What is the center x^2+y^2-8x-2y-19=0 ?

The center x^2+y^2-8x-2y-19=0 is (4,1)

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center x^2+y^2-8x-2y-19=0

center

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

(4, 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 1), r = 6

And the center is:

(4, 1)

(x - 5)^{2} + (y + 2)^{2} = 16

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

2 x^{2} + 2 y^{2} - 12 x + 4 y - 12 = 0

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

(x + 12)^{2} + (y - 4)^{2} - 30 = 0