What is the center x^2+y^2-8x+18y+16=0 ?

The center x^2+y^2-8x+18y+16=0 is (4,-9)

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center x^2+y^2-8x+18y+16=0

center

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

(4, - 9)

Solution steps

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

Rewrite

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(4, - 9)

(x - 3)^{2} + (y + 5)^{2} = 25

x^{2} + y^{2} - 6 x + 4 y - 3 = 0

(x + 12)^{2} + (y - 9)^{2} = 35

x^{2} + y^{2} = 10

(x + 1)^{2} + (y + 3)^{2} = 5

What is the center x^2+y^2-8x+18y+16=0 ?
The center x^2+y^2-8x+18y+16=0 is (4,-9)