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

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

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center x^2+y^2+2x=3y

center

x^{2} + y^{2} + 2 x = 3 y

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

Solution steps

x^{2} + y^{2} + 2 x = 3 y

Rewrite

x^{2} + y^{2} + 2 x = 3 y

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{3}{2})^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{3}{2}), r = \frac{13}{2}

And the center is:

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

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

x^{2} + 6 x + y^{2} - 3 y = - \frac{1}{4}

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

x^{2} + 4 x + 10 y + y^{2} + 3 = 0

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