Graphics Reference
In-Depth Information
Figure 6.20.
Circle through three points.
aba b
ab
-
r =
.
2
¥
In the special case where
p
1
,
p
2
, and
p
3
are points in the plane, these formulas can
be simplified. Let
a
= (a
1
,a
2
),
b
= (b
1
,b
2
), and
c
= (c
1
,c
2
). Then
2
2
2
2
a
b
-
-
b
a
b
a
-
-
a
b
2
2
1
1
c
=
and
c
=
.
(6.33)
1
2
(
)
(
)
2
ab
ba
2
ab
ba
12
12
12
12
Proof.
We shall give essentially two proofs for this formula. We give a geometric
argument in the general case and an algebraic one in the planar case. In either case,
the solution to this problem will be easier if we first move the point
p
1
to the origin
and solve the problem of finding the center
c
of the circle through the points
0
,
a
, and
b
. Consider the following planes and their point-normal equations:
plane
equation
the plane
X
containing the points
0
,
a
, and
b
(and
c
)
(
a
¥
b
) (
x
-
c
)
=
0
the plane
X
1
that is the perpendicular bisector of the segment [
0
,
a
]
a
•(
x
-
(1/2)
a
)
=
0
the plane
X
2
that is the perpendicular bisector of the segment [
0
,
b
]
b
•(
x
-
(1/2)
b
)
=
0
Basic geometric facts about circles imply that
c
is the intersection of these three planes
X
,
X
1
, and
X
2
. See Figure 6.20 where we have identified
X
with the xy-plane. Letting
x
be
0
in the equation for
X
and
c
in the other two equations gives us that
2
2
(
)
=
()
=
()
a
¥
b •c
=
0
,
c•a
1 2
a
,
and
c•b
1 2
b
.
Applying Formula 6.5.6 for the intersection of three planes now gives us our formula
(6.32). The formula for the radius r is gotten by substituting formula (6.32) into the
equation
r
2
=
c
•
c
and simplifying.
Next, assume that our points
p
i
lie in the xy-plane. The equation for the circle
with center
c
and radius r is
2
2
(
)
(
)
2
.
xc
-
+-
yc
=
r
(6.34)
1
2
