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

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

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x^2+y^2<= 2

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

(a, b) = (0, 0), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

\frac{( x - 4 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{2} = 1

3 y + 6 x - 3 x^{2} - 3 y^{2} = - 1

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

x^{2} + y^{2} - 12 x + 20 y + 15 = 0

\frac{x ^{2}}{49} + \frac{y ^{2}}{49} = 1

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