What is the center x^2+y^2+6x-8y+16=0 ?

The center x^2+y^2+6x-8y+16=0 is (-3,4)

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center x^2+y^2+6x-8y+16=0

center

x^{2} + y^{2} + 6 x - 8 y + 16 = 0

(- 3, 4)

Solution steps

x^{2} + y^{2} + 6 x - 8 y + 16 = 0

Rewrite

x^{2} + y^{2} + 6 x - 8 y + 16 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 3, 4), r = 3

And the center is:

(- 3, 4)

9 x^{2} + 9 y^{2} = 729

x^{2} + y^{2} + 3 x + 4 y + 4 = 0

(x - 9)^{2} + (y + 4)^{2} = 25

x^{2} + y^{2} - 6 x + 2 y + 4 = 0

(x + 3)^{2} + (y + 2)^{2} = 9