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

The radius x^2+y^2-2x-4y-31=0 is r=6

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

radius

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

r = 6

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 6

x^{2} + y^{2} + 4 x - 6 y + 12 = 0

x^{2} - 14 x + y^{2} + 6 y + 9 = 0

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

x^{2} + y^{2} - 14 x + 12 y = 36

x^{2} + y^{2} + 8 y - 48 = 0