Geometry with Trigonometry

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

.