What is the circumference 2x^2+2y^2-5x+4y-1=0 ?

The circumference 2x^2+2y^2-5x+4y-1=0 is C=(7pi)/2

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circumference 2x^2+2y^2-5x+4y-1=0

circumference

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

C = \frac{7 π}{2}

Solution steps

C = 2 π r

Find the radius

Radius given

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0 : r = \frac{7}{4}

C = 2 π \frac{7}{4}

Refine

C = \frac{7 π}{2}

(x + 1)^{2} + (y - 3)^{2} = 16

(x + 1)^{2} + (y + 1)^{2} = 13

x^{2} + y^{2} = 7 + x

x^{2} + y^{2} - 4 x - 5 = 0

x^{2} - 3 x + y^{2} + 1 = 0

What is the circumference 2x^2+2y^2-5x+4y-1=0 ?
The circumference 2x^2+2y^2-5x+4y-1=0 is C=(7pi)/2