What is the center x^2+y^2-16x+12y+64=0 ?

The center x^2+y^2-16x+12y+64=0 is (8,-6)

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center x^2+y^2-16x+12y+64=0

center

x^{2} + y^{2} - 16 x + 12 y + 64 = 0

(8, - 6)

Solution steps

x^{2} + y^{2} - 16 x + 12 y + 64 = 0

Rewrite

x^{2} + y^{2} - 16 x + 12 y + 64 = 0

in the form of the standard circle equation

(x - 8)^{2} + (y - (- 6))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (8, - 6), r = 6

And the center is:

(8, - 6)

x^{2} + 2 x - 4 y + y^{2} = 76

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

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

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

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

What is the center x^2+y^2-16x+12y+64=0 ?
The center x^2+y^2-16x+12y+64=0 is (8,-6)