Posted by math on Wednesday, December 28, 2011 at 12:33am.
The center of the circle also lies on the perpendicular bisector of any chord. We have the chord between the two given points.
The chord has slope (0-4)/(5-1) = -1.
The midpoint of the chord is (3,2).
The equation of the line containing (3,2) and perpendicular to the chord is
(y-2)/(x-3) = 1
or
-x + y = -1
This intersects the line
x + y = 3 at
(2,1)
So, our circle has equation
(x-2)^2 + (y-1)^2 = r^2
plug in either of the given points and we find r^2 = 10. so,
(x-2)^2 + (y-1)^2 = 10
x = 2 + √10 cos t
y = 1 + √10 sin t
the rest should be easy to figure out.
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