What is the center y^2+x^2-12x+18y+116=0 ?

The center y^2+x^2-12x+18y+116=0 is (6,-9)

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center y^2+x^2-12x+18y+116=0

center

(6, - 9)

Solution steps

y^{2} + x^{2} - 12 x + 18 y + 116 = 0

Rewrite

y^{2} + x^{2} - 12 x + 18 y + 116 = 0

in the form of the standard circle equation

(x - 6)^{2} + (y - (- 9))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (6, - 9), r = 1

And the center is:

(6, - 9)

What is the center y^2+x^2-12x+18y+116=0 ?
The center y^2+x^2-12x+18y+116=0 is (6,-9)