Open a new sketch draw circle c1 labeling its center

• 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
.