What is 3y^2+3x^2+24x+32=0 ?

The solution to 3y^2+3x^2+24x+32=0 is Circle with (a,b)=(-4,0),r= 4/(sqrt(3))

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3y^2+3x^2+24x+32=0

3 y^{2} + 3 x^{2} + 24 x + 32 = 0

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

Solution steps

3 y^{2} + 3 x^{2} + 24 x + 32 = 0

Rewrite

3 y^{2} + 3 x^{2} + 24 x + 32 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - x + 2 y - 30 = 0

(x - 6) (x - 6) + (y - 7) (y - 7) = 64

36 = x^{2} + 12 + y^{2}

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

(x + 2)^{2} + (y - 3)^{2} = 3.5

What is 3y^2+3x^2+24x+32=0 ?
The solution to 3y^2+3x^2+24x+32=0 is Circle with (a,b)=(-4,0),r= 4/(sqrt(3))