What is the diameter 3x^2+3y^2+2x-4y-1=0 ?

The diameter 3x^2+3y^2+2x-4y-1=0 is d=(4sqrt(2))/3

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diameter 3x^2+3y^2+2x-4y-1=0

diameter

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

d = \frac{4 2}{3}

Solution steps

d = 2 r

Find the radius

Radius given

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0 : r = \frac{2 2}{3}

d = 2 \cdot \frac{2 2}{3}

Refine

d = \frac{4 2}{3}

x^{2} + y^{2} - 16 = 0

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

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

x^{2} + y^{2} = 49

x^{2} + y^{2} = 9

What is the diameter 3x^2+3y^2+2x-4y-1=0 ?
The diameter 3x^2+3y^2+2x-4y-1=0 is d=(4sqrt(2))/3