What is the center (x)^2+(y-3/2)^2=4 ?

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

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center (x)^2+(y-3/2)^2=4

center

(x)^{2} + (y - \frac{3}{2})^{2} = 4

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

Solution steps

x^{2} + (y - \frac{3}{2})^{2} = 4

Rewrite

x^{2} + (y - \frac{3}{2})^{2} = 4

in the form of the standard circle equation

(x - 0)^{2} + (y - \frac{3}{2})^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (0, \frac{3}{2}), r = 2

And the center is:

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

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