(snip)
A sneaky way of finding the radius is to "notice" that the sum of the
squares of the coordinates of each of the three points is the same, namely
5. Thus, they are all the same distance, sqrt(5), away from the origin.
Since three points uniquely determine a circle, the circle containing all
three of them must have center at the origin, and radius sqrt(5).
A slightly less sleazy solution is the following:
For this problem, the symmetry of the points makes it easy to get the
answer. Let
A=(-2,1)
B=(2,1)
C=(-1,2)
Since all points are on the peripehery of the circle, the intersection of
the perpendicular bisectors of the chords AB and AC will pass through the
center of the circle. By symmetry (i.e., because A and B are reflections of
each other in the y axis), this center must be located on the y axis.
Said another way, the midpoint of the chord AB is on the y axis, and since
the line AB is horizontal, the perpendicular bisector of AB must be just
the y axis, so the center is located on the y axis.
Take the midpoint M of the chord AC. Its coordinates are (-1.5, 1.5). The
slope of the chord AC is 1, so the slope of the perpendicular bisector
which passes through M is -1. That is, the perpendicular bisector passes
through the origin, which is therefore the center of the circle.
Thus the radius is, from the Pythagorean Theorem, sqrt(2^2 + 1^2) = sqrt(5).
Dr. Logan's proof is more general, of course.
mark snyder
fitchburg state college
msnyder@fsc.edu,msnyder@tiac.net
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