What is the radius 3x^2+3y^2+2x-4y-1=0 ?

The radius 3x^2+3y^2+2x-4y-1=0 is r=(2sqrt(2))/3

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radius 3x^2+3y^2+2x-4y-1=0

radius

r = \frac{2 2}{3}

Solution steps

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

Rewrite

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, \frac{2}{3}), r = \frac{2 2}{3}

And the radius is:

r = \frac{2 2}{3}

What is the radius 3x^2+3y^2+2x-4y-1=0 ?
The radius 3x^2+3y^2+2x-4y-1=0 is r=(2sqrt(2))/3