What is the center x^2+y^2+14x-16y+64=0 ?

The center x^2+y^2+14x-16y+64=0 is (-7,8)

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center x^2+y^2+14x-16y+64=0

center

x^{2} + y^{2} + 14 x - 16 y + 64 = 0

(- 7, 8)

Solution steps

x^{2} + y^{2} + 14 x - 16 y + 64 = 0

Rewrite

x^{2} + y^{2} + 14 x - 16 y + 64 = 0

in the form of the standard circle equation

(x - (- 7))^{2} + (y - 8)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (- 7, 8), r = 7

And the center is:

(- 7, 8)

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

4 x^{2} + 4 y^{2} - 4 x + 2 y + 1 = 0

x^{2} + 2 x + y^{2} - 3 = 6

x^{2} + y^{2} + 14 x - 16 y = 87

4 x^{2} + 4 x + 4 y^{2} - 16 y - 19 = 0

What is the center x^2+y^2+14x-16y+64=0 ?
The center x^2+y^2+14x-16y+64=0 is (-7,8)