Geometry with Trigonometry

180 Complex coordinates; sensed angles; angles between lines Ch. 10

Now (10.11.7) is a circle whenc= 2 h=0andg=−f. The equation then becomes

x^2 +y^2 +

[g

h

(v 1 −v 2 )−u 1 −u 2

]

x+

[g

h

(u 2 −u 1 )−v 1 −v 2

]

y

+

g

h

(u 1 v 2 −u 2 v 1 )+u 1 u 2 +v 1 v 2 = 0. (10.11.9)

This is the set of circles which pass through the pointsW 1 andW 2 , a set of coaxal
circles. The corresponding equation for the second locus is

u^2 +v^2 +

[g

h

(y 1 −y 2 )−x 1 −x 2

]

u+

[g

h

(x 2 −x 1 )−y 1 −y 2

]

v

+

g

h

(x 1 y 2 −x 2 y 1 )+x 1 x 2 +y 1 y 2 = 0 , (10.11.10)

and this gives the set of coaxal circles passing throughZ 1 andZ 2.

We can take an arbitrary circle
from the first coaxal set and then
there is a unique one from the
second set corresponding to it. If
we takeZ 1 ,Z 2 ,W 1 ,W 2 to be con-
cyclic we get just one circle and
that is the classical case; it oc-
curs when the remaining coeffi-
cients in the two equations are
pairwise equal.

Z 1

Z 2

W 1

W 2

W

Z

Figure 10.17. Pascal result for two circles.

10.11.3 ....................................

Instead of using parallelism of lines as the basis of the relation in 10.11.2, we could
take instead a fixed lineowith equationlx+my+n=0, and let(Z 1 ,W 1 )∼(Z 2 ,W 2 )
if the linesZ 1 W 2 andW 1 Z 2 meet ono. The results are like those in 10.11.2 and the
transformation corresponding to (10.11.4) is

n(y 2 −v)−l(x 2 v−y 2 u)

n(x 2 −u)+m(x 2 v−y 2 u)

=

n(v 2 −y)−l(u 2 y−v 2 x)

n(u 2 −x)+m(u 2 y−v 2 x)

,

n(y 1 −v)−l(x 1 v−y 1 u)

n(x 1 −u)+m(x 1 v−y 1 u)

=

n(v 1 −y)−l(u 1 y−v 1 x)

n(u 1 −x)+m(u 1 y−v 1 x)

.

Exercises

10.1 Prove the result of Varignon (1731) that ifA,B,C,Dare the vertices of a convex

quadrilateral and

P=mp(A,B),Q=mp(B,C),R=mp(C,D),S=mp(D,A),