Graphics Reference
In-Depth Information
Since r
2
= c
1
2
+ c
2
2
, equation (6.34) reduces to
2
2
2
xc
+
2
yc
=
x
+
y
.
(6.35)
1
2
Substituting points
a
and
b
into equation (6.35) gives two equations in two unknowns
c
1
and c
2
:
2
2
ac
+
2
a c
=
a
11
2 2
2
2
bc
+
2
b c
=
b
.
(6.36)
12
22
The solution to this system of equations leads to the formula (6.33). This finished the
proof of Formula 6.8.1.
The next two formulas are useful in blending computations.
6.8.2
Formula.
Let
L
1
and
L
2
be intersecting lines in the plane defined by
equations
a x
++=
b y
c
0
and
ax by c
++=,
0
1
1
1
2
2
2
respectively. The circles of radius r that are tangent to
L
1
and
L
2
have centers (d,e)
defined by
(
)
2
2
2
2
bc
-
cb
±
rb
a
+
b
-
b
a
+
b
12
1 2
2
1
1
12
2
d
=
,
ab
-
a b
12
21
(
)
2
2
2
2
ac
-
ac
±
ra a
+
b
-
a
a
+
b
21
12
1
2
2
21
1
e
=
.
ab
-
a b
12
21
Proof.
From Figure 6.21(a) one can see that there are four solutions in general. Let
L
1
¢ and
L
2
¢ be lines that are parallel to and a distance r from lines
L
1
and
L
2
, respec-
tively. See Figure 6.21(b). There are four such pairs of lines and it is easy to see that
the intersection of these lines defines the centers (d,e) of the circles we seek. Now the
lines
L
1
¢ and
L
2
¢ are a distance r closer or further to the origin than the lines
L
1
and
L
2
. Therefore, it is easy to see from equation (6.24) that the equations for
L
1
¢ and
L
2
¢
are
2
2
ax by
+
-±
c
r a
+
b
(6.37a)
1
1
1
1
1
2
2
ax by
+
-±
c
r a
+.
b
(6.37b)
2
2
2
2
2
Solving equations (6.37) gives our answer.
