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

The solution to 3x^2+3y^2-2x+y-5=0 is Circle with (a,b)=(1/3 ,-1/6),r=(sqrt(65))/6

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

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

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

x^{2} - 6 x + y^{2} - 8 y = 0

(x - 1)^{2} + (y + 5)^{2} = 16

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

(x - 6)^{2} + (y + 3)^{2} = 16

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

What is 3x^2+3y^2-2x+y-5=0 ?
The solution to 3x^2+3y^2-2x+y-5=0 is Circle with (a,b)=(1/3 ,-1/6),r=(sqrt(65))/6