Geometry with Trigonometry

104 Circles; their basic properties Ch. 7

P=M

O

l

P

Q

M O

(i) Suppose that|O,M|=k,so
thatMis a point of the circle.
We writeP=M.ThenP∈l,P∈
C(O;k)andOP⊥l. Thus ifVis
any point ofl, other thanP,by
4.3.1 we have|O,V|>|O,P|=
k. HenceVis exterior to the cir-
cle, and so there is no point com-
mon toland the circle exceptP.

M

O l

Figure 7.1.

(ii) Suppose that|O,M|<k,sothatMis interior to the circle. Thenk^2 −|O,M|^2 >
0 so that its square root can be extracted as a positive real number. By A 4 (iv) choose

P∈lso that|M,P|=

√

k^2 −|O,M|^2. There is also a pointQ∈lon the other side of
MfromPandsuchthat|M,Q|=|M,P|. ClearlyMis the mid-point ofPandQ.
WhenM=O,thisgives|O,P|=kso thatP∈C(O;k). By 2.1.3 any pointX=P
of the half-line[O,P must satisfy eitherX∈[O,P]orP∈[O,X].IfX∈[O,P]then
by 3.1.2|O,X|<|O,P|=k,sothatXis interior to the circle. On the other hand
ifP∈[O,X]then|O,X|>|O,P|=k,andsoXis an exterior point for the circle.
MoreoverQis also on the circle and similar results hold whenX∈[O,Q.
WhenM=O,wehaveMP=l,MO=m,sothatMP⊥MOand then by Pythago-
ras’ theorem

|O,P|^2 =|O,M|^2 +|M,P|^2 =|O,M|^2 +[k^2 −|O,M|^2 ]=k^2 ;

thus again|O,P|=k,sothatPis on the circle. By 2.1.3 any pointX=Pof the
half-line[M,P must satisfy eitherX∈[M,P]orP∈[M,X].IfX∈[M,P],thenby
3.1.2|M,X|<|M,P|;whenX=M, clearlyXan interior point; whenX=M,by
Pythagoras’ theorem this gives

|O,X|^2 =|O,M|^2 +|M,X|^2 <|O,M|^2 +|M,P|^2 =k^2 ,

so that|O,X|<kandsoagainXis interior to the circle. If on the other handP∈
[M,X], while stillX=P,thenby3.1.2|M,X|>|M,P|; by Pythagoras’ theorem we
have
|O,X|^2 =|O,M|^2 +|M,X|^2 >|O,M|^2 +|M,P|^2 =k^2 ,

so that|O,X|>kand soXis exterior to the circle. Thus the points of[M,P]{P}are
interior to the circle, and the points of([M,P)[M,P]are exterior to the circle.