What is the radius x^2+y^2+4x-6y+12=0 ?

The radius x^2+y^2+4x-6y+12=0 is r=1

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radius x^2+y^2+4x-6y+12=0

radius

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

r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 3)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 1

And the radius is:

r = 1

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

(x + 5)^{2} + (y + 1)^{2} = 9