What is the radius x^2+y^2+8x-12y+27=0 ?

The radius x^2+y^2+8x-12y+27=0 is r=5

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radius x^2+y^2+8x-12y+27=0

radius

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

r = 5

Solution steps

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

Rewrite

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 6)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 4, 6), r = 5

And the radius is:

r = 5

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

(x + 2)^{2} + (y - 5)^{2} = 16

(x - 2)^{2} + (y + 4)^{2} = 9

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

(x + 1)^{2} + (y - 1)^{2} = 40