What is the center x^2+y^2-6x+10y+9=0 ?

The center x^2+y^2-6x+10y+9=0 is (3,-5)

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center x^2+y^2-6x+10y+9=0

center

x^{2} + y^{2} - 6 x + 10 y + 9 = 0

(3, - 5)

Solution steps

x^{2} + y^{2} - 6 x + 10 y + 9 = 0

Rewrite

x^{2} + y^{2} - 6 x + 10 y + 9 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 5))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (3, - 5), r = 5

And the center is:

(3, - 5)

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

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

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

(x - 3)^{2} + (y - 5)^{2} = 121

(x + 4)^{2} + (y - 3)^{2} = 16