What is the circumference x^2+y^2+8x+4y-3=40 ?

The circumference x^2+y^2+8x+4y-3=40 is C=6sqrt(7)pi

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circumference x^2+y^2+8x+4y-3=40

circumference

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

C = 67 π

Solution steps

C = 2 π r

Find the radius

Radius given

x^{2} + y^{2} + 8 x + 4 y - 3 = 40 : r = 37

C = 2 π 37

Refine

C = 67 π

x^{2} + y^{2} \leq 9

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

x^{2} + y^{2} + 3 x + 2 y - 5 = 0

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

(x - 4)^{2} + (y + 2)^{2} = 49

What is the circumference x^2+y^2+8x+4y-3=40 ?
The circumference x^2+y^2+8x+4y-3=40 is C=6sqrt(7)pi