What is the center (x-3)^2+(y-2)^2=49 ?

The center (x-3)^2+(y-2)^2=49 is (3,2)

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center (x-3)^2+(y-2)^2=49

center

(x - 3)^{2} + (y - 2)^{2} = 49

(3, 2)

Solution steps

(x - 3)^{2} + (y - 2)^{2} = 49

Rewrite

(x - 3)^{2} + (y - 2)^{2} = 49

in the form of the standard circle equation

(x - 3)^{2} + (y - 2)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (3, 2), r = 7

And the center is:

(3, 2)

(x - 1)^{2} + (y + 5)^{2} = 16

(x - 4)^{2} + (y + 2)^{2} = 1 0^{2}

(x - 7)^{2} + (y + 5)^{2} = 4

(x + 4)^{2} + (y - 2)^{2} = 121

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36