algebra2

name a point that lies on the circle:
(x-2)^2+(y+3)^2=16
I cannot use the points (2,1) or (6,-3) and I can't find another point that works.

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  1. Try x = 0 and y = 0 and solve for the other variable. Since the radius of the circle is 4, you will need to stay within 4 of x = 2 and y = -3, or you won't get a real answer.

    If y = 0,
    (x-2)^2 + (3)^2 =16
    (x-2)^2 = 7
    x-2 = + or - sqrt7
    x = 2 + or - sqrt7

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