Algebra II

Find the radius and center of the circle which passes through the points (-2,3), (0,-3), (4,-3).

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  1. notice the points (0,-3) and (4,-3) form a horizontal line, so the centre must be on the line x = 2
    it must also be on the right-bisector of the line between (-2,3) and (0,-3)
    midpoint of that line is (-1,0)
    slope of the right - bisector is 1/3
    equation:
    y = (1/3)(x+1)
    3y = x+1 , but x = 2
    3y = 3
    y = 1
    centre is (2,1)
    equation:
    (x-2)^2 + (y-1)^2 = r^2
    point (4,-3) is on it
    2^2 + (-4)^2 = r^2 = 20

    centre (2,1)
    radius : √20 or 2√5

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