What is the center x^2+y^2-4x-2y-31=0 ?

The center x^2+y^2-4x-2y-31=0 is (2,1)

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center x^2+y^2-4x-2y-31=0

center

x^{2} + y^{2} - 4 x - 2 y - 31 = 0

(2, 1)

Solution steps

x^{2} + y^{2} - 4 x - 2 y - 31 = 0

Rewrite

x^{2} + y^{2} - 4 x - 2 y - 31 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 1), r = 6

And the center is:

(2, 1)

(x - 3)^{2} + (y - 2)^{2} = 16

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

x^{2} + y^{2} - 14 x + 6 y + 49 = 0

X^{2} + Y^{2} = 4

(x + 4)^{2} + (y - 1)^{2} = 18