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

The solution to y^2+x^2<= 1 is Circle with (a,b)=(0,0),r=1

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y^2+x^2<= 1

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

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

Solution steps

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

Rewrite

y^{2} + x^{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

x^{2} + y^{2} - 6 x - 8 y - 14 = 0

x^{2} + y^{2} + 10 x + 12 y - 3 = 0

x^{2} + y^{2} - 6 x - 4 y - 7 = 0

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

x^{2} + y^{2} + 8 x - 6 y + 5 = 0

What is y^2+x^2<= 1 ?
The solution to y^2+x^2<= 1 is Circle with (a,b)=(0,0),r=1