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

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

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

center

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

(- 7, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 7))^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 7, 0), r = 1

And the center is:

(- 7, 0)

x^{2} + 8 x + y^{2} + 4 y = 4

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

(x - 3)^{2} + y^{2} = 4

x^{2} + y^{2} = 21

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