What is the radius x^2+y^2+12y+32=0 ?

The radius x^2+y^2+12y+32=0 is r=2

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radius x^2+y^2+12y+32=0

radius

x^{2} + y^{2} + 12 y + 32 = 0

r = 2

Solution steps

x^{2} + y^{2} + 12 y + 32 = 0

Rewrite

x^{2} + y^{2} + 12 y + 32 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, - 6), r = 2

And the radius is:

r = 2

2 x^{2} + 2 y^{2} + 10 x - 6 y - 55 = 0

x^{2} - 6 x + 9 + y^{2} - 4 y + 4 = 16

x^{2} + y^{2} - 4 x + 6 y - 3 = 0

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

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