What is the center 16x^2+16y^2+32x+8y=47 ?

The center 16x^2+16y^2+32x+8y=47 is (-1,-1/4)

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center 16x^2+16y^2+32x+8y=47

center

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

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

Solution steps

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

Rewrite

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

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

4 x^{2} + 4 y^{2} - 14 x - 18 y + 15 = 0

4 x^{2} + 4 y^{2} + 40 y + 75 = 0

(x + \frac{8}{3})^{2} + y^{2} = 1

(x - 3)^{2} + (y + 4)^{2} = \frac{16}{9}

x^{2} + y^{2} + 4 y - 5 = 0

What is the center 16x^2+16y^2+32x+8y=47 ?
The center 16x^2+16y^2+32x+8y=47 is (-1,-1/4)