
614 Chapter 13 Containment Methods
where b
i
∈ [0, 1] and b
0
+ b
1
+ b
2
= 1. The center is equidistant from the three
points, so
|C − S
0
|=|C − S
1
|=|C − S
2
|=r
2
where r is the (as of yet unknown) radius of the circle. From these conditions, we
have
C − S
0
= b
0
E
0
+ b
1
E
1
− E
0
C − S
1
= b
0
E
0
+ b
1
E
1
− E
1
C − S
2
= b
0
E
0
+ b
1
E
1
where E
0
= S
0
− S
2
and E
1
= S
1
− S
2
. These lead to
r
2
=|b
0
E
0
+ b
1
E
1
|
2
− 2E
0
.
(b
0
E
0
+ b
1
E
1
) +|E
0
|
2
r
2
=|b
0
E
0
+ b
1
E
1
|
2
− 2E
1
.
(b
0
E
0
+ b
1
E
1
) +|E
1
|
2
r
2
=|b
0
E
0
+ b
1
E
1
|
2
Subtracting the last equation from the first two and writing the equations as a linear
system:
E
0
.
E
0
E
0
.
E
1
E
1
.
E
0
E
1
.
E
1
b
0
b
1
=
1
2
E
0
.
E
0
E
1
.
E
1
Thesystemissolvedforb
0
and b
1
, and then b
2
= 1 − b
0
− b
1
. One of the equations ...