What is the center (x-13)^2+(y+12)^2=64 ?

The center (x-13)^2+(y+12)^2=64 is (13,-12)

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center (x-13)^2+(y+12)^2=64

center

(13, - 12)

Solution steps

(x - 13)^{2} + (y + 12)^{2} = 64

Rewrite

(x - 13)^{2} + (y + 12)^{2} = 64

in the form of the standard circle equation

(x - 13)^{2} + (y - (- 12))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (13, - 12), r = 8

And the center is:

(13, - 12)

What is the center (x-13)^2+(y+12)^2=64 ?
The center (x-13)^2+(y+12)^2=64 is (13,-12)