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circumference x^2+y^2<= 1

Solution

circumference

x^{2} + y^{2} \leq 1

Solution

C = 2 π

Solution steps

C = 2 π r

Find the radius

Radius given

x^{2} + y^{2} \leq 1 : r = 1

C = 2 π 1

Refine

C = 2 π

Popular Examples

circumference x^2+y^2<= 4 circumference

x^{2} + y^{2} \leq 4

circumference x^2+y^2+6x-2y-15=0 circumference

x^{2} + y^{2} + 6 x - 2 y - 15 = 0

circumference (x-4)^2+(y-(-4))^2=16 circumference

(x - 4)^{2} + (y - (- 4))^{2} = 16

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

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

circumference x^2+y^2+8x-2y+15=0 circumference

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

Frequently Asked Questions (FAQ)

What is the circumference x^2+y^2<= 1 ?
The circumference x^2+y^2<= 1 is C=2pi