What is the center 16x^2+16y^2+8x-3y=-1 ?

The center 16x^2+16y^2+8x-3y=-1 is (-1/4 , 3/32)

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center 16x^2+16y^2+8x-3y=-1

center

16 x^{2} + 16 y^{2} + 8 x - 3 y = - 1

(- \frac{1}{4}, \frac{3}{32})

Solution steps

16 x^{2} + 16 y^{2} + 8 x - 3 y = - 1

Rewrite

16 x^{2} + 16 y^{2} + 8 x - 3 y = - 1

in the form of the standard circle equation

(x - (- \frac{1}{4}))^{2} + (y - \frac{3}{32})^{2} = (\frac{3}{32})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{4}, \frac{3}{32}), r = \frac{3}{32}

And the center is:

(- \frac{1}{4}, \frac{3}{32})

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

(x + 2)^{2} + (y - 1)^{2} = 9

(x - 0)^{2} + (y - 0)^{2} = 25

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

x^{2} + y^{2} + 8 y = 2 x + 20

What is the center 16x^2+16y^2+8x-3y=-1 ?
The center 16x^2+16y^2+8x-3y=-1 is (-1/4 , 3/32)