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

The radius 3y^2+3x^2+24x+32=0 is r= 4/(sqrt(3))

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

radius

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

And the radius is:

r = \frac{4}{3}

(x + 4)^{2} + (y - 3)^{2} = 25

(x - 1)^{2} + (y - 4)^{2} = 25

(x + 2)^{2} + (y - 1)^{2} = 9

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

(x + 4)^{2} + (y - 3)^{2} = 16

What is the radius 3y^2+3x^2+24x+32=0 ?
The radius 3y^2+3x^2+24x+32=0 is r= 4/(sqrt(3))