What is the center x^2+y^2-18x+8y+81=0 ?

The center x^2+y^2-18x+8y+81=0 is (9,-4)

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center x^2+y^2-18x+8y+81=0

center

x^{2} + y^{2} - 18 x + 8 y + 81 = 0

(9, - 4)

Solution steps

x^{2} + y^{2} - 18 x + 8 y + 81 = 0

Rewrite

x^{2} + y^{2} - 18 x + 8 y + 81 = 0

in the form of the standard circle equation

(x - 9)^{2} + (y - (- 4))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (9, - 4), r = 4

And the center is:

(9, - 4)

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

x^{2} + y^{2} - 6 x + 6 y + 2 = 0

2 x^{2} + 2 y^{2} - 12 x + 4 y - 30 = 0

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

x^{2} - 4 x + y^{2} - 8 y = 20

What is the center x^2+y^2-18x+8y+81=0 ?
The center x^2+y^2-18x+8y+81=0 is (9,-4)