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

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

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

radius

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

r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 1

(x - 2)^{2} + (y - (- 5))^{2} = 97

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

x^{2} - 4 x + 12 y + y^{2} + 14 = 0

x^{2} + y^{2} - 16 x - 4 = 0

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