What is the circumference x^2+y^2+8x+4y+2=0 ?

The circumference x^2+y^2+8x+4y+2=0 is C=6sqrt(2)pi

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circumference x^2+y^2+8x+4y+2=0

circumference

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

C = 62 π

Solution steps

C = 2 π r

Find the radius

Radius given

x^{2} + y^{2} + 8 x + 4 y + 2 = 0 : r = 32

C = 2 π 32

Refine

C = 62 π

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

2 (x - 3)^{2} + 2 y^{2} = 8

x^{2} + y^{2} \leq 25

What is the circumference x^2+y^2+8x+4y+2=0 ?
The circumference x^2+y^2+8x+4y+2=0 is C=6sqrt(2)pi