What is -16+x^2+y^2>= 0 ?

The solution to -16+x^2+y^2>= 0 is Circle with (a,b)=(0,0),r=4

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-16+x^2+y^2>= 0

- 16 + x^{2} + y^{2} \geq 0

(a, b) = (0, 0), r = 4

Solution steps

- 16 + x^{2} + y^{2} \geq 0

Rewrite

- 16 + x^{2} + y^{2} \geq 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 4

x^{2} + y^{2} - 50 x - 24 y + 669 = 0

(x + \frac{3}{2})^{2} + (y - 1)^{2} = \frac{17}{4}

X^{2} + Y^{2} - 20 X - 35 Y + 28 = 0

x^{2} - 4 x + y^{2} - 5 = 0

25 - x^{2} - y^{2} > 0

What is -16+x^2+y^2>= 0 ?
The solution to -16+x^2+y^2>= 0 is Circle with (a,b)=(0,0),r=4