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Popular >

diameter x^2+y^2<= 1

Solution

diameter

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

Solution

d = 2

Solution steps

d = 2 r

Find the radius

Radius given

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

d = 2 \cdot 1

Refine

d = 2

Popular Examples

diameter x^2+y^2+6x-47=387 diameter

x^{2} + y^{2} + 6 x - 47 = 387

diameter x^2+y^2-4x-2y-36=0 diameter

x^{2} + y^{2} - 4 x - 2 y - 36 = 0

diameter x^2+y^2-14x+8y-16=0 diameter

x^{2} + y^{2} - 14 x + 8 y - 16 = 0

diameter (x-3)^2+(y+3)^2=6^2 diameter

(x - 3)^{2} + (y + 3)^{2} = 6^{2}

diameter x^2+y^2<= 25 diameter

x^{2} + y^{2} \leq 25

Frequently Asked Questions (FAQ)

What is the diameter x^2+y^2<= 1 ?
The diameter x^2+y^2<= 1 is d=2