What is the circumference 2x^2+2y^2-4x+8y-22=0 ?

The circumference 2x^2+2y^2-4x+8y-22=0 is C=8pi

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circumference 2x^2+2y^2-4x+8y-22=0

circumference

2 x^{2} + 2 y^{2} - 4 x + 8 y - 22 = 0

C = 8 π

Solution steps

C = 2 π r

Find the radius

Radius given

2 x^{2} + 2 y^{2} - 4 x + 8 y - 22 = 0 : r = 4

C = 2 π 4

Refine

C = 8 π

x^{2} + y^{2} - 0 x - 4 y - 64 = 0

(x + 2)^{2} + (y + 5)^{2} = 16

x^{2} + y^{2} - x - 6 y - 5 = 0

(x - 6)^{2} + (y + 5)^{2} = 64

x^{2} - 6 x + 2 y + y^{2} + 1 = 0

What is the circumference 2x^2+2y^2-4x+8y-22=0 ?
The circumference 2x^2+2y^2-4x+8y-22=0 is C=8pi