• Open a new sketch. Draw circle C 1 , labeling its center O , and construct point A not on the circle as well as a point D on the circle. • Construct the tangent line to the circle C 1 at D and then the segment AD . • Construct the perpendicular bisector of AD . The intersection of this perpendicular bisector with the tangent line to the circle at D will be the center of a circle passing through both A and D and intersecting the circle C 1 orthogonally at D . Why? The figure below illustrates the construction when A is outside circle C 1.
27 C 1 C 2 O D A What turns out to be of critical importance is the locus of circle C 2 passing through A and D and intersecting the given circle C 1 orthogonally at D , as D moves. Use Sketchpad to explore the locus. • Select the circle C 2 , and under the Display menu select trace circle. Drag D . • Alternatively you can select the circle C 2 , then select the point D and under the Construct menu select locus. The following figure was obtained by choosing different D on the circle C 1 and using a script to construct the circle through A (outside C 1 ) and D orthogonal to C 2 . The figure you obtain should look similar to this one, but perhaps more cluttered if you have traced the circle.
28 C 1 C 2 A O D B Your figure should suggest that all the circles orthogonal to the given circle C 1 that pass through A have a second common point on the line through O (the center of C 1 ) and A . In the figure above this second common point is labeled by B . [Does the figure remind you of anything in Physics - the lines of magnetic force in which the points A and B are the poles of the magnet. say?] Repeat the previous construction when A is inside C 1 and you should see the same result. End of Demonstration 3.5.2. At this moment, Theorem 2.9.2 and its converse 2.9.4 will come into play. 3.5.3 Theorem. Fix a circle C 1 with center O, a point A not on the circle, and point D on the circle, Now let B be the point of intersection of the line through O with the circle through A and D that is orthogonal to C 1 . Then B satisfies OA . OB = OD 2 . In particular, the point B is independent of the choice of point D . The figure below illustrates the theorem when A is outside C 1 .
29 C 1 A O D B OA = 1.65 inches OD 2 = 0.86 inches 2 OB = 0.52 inches OA OB = 0.86 inches 2 Proof. By construction the segment OD is tangential to the orthogonal circle. Hence OA . OB = OD 2 by Theorem 2.9.2. QED Theorem 3.5.3 has an important converse. 3.5.4 Theorem. Let C 1 be a circle of radius r centered at O. Let A and B be points on a line through O (neither A or B on C 1 ). If OA . OB = r 2 then any circle through A and B will intersect the circle C 1 orthogonally. Proof. Let D denote a point of intersection of the circle C 1 with any circle passing through A and B . Then OA . OB = OD 2 . So by Theorem 2.9.4, the line segment OD will be tangential to the circle passing through A , B , and D . Thus the circle centered at O will be orthogonal to the circle passing through A , B , and D . QED Theorems 3.5.3 and 3.5.4 can be used to construct a circle orthogonal to a given circle C 1 and passing through two given points P , Q inside C 1 .
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