What is the center x^2+y^2<= 1 ?

The center x^2+y^2<= 1 is (0,0)

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center x^2+y^2<= 1

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 1

And the center is:

(0, 0)

x^{2} + y^{2} - 4 x - 8 y - 16 = 0

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

x^{2} - 9 = - y^{2}

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

(x - 3)^{2} + (y + 2)^{2} = 29