What is the radius 3x^2+3y^2-12x+6y-12=0 ?

The radius 3x^2+3y^2-12x+6y-12=0 is r=3

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radius 3x^2+3y^2-12x+6y-12=0

radius

3 x^{2} + 3 y^{2} - 12 x + 6 y - 12 = 0

r = 3

Solution steps

3 x^{2} + 3 y^{2} - 12 x + 6 y - 12 = 0

Rewrite

3 x^{2} + 3 y^{2} - 12 x + 6 y - 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 3

x^{2} + y^{2} - 9 x + 4 y - 127 = 0

x^{2} + y^{2} + 14 x - 18 y + 49 = 0

4 x^{2} + 4 y^{2} - 8 x + 4 y - 11 = 0

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

What is the radius 3x^2+3y^2-12x+6y-12=0 ?
The radius 3x^2+3y^2-12x+6y-12=0 is r=3