What is the circumference (x-0)^2+(y-0)^2= 52/4 ?

The circumference (x-0)^2+(y-0)^2= 52/4 is C=2sqrt(13)pi

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circumference (x-0)^2+(y-0)^2= 52/4

circumference

(x - 0)^{2} + (y - 0)^{2} = \frac{52}{4}

C = 213 π

Solution steps

C = 2 π r

Find the radius

Radius given

(x - 0)^{2} + (y - 0)^{2} = \frac{52}{4} : r = \frac{52}{4}

C = 2 π \frac{52}{4}

Refine

C = 213 π

x^{2} + y^{2} = 10

x^{2} + 2 x + 10 + y^{2} - 6 y - 10 = 0

x^{2} + y^{2} + 18 x - 24 y - 69 = 0

x^{2} - 10 x + y^{2} + 6 y - 1 = 0

(x - 0)^{2} + (y + 0)^{2} = 7^{2}

What is the circumference (x-0)^2+(y-0)^2= 52/4 ?
The circumference (x-0)^2+(y-0)^2= 52/4 is C=2sqrt(13)pi