What is the center x^2+y^2-10x+9y+1=0 ?

The center x^2+y^2-10x+9y+1=0 is (5,-9/2)

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center x^2+y^2-10x+9y+1=0

center

x^{2} + y^{2} - 10 x + 9 y + 1 = 0

(5, - \frac{9}{2})

Solution steps

x^{2} + y^{2} - 10 x + 9 y + 1 = 0

Rewrite

x^{2} + y^{2} - 10 x + 9 y + 1 = 0

in the form of the standard circle equation

(x - 5)^{2} + (y - (- \frac{9}{2}))^{2} = (\frac{177}{2})^{2}

Therefore the circle properties are:

(a, b) = (5, - \frac{9}{2}), r = \frac{177}{2}

And the center is:

(5, - \frac{9}{2})

(x - 1)^{2} + (y + 3)^{2} = 9

x^{2} + y (y + 4) = 10 x + 7

x^{2} + y^{2} - 12 x + 6 y + 36 = 0

9 = 2 y - y^{2} - 6 x - x^{2}

(x + 1)^{2} + (y - 3)^{2} = 4

What is the center x^2+y^2-10x+9y+1=0 ?
The center x^2+y^2-10x+9y+1=0 is (5,-9/2)