What is the center x^2+y^2-3x+4y=0 ?

The center x^2+y^2-3x+4y=0 is (3/2 ,-2)

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center x^2+y^2-3x+4y=0

center

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{3}{2}, - 2), r = \frac{5}{2}

And the center is:

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

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x^{2} + y^{2} + 14 x - 12 y + 4 = 0