What is 16x^2+16y^2+8x-32=-1 ?

The solution to 16x^2+16y^2+8x-32=-1 is Circle with (a,b)=(-1/4 ,0),r=sqrt(2)

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16x^2+16y^2+8x-32=-1

16 x^{2} + 16 y^{2} + 8 x - 32 = - 1

(a, b) = (- \frac{1}{4}, 0), r = 2

Solution steps

16 x^{2} + 16 y^{2} + 8 x - 32 = - 1

Rewrite

16 x^{2} + 16 y^{2} + 8 x - 32 = - 1

in the form of the standard circle equation

(x - (- \frac{1}{4}))^{2} + (y - 0)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{4}, 0), r = 2

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

4 x^{2} + 4 y^{2} = 128

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

x^{2} + y^{2} \geq 5

x^{2} + y^{2} - 6 x + 6 y + 14 = 0

What is 16x^2+16y^2+8x-32=-1 ?
The solution to 16x^2+16y^2+8x-32=-1 is Circle with (a,b)=(-1/4 ,0),r=sqrt(2)