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} > 0

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

Solution steps

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

Rewrite

16 - x^{2} - y^{2} > 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)^{2} + (y + 4)^{2} = 25

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

x^{2} + 10 x + y^{2} - 16 - 11 = 0

x^{2} + y^{2} + 10 x - 8 y = - 32

x^{2} - 2 x + y^{2} + 4 y - 20 = 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