•
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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- Fall '16
- Mr. B
- Geometry, The Bible, The Land, triangle, Euclidean geometry