What is x^2+y^2<= 5 ?

The solution to x^2+y^2<= 5 is Circle with (a,b)=(0,0),r=sqrt(5)

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x^2+y^2<= 5

x^{2} + y^{2} \leq 5

(a, b) = (0, 0), r = 5

Solution steps

x^{2} + y^{2} \leq 5

Rewrite

x^{2} + y^{2} \leq 5

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 5

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

(x - 2)^{2} + (y - 1)^{2} = 5^{2}

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

(x + 2)^{2} + (y + 2)^{2} = 25

(x - 5)^{2} + (y - 4)^{2} = 3^{2}

What is x^2+y^2<= 5 ?
The solution to x^2+y^2<= 5 is Circle with (a,b)=(0,0),r=sqrt(5)