Geometry with Trigonometry

Sec. 10.9 Some results on circles 165

10.9.2 A sufficient condition to lie on a circle ...............

Let Z 1 ,Z 2 be fixed distinct points
and Z a variable point. As Z
varies in one of the half-planes
with edge Z 1 Z 2 , for the sensed
angleθ=FZ 1 ZZ 2 let|θ|◦=
|γ|◦ where γ is a fixed non-
null and non-straight angle in
A(F′), while as Z varies in the
other half-plane with edge Z 1 Z 2 ,

let|θ|◦=|γ+ (^180) F′|◦.ThenZ
lies on a circle which passes
through Z 1 and Z 2.

O I

J H^1

H 2

H 4 H 3

Z 0 I 0

J 0

Z 1

Z 2

Z

Figure 10.9.

Proof.Wehave

z 2 −z

z 1 −z

=tcisγ

for somet∈R{ 0 }.Then

z=

z 2 −tz 1 cisγ

1 −tcisγ

so that with cotγ=cosγ/sinγ,

z−

1

2

(z 1 +z 2 )−

1

2

ıcotγ.(z 2 −z 1 )

=

z 2 −tz 1 cisγ

1 −tcisγ

−

1

2

(z 1 +z 2 )−

1

2

icotγ.(z 2 −z 1 )

=

1

2 (z^2 −z^1 )[^1 +tcisγ−ıcotγ(^1 −tcisγ)]

1 −tcisγ

=

1

2 (z^2 −z^1 )[sinγ(^1 +tcisγ)−ıcosγ(^1 −tcisγ)]

sinγ( 1 −tcisγ)

=

1

2 (z^2 −z^1 )[sinγ+ı(t−cosγ)]

sinγ( 1 −tcisγ)

and this has absolute value
|z 2 −z 1 |
2 |sinγ|

This shows thatZlies on a circle, the centre and length of radius of which are evident.

10.9.3Complexcross-ratio ..........................

Let Z 2 ,Z 3 ,Z 4 be non-collinear points andCthe circle that contains them. Then Z∈
Z 3 Z 4 lies inCif and only if

ℑ

(z−z 3 )(z 2 −z 4 )

(z−z 4 )(z 2 −z 3 )

= 0.