# 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)

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