What is 3x^2+6x+3y^2-8y=48 ?

The solution to 3x^2+6x+3y^2-8y=48 is Circle with (a,b)=(-1, 4/3),r= 13/3

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3x^2+6x+3y^2-8y=48

3 x^{2} + 6 x + 3 y^{2} - 8 y = 48

(a, b) = (- 1, \frac{4}{3}), r = \frac{13}{3}

Solution steps

3 x^{2} + 6 x + 3 y^{2} - 8 y = 48

Rewrite

3 x^{2} + 6 x + 3 y^{2} - 8 y = 48

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{4}{3})^{2} = (\frac{13}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{4}{3}), r = \frac{13}{3}

(x - 1)^{2} + (y - 2)^{2} = \frac{49}{16}

x^{2} + y^{2} - 2 x - 10 y - 29 = 0

- 16 + x^{2} + y^{2} \geq 0

x^{2} + y^{2} - 50 x - 24 y + 669 = 0

(x + \frac{3}{2})^{2} + (y - 1)^{2} = \frac{17}{4}

What is 3x^2+6x+3y^2-8y=48 ?
The solution to 3x^2+6x+3y^2-8y=48 is Circle with (a,b)=(-1, 4/3),r= 13/3