108 Circles; their basic properties Ch. 7
With the notation of the last result, let Q≡F( 1 , 0 )and sl(Q)=P 3 where P 3 ≡F
(a 3 ,b 3 ).Then
a 3 =(a 1 +a 2 )^2 −(b 1 +b 2 )^2
(a 1 +a 2 )^2 +(b 1 +b 2 )^2,b 3 =2 (a 1 +a 2 )(b 1 +b 2 )
(a 1 +a 2 )^2 +(b 1 +b 2 )^2,
when P 1 and P 2 are not diametrically opposite, and
a 3 =b^21 −a^21 ,b 3 =− 2 a 1 b 1 ,when they are.
Proof.Forl≡ax+by+c= 0 ,we recall from 6.6.1 that
sl(Z 0 )≡(
x 0 −2 a
a^2 +b^2(ax 0 +by 0 +c),y 0 −2 b
a^2 +b^2(ax 0 +by 0 +c))
WhenP 1 andP 2 are not diametrically opposite,l≡(b 1 +b 2 )x−(a 1 +a 2 )y= 0 .Thus
for itx 0 = 1 ,y 0 = 0 ,a=b 1 +b 2 ,b=−(a 1 +a 2 ),c=0andso
a 3 =(a 1 +a 2 )^2 −(b 1 +b 2 )^2
(a 1 +a 2 )^2 +(b 1 +b 2 )^2,b 3 =2 (a 1 +a 2 )(b 1 +b 2 )
(a 1 +a 2 )^2 +(b 1 +b 2 )^2WhenP 1 andP 2 are diametrically opposite,l≡a 1 x+b 1 y=0. Thus for itx 0 =
1 ,y 0 = 0 ,a=a 1 ,b=b 1 ,c=0, so we have
a 3 =b^21 −a^21 ,b 3 =− 2 a 1 b 1 ,asa^21 +b^21 = 1.
7.4 Polar properties of a circle .......................
7.4.1 Tangentsfromanexteriorpoint ....................
Let P be a point exterior to a circleC. Then two tangents to the circle pass through
P. Their points of contact are equidistant from P.
Proof. Let the circle have cen-
treOand length of radiusa.Let
|O,P|=b,sothatb>a. Choose
the pointU∈[O,P so thatx=
|O,U|=a^2 /b.Asb>a,then
x<a<bsoU∈[O,P]. Erect
a perpendicular toOPatUand
mark off on it a distance
y=|U,T 1 |=a√
1 −
(a
b) 2
.
OP
U V
T 1
T 2
Figure 7.4.