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

The radius 6x^2+6y^2+12x-y=0 is r=(sqrt(145))/(12)

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

radius

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

r = \frac{145}{12}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{1}{12})^{2} = (\frac{145}{12})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{1}{12}), r = \frac{145}{12}

And the radius is:

r = \frac{145}{12}

x^{2} + y^{2} + 10 x - 16 y = - 80

(x + 2)^{2} + (y - 1)^{2} = 16

(x - 3)^{2} + (y - 1)^{2} = 4

(x + 0)^{2} + (y - 0)^{2} = 22

(x - 3)^{2} + (y - 1)^{2} = 9

What is the radius 6x^2+6y^2+12x-y=0 ?
The radius 6x^2+6y^2+12x-y=0 is r=(sqrt(145))/(12)