210 Vector and complex-number methods Ch. 11
By Pythagoras’ theorem
|z 16 −z 2 |^2 =|z 4 −z 2 |^2 +|z 4 −z 16 |^2
={^12 |z 3 −z 2 |^2 }^2 +{|q|^2 |z 3 −z 2 |}^2
={^14 +[
q^21 +p^21 −p 1
2 q 1
]^2 }|z 3 −z 2 |^2
=
a^2
4 q^21
{(q^21 +p^21 −p 1 )^2 +q^21 }.
Thus the radius length of the circumcircle is
|z 16 −z 2 |=
a
2 |q 1 |
√
q^21 +[p^21 +q^21 −p 1 ]^2. (11.6.5)
In fact we can also derive the formula (11.6.5) slightly differently as follows.
Directly from (11.6.4) we have that
|z 16 −z 2 |^2 =
1
4
| 1 +
(
p^21 +q^21
q 1
−
p 1
q 1
)
ı|^2 a^2
=^14
[
1 +
(
p^21 +q^21
q 1
−
p 1
q 1
) 2 ]
a^2
=
a^2
4 q^21
[
q^21 +[p^21 +q^21 −p 1 ]^2
]
.
We can deduce from this material a formula for the circumcentre in terms of areal
coordinates. For by (11.6.2)
p^21 −p 1 +q^21 =
bc
a^2
cosα,q 1 =
bc
a^2
sinα,
so thatz=z 2 +^12 ( 1 +cotαı)(z 3 −z 2 ). Then by (11.6.1) we have
x−x 2 =^12 (x 3 −x 2 )−^12 cotα(y 3 −y 2 ),y−y 2 =^12 cotα(x 3 −x 2 )+^12 (y 3 −y 2 ).
From this we have that
δF(Z,Z 2 ,Z 3 )=δF(Z−Z 2 ,O,Z 3 −Z 2 )
=^14 cotα
[
(x 3 −x 2 )^2 +(y 3 −y 2 )^2
]
=^14 a^2 cotα.
AsδF(Z 1 ,Z 2 ,Z 3 )=^12 q 1 a^2 =^12 bcsinαwe have
δF(Z,Z 2 ,Z 3 )
δF(Z 1 ,Z 2 ,Z 3 )
=
1
2
a^2 cotα
bcsinα
,
and by use of the sine rule this is seen to be equal to
1
2
cosα
sinβsinγ