What is 3x^2+3y^2+8y-16=0 ?

The solution to 3x^2+3y^2+8y-16=0 is Circle with (a,b)=(0,-4/3),r= 8/3

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3x^2+3y^2+8y-16=0

3 x^{2} + 3 y^{2} + 8 y - 16 = 0

(a, b) = (0, - \frac{4}{3}), r = \frac{8}{3}

Solution steps

3 x^{2} + 3 y^{2} + 8 y - 16 = 0

Rewrite

3 x^{2} + 3 y^{2} + 8 y - 16 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{4}{3}))^{2} = (\frac{8}{3})^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{4}{3}), r = \frac{8}{3}

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

x^{2} + y^{2} + 9 x - 8 y - 62 = 0

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

x^{2} + y^{2} - 8 x + 8 y + 22 = 0

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

What is 3x^2+3y^2+8y-16=0 ?
The solution to 3x^2+3y^2+8y-16=0 is Circle with (a,b)=(0,-4/3),r= 8/3