What is the radius 3x^2+3y^2-2x+y-5=0 ?

The radius 3x^2+3y^2-2x+y-5=0 is r=(sqrt(65))/6

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radius 3x^2+3y^2-2x+y-5=0

radius

3 x^{2} + 3 y^{2} - 2 x + y - 5 = 0

r = \frac{65}{6}

Solution steps

3 x^{2} + 3 y^{2} - 2 x + y - 5 = 0

Rewrite

3 x^{2} + 3 y^{2} - 2 x + y - 5 = 0

in the form of the standard circle equation

(x - \frac{1}{3})^{2} + (y - (- \frac{1}{6}))^{2} = (\frac{65}{6})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{3}, - \frac{1}{6}), r = \frac{65}{6}

And the radius is:

r = \frac{65}{6}

3 x^{2} - 12 x - 18 y + 3 y^{2} = 0

(x - 1)^{2} + (y + 4)^{2} = 9

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

y^{2} = 4 - x^{2}

(x - 17)^{2} + (y + 5)^{2} = 144

What is the radius 3x^2+3y^2-2x+y-5=0 ?
The radius 3x^2+3y^2-2x+y-5=0 is r=(sqrt(65))/6