What is the circumference 3x^2+3y^2+6x-4y=0 ?

The circumference 3x^2+3y^2+6x-4y=0 is C=(2sqrt(13)pi)/3

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circumference 3x^2+3y^2+6x-4y=0

circumference

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

C = \frac{2 13 π}{3}

Solution steps

C = 2 π r

Find the radius

Radius given

3 x^{2} + 3 y^{2} + 6 x - 4 y = 0 : r = \frac{13}{3}

C = 2 π \frac{13}{3}

Refine

C = \frac{2 13 π}{3}

(x + 3)^{2} + (y - 2)^{2} = \frac{9}{4}

(x - 2)^{2} + (y - 9)^{2} = 1 6^{2}

x^{2} + y^{2} + 6 x - 4 y - 51 = 0

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

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

What is the circumference 3x^2+3y^2+6x-4y=0 ?
The circumference 3x^2+3y^2+6x-4y=0 is C=(2sqrt(13)pi)/3