What is the center x^2-6x+y^2-8y=0 ?

The center x^2-6x+y^2-8y=0 is (3,4)

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center x^2-6x+y^2-8y=0

center

x^{2} - 6 x + y^{2} - 8 y = 0

(3, 4)

Solution steps

x^{2} - 6 x + y^{2} - 8 y = 0

Rewrite

x^{2} - 6 x + y^{2} - 8 y = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 4)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (3, 4), r = 5

And the center is:

(3, 4)

3 x^{2} + 3 y^{2} - 12 x + 6 y - 12 = 0

+ x^{2} + y^{2} - 14 x + 2 y + 41 = 0

x^{2} + 4 x + y^{2} - 2 y + 1 = 0

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

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