What is the center x^2+(y-3)^2=36 ?

The center x^2+(y-3)^2=36 is (0,3)

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center x^2+(y-3)^2=36

center

x^{2} + (y - 3)^{2} = 36

(0, 3)

Solution steps

x^{2} + (y - 3)^{2} = 36

Rewrite

x^{2} + (y - 3)^{2} = 36

in the form of the standard circle equation

(x - 0)^{2} + (y - 3)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (0, 3), r = 6

And the center is:

(0, 3)

(x - \frac{7}{4})^{2} + (y + \frac{3}{4})^{2} = \frac{169}{16}

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