Graphics Reference
In-Depth Information
AB
AB
=
(
)
=
(
)
u
=
u
,
u
and
v
u
,
u
.
12
21
Proof.
Note that the associated pairs of lines (
L
1
,
L
2
) and (
L
3
,
L
4
) intersect on the line
through the points
A
and
B
. We again switch to the coordinate system defined by the
frame (
A
,
u
,
v
) and assume that
A
= (0,0) and
B
= (b,0), b > 0. See Figure 6.23(b). Let
F
=
A
+ s
AB
be the point where
L
1
intersects the x-axis. Then
AF
=
=-
s
s1
AB
BF
AB
.
(6.38)
Because the triangles
ADF
and
BEF
are similar, we get that
r
r
AF
BF
s
AB
AB
s
1
2
=
=
=
.
(6.39)
s
-
1
s
-
1
Case 1:
s > 1. In this case equation (6.39) implies that
r
rr
1
12
s
=
.
-
Case 2:
0 < s < 1. In this case equation (6.38) implies that
r
rr
1
12
s
=
.
+
In either case, since we now know
F
, we can now use Formula 6.8.3 to compute
D
i,±
and
E
i,±
. For example, in Case 1,
2
2
2
(
)
±
=-
-
s
1
AB
AB
-
r
r
2
2
2
2
2
(
)
DF
u
±
s
-
1
AB
-
r
v
.
2
,
2
s
-
1
s
-
1
AB
Note that to use Formula 6.8.3 we must use the frame (
F
,-
u
,
v
). It is now a simple
matter to rewrite this formula in the form stated earlier.
6.8.5 Formula.
Consider two circles in the plane centered at points
A
and
B
with
radii r
1
and r
2
, respectively. Assume that r
1
+ r
2
<|
AB
|<r
1
+ r
2
+ 2r. The circles of
radius r that are tangent to these circles have center
C
defined by
2
2
(
)
CA u
=+ ±
e
r
+
r
-
e
v
,
1
where
