What is the radius 3x^2+3y^2+4y-7=0 ?

The radius 3x^2+3y^2+4y-7=0 is r= 5/3

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radius 3x^2+3y^2+4y-7=0

radius

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

r = \frac{5}{3}

Solution steps

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

Rewrite

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{2}{3}))^{2} = (\frac{5}{3})^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{2}{3}), r = \frac{5}{3}

And the radius is:

r = \frac{5}{3}

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