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

The center 16x^2+16y^2+8x-32y=-1 is (-1/4 ,1)

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

center

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

(- \frac{1}{4}, 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

And the center is:

(- \frac{1}{4}, 1)

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

(x - 5)^{2} + (y + 4)^{2} = 4

x^{2} + y^{2} - 20 x - 18 y + 144 = 0

5 x^{2} + 40 x + 5 y^{2} = 0

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

What is the center 16x^2+16y^2+8x-32y=-1 ?
The center 16x^2+16y^2+8x-32y=-1 is (-1/4 ,1)