x - 5)^{2} + (y + 4)^{2} = 4

(5, - 4)

Solution steps

(x - 5)^{2} + (y + 4)^{2} = 4

Rewrite

(x - 5)^{2} + (y + 4)^{2} = 4

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 4))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (5, - 4), r = 2

And the center is:

(5, - 4)

x^{2} + y^{2} - 20 x - 18 y + 144 = 0

5 x^{2} + 40 x + 5 y^{2} = 0

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

(x - 17)^{2} + (y + 5)^{2} = 144

(x + 5)^{2} + (y + 5)^{2} = 64