What is the center x^2+y^2+10x-16y+75=0 ?

The center x^2+y^2+10x-16y+75=0 is (-5,8)

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center x^2+y^2+10x-16y+75=0

center

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

(- 5, 8)

Solution steps

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

Rewrite

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 8)^{2} = (14)^{2}

Therefore the circle properties are:

(a, b) = (- 5, 8), r = 14

And the center is:

(- 5, 8)

- 20 - x^{2} = - 16 x - 20 y + y^{2}

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

x^{2} + y^{2} - 8 x - 4 y - 16 = 0

x^{2} + y^{2} = 72

(x - 3)^{2} + (y - 2)^{2} = 49

What is the center x^2+y^2+10x-16y+75=0 ?
The center x^2+y^2+10x-16y+75=0 is (-5,8)