Sec. 7.4 Polar properties of a circle 111and this is fixed.
We continue with the case wherePis exterior to the circle, and may suppose that
|P,R|<|P,S|, as otherwise we can just interchange the pointsRandS.AsPis outside
the circle, by 7.1.1 it is outside the segment[R,S]on the lineRS.Thenwehave|P,R||P,S|=(|P,M|−|M,R|)(|P,M|+|M,S|)=|P,M|^2 −|M,R|^2=(
|P,O|^2 −|O,M|^2
)
−|M,R|^2 =|P,O|^2 −
(
|O,M|^2 +|M,R|^2
)
=|P,O|^2 −|O,R|^2 =|P,O|^2 −|O,T 1 |^2 =|P,T 1 |^2.
7.4.3 Aharmonicrange ...........................
Let T 1 and T 2 be the points of contact of the tangents from an exterior point P to a
circleCwith centre O. If a line l through P cutsCin the points R and S, and cuts
T 1 T 2 in Q, then P and Q divide{R,S}internally and externally in the same ratio.
Proof. We use the notation of
7.4.1 and first recall thatT 1 T 2
cutsOPat right-angles at a point
U. Then, by 7.2.1(ii), the circle
C 2 on[O,Q]as a diameter passes
throughU.WeletMbe the mid-
point ofRandS;thenby4.1.1
OM⊥MQ,andsoMalso lies
on the circleC 2.
O
UR P
T 1
T 2
Q
M
S
Figure 7.6.
We have |P,R||P,S|=|P,T 1 |^2 by 7.4.2,|P,T 1 |^2 =|P,U||P,O|by the proof of
Pythagoras’ theorem in 5.4.1, and|P,U||P,O|=|P,Q||P,M|, by the 7.4.2 applied to
the circleC 2. On combining these we have|P,R||P,S|=|P,Q||P,M|.
We cannot havel⊥OPas that would makel‖T 1 T 2 whereaslmeetsT 1 T 2. Then,
with the notation of 7.4.1,lis not a tangent toC 1 atPso, by 7.1.1lmust meetC 1 at a
pointH. We are supposing thatlis not either ofPT 1 ,PT 2 and soHis notT 1 orT 2 .We
letKbe the foot of the perpendicular fromHtoOP. Then by 4.3.3K∈[P,O]and by
the proof of Pythagoras’ theorem in 5.4.1|P,H|^2 =|P,K||P,O|.IfwehadK∈[P,U]
we would have|P,K|<|P,U|and so|P,H|^2 =|P,O||P,K|<|P,O||P,U|=|P,T 1 |^2.From this it would follow that|O,H|^2 =|O,P|^2 −|P,H|^2 >|O,P|^2 −|P,T 1 |^2 =a^2 ,and makeHexterior to the circle. ButHis the foot of the perpendicular fromOto
l, and by 7.1.1 this would causelto have no point in common with the circle. This
cannot occur and so we must haveK∈[O,U]. By a similar argument it then follows
thatHis interior to the circleCand solmeetsCin two pointsRandS.
By 7.2.1(iv) every point of the circleCis in the closed half-planeH 1 with edge
PT 1 and which containsO. By 2.2.3H 1 containsU∈[P,Oand then it also contains