What is the center (x-2)^2+(y+5)^2=16 ?

The center (x-2)^2+(y+5)^2=16 is (2,-5)

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center (x-2)^2+(y+5)^2=16

center

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

(2, - 5)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(2, - 5)

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

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

(x + 5)^{2} + y^{2} = 9

x^{2} + (y^{2} - 3) = 7

(x - 4)^{2} + (y + 5)^{2} = 49