What is 16-x^2+2x-y^2-4y-5>0 ?

The solution to 16-x^2+2x-y^2-4y-5>0 is Circle with (a,b)=(1,-2),r=4

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16-x^2+2x-y^2-4y-5>0

16 - x^{2} + 2 x - y^{2} - 4 y - 5 > 0

(a, b) = (1, - 2), r = 4

Solution steps

16 - x^{2} + 2 x - y^{2} - 4 y - 5 > 0

Rewrite

16 - x^{2} + 2 x - y^{2} - 4 y - 5 > 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 2), r = 4

(x + 3)^{2} + (y + 4)^{2} = 36

(x - 3)^{2} + (y + 6)^{2} = 16

5 x^{2} + 30 x + 5 y^{2} = 0

x^{2} + 6 x + y^{2} - 10 y = 15

x^{2} + y^{2} + 12 x + 6 y = 7

What is 16-x^2+2x-y^2-4y-5>0 ?
The solution to 16-x^2+2x-y^2-4y-5>0 is Circle with (a,b)=(1,-2),r=4