What is y^2=(-3x^2+1)/3 ?

The solution to y^2=(-3x^2+1)/3 is Circle with (a,b)=(0,0),r=sqrt(1/3)

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y^2=(-3x^2+1)/3

y^{2} = \frac{- 3 x ^{2} + 1}{3}

(a, b) = (0, 0), r = \frac{1}{3}

Solution steps

y^{2} = \frac{- 3 x ^{2} + 1}{3}

Rewrite

y^{2} = \frac{- 3 x ^{2} + 1}{3}

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (\frac{1}{3})^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = \frac{1}{3}

4 x^{2} + 4 x + 4 y^{2} - 24 y - 27 = 0

\frac{y ^{2}}{2} = - \frac{x ^{2}}{2} + 2

x^{2} + y^{2} + 10 x - 56 = 0

(x + 5)^{2} + (y + 2)^{2} = 21

0 = 3 - 6 (x^{2} + y^{2})

What is y^2=(-3x^2+1)/3 ?
The solution to y^2=(-3x^2+1)/3 is Circle with (a,b)=(0,0),r=sqrt(1/3)