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Let P,Q,R be three points in the plane which do not all lie on the same line. Then there is a unique circle that passes through all three of them. See Figure 3.20. There are several ways to confirm this assertion.
(a) A general circle has equation
x2 + ax + y2 + by = c.
Thus there are three undetermined parameters. And the three pieces of information provided by the fact that the circle must pass through P =( p1,p2), Q =( q1,q2), R =( r1,r2) (and therefore these three points must satisfy the equation) will determine those parameters. Use this idea to find the unique circle that passes through (1,2), (2,3), and (4,9).

(b) There is a well-defined perpendicular bisector to the segment PQ. This line represents the set of all points that are equidistant from P and Q. There is also a welldefined perpendicular bisector to the segment QR. This line represents the set of all points that are equidistant from Q and R. The intersection of these two lines— which will be a single point C—will be the unique point that is equidistant from all three of P,Q,R. That must be thecenter of the circle. See Figure 3.21. The distance of C to P will be the radius. Use this idea to find the unique circle that passes through (1,0), (0,1), (1,1).

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Tod Thiel
Tod ThielLv2
11 Oct 2019

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