What is the center (x-2)^2+(y+4)^2=25 ?

The center (x-2)^2+(y+4)^2=25 is (2,-4)

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center (x-2)^2+(y+4)^2=25

center

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

(2, - 4)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(2, - 4)

x^{2} + y^{2} - 4 x - 8 y - 5 = 0

x^{2} + (y - 5)^{2} = 36

x^{2} + y^{2} + 12 x - 12 y + 36 = 0

x^{2} + (y - 5)^{2} = 16

(x + 1)^{2} + y^{2} = 25