What is the radius x^2+y^2+12y-108=0 ?

The radius x^2+y^2+12y-108=0 is r=12

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radius x^2+y^2+12y-108=0

radius

x^{2} + y^{2} + 12 y - 108 = 0

r = 12

Solution steps

x^{2} + y^{2} + 12 y - 108 = 0

Rewrite

x^{2} + y^{2} + 12 y - 108 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 6))^{2} = 1 2^{2}

Therefore the circle properties are:

(a, b) = (0, - 6), r = 12

And the radius is:

r = 12

x^{2} + 6 x + y^{2} - 2 y - 15 = 0

x^{2} + y^{2} - 8 x + 10 y - 8 = 0

X^{2} + Y^{2} = 49

x^{2} + (y - 2)^{2} = 36

x^{2} + y^{2} \leq 25