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

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

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

center

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

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

2 x^{2} + 2 y^{2} + 4 x - 8 y - 22 = 0

x^{2} + y^{2} - 4 x + 2 y - 15 = 0

(x - 1)^{2} + (y + 4)^{2} = 9

x^{2} + y^{2} - x - 6 y + 9 = 0

(x + 4)^{2} + (y + 3)^{2} = 17