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

The center x^2-8x+y^2-2y+8=0 is (4,1)

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

center

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

(4, 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 4)^{2} + (y - 1)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (4, 1), r = 3

And the center is:

(4, 1)

(x + 1)^{2} + (y + 6)^{2} = 4

x^{2} + y^{2} - 2 x + 3 y - 1 = 0

x^{2} + y^{2} - 4 x + 6 y + 9 = 0

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

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