What is 4x^2-8>=-4y^2 ?

The solution to 4x^2-8>=-4y^2 is Circle with (a,b)=(0,0),r=sqrt(2)

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4x^2-8>=-4y^2

4 x^{2} - 8 \geq - 4 y^{2}

(a, b) = (0, 0), r = 2

Solution steps

4 x^{2} - 8 \geq - 4 y^{2}

Rewrite

4 x^{2} - 8 \geq - 4 y^{2}

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

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

(a + b) 2 = a^{2} + b^{2}

x^{2} - 10 x + 25 + y^{2} - 8 y + 16 = 64

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

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

What is 4x^2-8>=-4y^2 ?
The solution to 4x^2-8>=-4y^2 is Circle with (a,b)=(0,0),r=sqrt(2)