What is the center x^2+y^2-8x+12=0 ?

The center x^2+y^2-8x+12=0 is (4,0)

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center x^2+y^2-8x+12=0

center

x^{2} + y^{2} - 8 x + 12 = 0

(4, 0)

Solution steps

x^{2} + y^{2} - 8 x + 12 = 0

Rewrite

x^{2} + y^{2} - 8 x + 12 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (4, 0), r = 2

And the center is:

(4, 0)

x^{2} + y^{2} - 6 x - 8 y - 11 = 0

x^{2} - 2 x + 3 y + y^{2} - 10 = 0

x^{2} + y^{2} + 8 x - 14 y + 1 = 0

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

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