What is the center 16x^2+16y^2+96x-40y=315 ?

The center 16x^2+16y^2+96x-40y=315 is (-3, 5/4)

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center 16x^2+16y^2+96x-40y=315

center

16 x^{2} + 16 y^{2} + 96 x - 40 y = 315

(- 3, \frac{5}{4})

Solution steps

16 x^{2} + 16 y^{2} + 96 x - 40 y = 315

Rewrite

16 x^{2} + 16 y^{2} + 96 x - 40 y = 315

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{5}{4})^{2} = (\frac{11}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{5}{4}), r = \frac{11}{2}

And the center is:

(- 3, \frac{5}{4})

x^{2} + y^{2} - 2 x + 6 y + 9 = 0

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

x^{2} + y^{2} + 6 x - 2 y - 15 = 0

8 x + x^{2} - 2 y = 64 - y^{2}

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

What is the center 16x^2+16y^2+96x-40y=315 ?
The center 16x^2+16y^2+96x-40y=315 is (-3, 5/4)