What is the center x^2+16x+y^2-12y=0 ?

The center x^2+16x+y^2-12y=0 is (-8,6)

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center x^2+16x+y^2-12y=0

center

(- 8, 6)

Solution steps

x^{2} + 16 x + y^{2} - 12 y = 0

Rewrite

x^{2} + 16 x + y^{2} - 12 y = 0

in the form of the standard circle equation

(x - (- 8))^{2} + (y - 6)^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (- 8, 6), r = 10

And the center is:

(- 8, 6)