What is 4x^2+8x+4y^2+4=16 ?

The solution to 4x^2+8x+4y^2+4=16 is Circle with (a,b)=(-1,0),r=2

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4x^2+8x+4y^2+4=16

4 x^{2} + 8 x + 4 y^{2} + 4 = 16

(a, b) = (- 1, 0), r = 2

Solution steps

4 x^{2} + 8 x + 4 y^{2} + 4 = 16

Rewrite

4 x^{2} + 8 x + 4 y^{2} + 4 = 16

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (- 1, 0), r = 2

x^{2} - \frac{5}{2} x + (y + 2)^{2} = 50

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

(x - 12)^{2} + (y + 7)^{2} = 961

(x + 4)^{2} + (y + 1)^{2} = 100

(x - 8)^{2} + (y - 6)^{2} = 1^{2}

What is 4x^2+8x+4y^2+4=16 ?
The solution to 4x^2+8x+4y^2+4=16 is Circle with (a,b)=(-1,0),r=2