What is x^2+y^2<= 3 ?

The solution to x^2+y^2<= 3 is Circle with (a,b)=(0,0),r=sqrt(3)

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x^2+y^2<= 3

x^{2} + y^{2} \leq 3

(a, b) = (0, 0), r = 3

Solution steps

x^{2} + y^{2} \leq 3

Rewrite

x^{2} + y^{2} \leq 3

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (3)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 3

x^{2} + y^{2} + x - 2 y = - 1

x^{(2)} - 5 x + y^{(2)} + 4 y = 46

x^{2} + y^{2} + \frac{4}{3} x - \frac{4}{3} y + \frac{4}{9} = 0

3 x^{2} - 9 x + 3 y^{2} + 3 y = \frac{39}{2}

x^{2} + y^{2} + x - 2 y = 16

What is x^2+y^2<= 3 ?
The solution to x^2+y^2<= 3 is Circle with (a,b)=(0,0),r=sqrt(3)