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

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

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

center

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

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

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

3 (x - 5)^{2} + 3 (y + 9)^{2} = 48

x^{2} + y^{2} + 8 y = 32

x^{2} + y^{2} - 6 x + 8 y - 11 = 0

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

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