Geometry with Trigonometry

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)