What is 3x^2+3y^2-4x+y-12=0 ?

The solution to 3x^2+3y^2-4x+y-12=0 is Circle with (a,b)=(2/3 ,-1/6),r=(sqrt(161))/6

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3x^2+3y^2-4x+y-12=0

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

(a, b) = (\frac{2}{3}, - \frac{1}{6}), r = \frac{161}{6}

Solution steps

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

Rewrite

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

in the form of the standard circle equation

(x - \frac{2}{3})^{2} + (y - (- \frac{1}{6}))^{2} = (\frac{161}{6})^{2}

Therefore the circle properties are:

(a, b) = (\frac{2}{3}, - \frac{1}{6}), r = \frac{161}{6}

x^{2} + y^{2} \leq 10

x^{2} + y^{2} \leq 36

4 x^{2} - 4 x + 4 y^{2} - 35 = 0

y^{2} = 10 - z^{2}

3 x^{2} + 3 y^{2} - 6 x + 5 y = 0

What is 3x^2+3y^2-4x+y-12=0 ?
The solution to 3x^2+3y^2-4x+y-12=0 is Circle with (a,b)=(2/3 ,-1/6),r=(sqrt(161))/6