What is the circumference 16x^2+16y^2-48x-16y-41=0 ?

The circumference 16x^2+16y^2-48x-16y-41=0 is C=(9pi)/2

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circumference 16x^2+16y^2-48x-16y-41=0

circumference

16 x^{2} + 16 y^{2} - 48 x - 16 y - 41 = 0

C = \frac{9 π}{2}

Solution steps

C = 2 π r

Find the radius

Radius given

16 x^{2} + 16 y^{2} - 48 x - 16 y - 41 = 0 : r = \frac{9}{4}

C = 2 π \frac{9}{4}

Refine

C = \frac{9 π}{2}

x^{2} + y^{2} + 8 x + 2 y + 1 = 0

x^{2} + y^{2} + 14 y + 13 = 0

(x - 3)^{2} + (y - 2)^{2} = 25

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

c i rc l e, f in d C i rc u m f ere n ce, g i v e n r = 2 x - 4

What is the circumference 16x^2+16y^2-48x-16y-41=0 ?
The circumference 16x^2+16y^2-48x-16y-41=0 is C=(9pi)/2