What is 2x-4y-x^2-y^2-1>= 0 ?

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

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2x-4y-x^2-y^2-1>= 0

2 x - 4 y - x^{2} - y^{2} - 1 \geq 0

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

Solution steps

2 x - 4 y - x^{2} - y^{2} - 1 \geq 0

Rewrite

2 x - 4 y - x^{2} - y^{2} - 1 \geq 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - (- 15))^{2} + (y - 37)^{2} = 2^{2}

(x + 1)^{2} + (y + 3)^{2} = 25

(x - 2)^{2} + (y + 5)^{2} = 100

(x + 3.2625400000000 \dots)^{2} + (y + 8)^{2} = + 1936

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

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