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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

- 3 x^{2} - 3 y^{2} + 3 x + 6 = 0

3 x^{2} - 5 x + 3 y^{2} + 5 y - 8 = 0

(x + 4)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} + 6 x + 8 x = - 16

x^{2} + y^{2} - 10 x - 6 y + 25 = 0

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