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

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

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

center

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

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

Solution steps

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

Rewrite

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

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} - 4 x + 6 y + 9 = 0