What is the center x^2-2x+3y+y^2-10=0 ?

The center x^2-2x+3y+y^2-10=0 is (1,-3/2)

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center x^2-2x+3y+y^2-10=0

center

x^{2} - 2 x + 3 y + y^{2} - 10 = 0

(1, - \frac{3}{2})

Solution steps

x^{2} - 2 x + 3 y + y^{2} - 10 = 0

Rewrite

x^{2} - 2 x + 3 y + y^{2} - 10 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{53}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, - \frac{3}{2}), r = \frac{53}{2}

And the center is:

(1, - \frac{3}{2})

x^{2} + y^{2} + 8 x - 14 y + 1 = 0

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

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

x^{2} + y^{2} + 6 y + 4 x + 8 = 0

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

What is the center x^2-2x+3y+y^2-10=0 ?
The center x^2-2x+3y+y^2-10=0 is (1,-3/2)