Geometry with Trigonometry

Sec. 7.4 Polar properties of a circle 109

By Pythagoras’ theorem,

|O,T 1 |^2 =|O,U|^2 +|U,T 1 |^2 =x^2 +y^2 =

a^4

b^2

+a^2 −

a^4

b^2

=a^2 ,

so thatT 1 ∈C.
LetVbe the mid-point ofOandP,sothatV∈[O,P]and|O,V|=b 2 .Then

|U,V|=±(|O,V|−|O,U|)=±(^12 b−x).

Again by Pythagoras’ theorem,

|V,T 1 |^2 =|U,V|^2 +|U,T 1 |^2 =(^12 b−x)^2 +y^2

=

(

1

2 b−

a^2

b

) 2

+a^2

(

1 −

a^2

b^2

)

=^14 b^2.

ThusT 1 is on the circleC 1 with centreVand radius length^12 b. Note thatC 1 also
passes throughOandP.Then∠OT 1 Pis an angle in a semi-circle ofC 1 ,sothatby
7.2.1 it is a right-angle. Thus by 7.1.1PT 1 is a tangent toCatT 1.
By a similar argument, if we takeT 2 so thatUis the mid-point ofT 1 andT 2 ,then
PT 2 is also a tangent toCatT 2. We note thatT 1 andT 2 are both on the line which is
perpendicular toOPat the pointU.
By Pythagoras’ theorem

|P,T 1 |^2 =|O,P|^2 −|O,T 1 |^2 =|O,P|^2 −|O,T 2 |^2 =|P,T 2 |^2 ,

and so|P,T 1 |=|P,T 2 |.
There cannot be a third tangentPT 3 as thenT 3 would be onCandC 1 , whereas by
7.2.3 these circles have only two points in common.

7.4.2 Thepowerpropertyofacircle.....................

For a fixed circleC(O;k)and fixed point P∈C(O;k), let a variable line l through P
meetC(O;k)at R and S. Then the product of distances|P,R||P,S|is constant. When
P is exterior to the circle,
|P,R||P,S|=|P,T 1 |^2 ,

where T 1 is the point of contact of a tangent from P to the circle.
Proof. By the distance formulaZ≡(x,y)is onC(O;k)if and only ifx^2 +y^2 −
k^2 =0. IfP≡(x 0 ,y 0 )andlhas Cartesian equationax+by+c=0, by 6.4.1 points
Zonlhave parametric equations of the formx=x 0 +bt,y=y 0 −at(t∈R).Nowl
also has Cartesian equation

a

√

a^2 +b^2

x+

b

√

a^2 +b^2

y+

c

√

a^2 +b^2

= 0.