What is the center x^2+y^2-8x-3=0 ?

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

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center x^2+y^2-8x-3=0

center

x^{2} + y^{2} - 8 x - 3 = 0

(4, 0)

Solution steps

x^{2} + y^{2} - 8 x - 3 = 0

Rewrite

x^{2} + y^{2} - 8 x - 3 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 0)^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (4, 0), r = 19

And the center is:

(4, 0)

4 x^{2} + 4 y^{2} = 100

(x - 2)^{2} + (y + 1)^{2} = 9

x^{2} + y^{2} - 2 x - 8 y - 19 = 0

x^{2} + y^{2} + 2 x - 4 y - 15 = 0

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