What is the circumference x^2+y^2+x+3y-104=0 ?

The circumference x^2+y^2+x+3y-104=0 is C=2pisqrt(213/2)

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circumference x^2+y^2+x+3y-104=0

circumference

x^{2} + y^{2} + x + 3 y - 104 = 0

C = 2 π \frac{213}{2}

Solution steps

C = 2 π r

Find the radius

Radius given

x^{2} + y^{2} + x + 3 y - 104 = 0 : r = \frac{213}{2}

C = 2 π \frac{213}{2}

6 x^{2} + 6 y^{2} = 12 x + 12 y + 6

x^{2} + y^{2} + 8 x + 4 y + 2 = 0

x^{2} + y^{2} + 6 x - 4 y - 12 = 0

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

(x - \frac{1}{2})^{2} + (y + \frac{2}{3})^{2} = 5

What is the circumference x^2+y^2+x+3y-104=0 ?
The circumference x^2+y^2+x+3y-104=0 is C=2pisqrt(213/2)