Geometry with Trigonometry

Sec. 10.11 A case of Pascal’s theorem, 1640 179

and so our condition for transitivity is got by equating the two right- hand sides here.
This turns out to beδF(Z 1 ,Z 2 ,Z)=δF(W 1 ,W 2 ,W).
Now (10.11.2) and (10.11.3) simultaneously give a transformation under which
Z→Was we see by writing them as

v−y 2

u−x 2

=

y−v 2

x−u 2

,

v−y 1

u−x 1

=

y−v 1

x−u 1

. (10.11.4)

On solving foruandvin this we obtain

u =

y 2 −y 1 +x (^1) xy−−uv^11 −x (^2) xy−−vu^22
y−v 1
x−u 1 −
y−v 2
x−u 2

,

v =

x 2 −x 1 +y 1 xy−−vu^11 −y 2 xy−−uv^22

x−u 1

y−v 1 −

x−u 2

y−v 2

.

To utilize this transformation we consider loci with equations of the form

2 h

y−v 2

x−u 2

y−v 1

x−u 1

+ 2 g

y−v 2

x−u 2

+ 2 f

y−v 1

x−u 1

+c= 0. (10.11.5)

Under the transformation this maps into the locus with equation

2 h

v−y 2

u−x 2

v−y 1

u−x 1

+ 2 g

v−y 2

u−x 2

+ 2 f

v−y 1

u−x 1

+c= 0. (10.11.6)

On clearing the equation (10.11.5) of fractions we obtain

2 h(y−v 2 )(y−v 1 )+ 2 g(y−v 2 )(x−u 1 )+ 2 f(y−v 1 )(x−u 2 )

+c(x−u 1 )(x−u 2 )= 0 ,

from which we see thatW 1 andW 2 are both on this locus. This equation can be re-
arranged as

cx^2 + 2 (g+f)xy+ 2 hy^2 −( 2 gv 2 + 2 fv 1 +cu 1 +cu 2 )x−

( 2 gu 1 + 2 fu 2 + 2 hv 1 + 2 hv 2 )y+ 2 hv 1 v 2 + 2 gu 1 v 2 + 2 fu 2 v 1 +cu 1 u 2 = 0. (10.11.7)

Similarly we see thatZ 1 andZ 2 are on the locus given by (10.11.6), and the equation
for it becomes

cu^2 + 2 (g+f)uv+ 2 hv^2 −( 2 gy 2 + 2 fy 1 +cx 1 +cx 2 )u−

( 2 gx 1 + 2 fx 2 + 2 hy 1 + 2 hy 2 )v+ 2 hy 1 y 2 + 2 gx 1 y 2 + 2 fx 2 y 1 +cx 1 x 2 = 0. (10.11.8)

We note thatW 1 maps toZ 1 andW 2 maps toZ 2 under the transformation in whichZ
maps toW
To identify all the loci that can occur in (10.11.7) and (10.11.8) would take us
beyond the concepts of the present course, so we concentrate on when they represent
circles.