What is the center x^2+y^2-10x+12y+25=0 ?

The center x^2+y^2-10x+12y+25=0 is (5,-6)

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center x^2+y^2-10x+12y+25=0

center

x^{2} + y^{2} - 10 x + 12 y + 25 = 0

(5, - 6)

Solution steps

x^{2} + y^{2} - 10 x + 12 y + 25 = 0

Rewrite

x^{2} + y^{2} - 10 x + 12 y + 25 = 0

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 6))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (5, - 6), r = 6

And the center is:

(5, - 6)

y^{2} + x^{2} - 12 x + 18 y + 116 = 0

16 x^{2} + 16 y^{2} + 8 x - 3 y = - 1

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

(x + 2)^{2} + (y - 1)^{2} = 9

(x - 0)^{2} + (y - 0)^{2} = 25

What is the center x^2+y^2-10x+12y+25=0 ?
The center x^2+y^2-10x+12y+25=0 is (5,-6)