106 Circles; their basic properties Ch. 7
P
S Q
O
Figure 7.2.P
S Q
O
M
P
S Q
O U
Proof.
(i) By 5.2.2
|∠OSP|◦+|∠SPO|◦+|∠POS|◦= 180 ,|∠OQP|◦+|∠QPO|◦+|∠POQ|◦= 180.But by 4.1.1,
|∠OSP|◦=|∠SPO|◦,|∠OQP|◦=|∠QPO|◦,
and so
2 |∠SPO|◦+ 2 |∠QPO|◦+|∠POS|◦+|∠POQ|◦= 360.
NowO∈[Q,S]so|∠POS|◦+|∠POQ|◦= 180 ,and as[P,O⊂IR(|QPS)we have
|∠SPO|◦+|∠OPQ|◦=|∠QPS|◦. Thus|∠QPS|◦= 90.
(ii) LetObe the mid-point ofQandSand throughOdraw the line parallel to
PQ. It will meet[P,S]in a pointM.Thenby5.3.1Mis the mid-point ofPandS.But
PQ⊥PSandPQ‖MOso by 5.1.1MO⊥PS.Then[O,P,M]≡[O,S,M]by the SAS
principle of congruence. It follows that|O,P|=|O,S|.
(iii) IfP∈QR, then by (i) of the present theorem and 4.3.3U∈[Q,S].IfP∈QS
thenUis eitherQorSand soU∈[Q,S].Thenby3.1.2|Q,U|≤|Q,S|.ButasO=
mp(Q,S), by 3.2.1|Q,O|=^12 |Q,S|,andso|Q,S|= 2 k.
(iv) Let[Q,S]be the diameter containingQandH 1 the closed half-plane with
edgelwhich containsO. Then by 2.2.3 every point of[Q,O lies inH 1 .IfPis any
point of the circle andU=πQO(P)then by (iii) aboveU∈[Q,S]⊂[Q,OsoU∈H 1.
Butl⊥QS,UP⊥QSsoUP‖l. Then by 4.3.2P∈H 1.
7.2.2 Equationofacircle ..........................
Let Z 0 ≡(x 0 ,y 0 )and k> 0 .ThenZ≡(x,y)is onC(Z 0 ;k)if and only if
(x−x 0 )^2 +(y−y 0 )^2 =k^2.Proof. This is immediate by the distance formula in 6.1.1.7.2.3 Circle through three points ......................
Given any three non-collinear points A,B and C, there is a unique circle which passes
through them.