What is the center x^2+y^2-4x-8y+18=0 ?

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

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center x^2+y^2-4x-8y+18=0

center

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

(2, 4)

Solution steps

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

Rewrite

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 4), r = 2

And the center is:

(2, 4)

x^{2} + y^{2} - 10 x + 2 y = 0

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

x^{2} + y^{2} + 4 x + 6 y = 23

(x - 5)^{2} + (y + 2)^{2} = 15

x^{2} + (y + 5)^{2} = 100