Sec. 11.6 Mobile coordinates 209
Note too that if
z−z 2 =(p+qı)(z 3 −z 2 ),z′−z 2 =(p′+q′ı)(z 3 −z 2 ),
then
|z′−z|=|p′−p+(q′−q)ı||z 3 −z 2 |,
and so
|Z,Z′|=
√
(p′−p)^2 +(q′−q)^2 |Z 2 ,Z 3 |=a
√
(p′−p)^2 +(q′−q)^2. (11.6.3)
11.6.4 Circumcentre of a triangle .......................
To make our notation consis-
tent we now use the notation
[Z 1 ,Z 2 ,Z 3 ] for the triangle
[A,B,C] in 7.2.3 and Z 16
for its circumcentre D.As
usual we denote by Z 4 ,Z 5 ,Z 6
the mid-points of the sides
[Z 2 ,Z 3 ],[Z 3 ,Z 1 ],[[Z 1 ,Z 2 ],
respectively. We note that
points Z on the perpendicular
bisector of[Z 2 ,Z 3 ] have com-
plex coordinates of the form
z=z 2 +(^12 +qı)(z 3 −z 2 ),where
q∈R.
Z 1
Z 2
Z 3
Z 16
Z 4
Z 5
Z 6
Figure 11.13. Circumcentre of triangle.
We also havez 6 =^12 (z 1 +z 2 ).AswearetohaveZZ 6 ⊥Z 1 Z 2 we must have
z 2 +(^12 +qı)(z 3 −z 2 )−^12 (z 1 +z 2 )=ır(z 1 −z 2 ),
for some real numberr.Then
z 2 +(^12 +qı)(z 3 −z 2 )−^12 [ 2 z 2 +(p 1 +q 1 ı(z 3 −z 2 )] =ır(z 1 −z 2 ),
[^12 +qı−^12 (p 1 +q 1 ı)−ır(p 1 +q 1 ı)](z 3 −z 2 )= 0 ,
1
2 p^1 −rq^1 −
1
2 +(
1
2 q^1 +rp^1 )ı=qı
Thus we must have
q=^12 q 1 +rp 1 ,rq 1 =^12 p 1 −^12 ,
from which we obtain the solutionsr=p 21 q− 11 and thenq= 2 q^11 [p^21 +q^21 −p 1 ].
Thus the circumcentre has complex coordinate
z 16 =z 2 +
1
2
[
1 +
p^21 +q^21 −p 1
q 1
ı
]
(z 3 −z 2 ). (11.6.4)