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} + 8 x - 2 y - 19 = 0

x^{2} + y^{2} - 6 x + 4 y + 9 = 0

100 x^{2} + 100 y^{2} - 100 x + 400 y + 409 = 0

x^{2} + y^{2} + 6 x - 4 y - 23 = 0

x^{2} + y^{2} - 4 x + 8 y - 5 = 0