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

(- 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})

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