8. POLE AND POLAR

Inverse Points  

We are going to look at some special related points associated with the circle. You might remember we have already looked at the power of a point with respect to a circle. Revise this.

Now suppose we have a circle centre O and radius a. We say that points P, Q are inverse points with respect to the circle if they lie on the same radius and if .

We observe that because the product is positive, P and Q must lie on the same side of O. So if P lies at A then so does Q, and similarly for B. Otherwise, one point must lie inside the circle and the other outside.

Obviously , so by Lemma 7.1, points P, Q separate A and B harmonically.

Suppose we now consider a pencil of lines through P. On each such line intercepting the circle, there will be a point which is the harmonic conjugate of P with respect to the circle intercepts. We seek the locus of such harmonic conjugates.
                                                                                                                                                                                                                                                                                                    


































Pole and Polar

Let P and Q be inverse points with respect to a given circle. We draw through Q a line p perpendicular to line PQ. Then p is called the polar of the point P. The point P is called the pole of the line p.

We notice that this definition of polar is valid whether P lies inside or outside the circle.   

If P lies outside the circle, the polar p of P is the line joining the two points of contact T, T' of the tangents from P to the circle.

Proof   Let TT' meet AB in Q. We show that Q is in fact the inverse of P. Since PT is a tangent and TT' AB, we have right angles at T and Q. As well, triangles OTQ, OPT have a common angle P, and so are similar. Hence
OT / OP = OQ / OT. So   , and Q is the inverse of P as required.


If P lies on the circle, the definition tells us that the polar p of P is the tangent at P.

                                                                                                                                                                                



































Polar Locus

Theorem 8.1 Any line through P cutting the circle in points U, V meets the polar p of P in the point R which is the harmonic conjugate of P with respect to points U, V.

Proof   To show that (P, R; U, V) = –1, it will be sufficient to show that QP and QR are angle bisectors of UQV. That is, we could show QU / QV = UR / RV, or (externally), QU / QV = UP / PV. (*)

Now, OP.OQ = a2 = OU2 (since P, Q are inverse points), so
OU / OP = OQ / OU. It follows that triangles UOP, QOU are similar, since they have a common angle at O.

Hence UP / UO = QU / QO (#). We can argue similarly with triangles VOP, VOQ to obtain VP / VO = QV / QO (‡).

Noting that UO = VO, dividing equations (#) and (‡) gives our required result (*).
Thus p is an angle bisector of UQV, and hence so is the perpendicular line OP. It follows that (P, R; U, V) = –1. That is, the harmonic conjugate R of P lies on the polar p of P.

                                 















































Reciprocal Property

We now establish the interesting

Reciprocal Property If P, R are points for which the polar of P passes through R, then the polar of R passes through P. Such points are said to be conjugate with respect to the circle.

You might think we have already established this with Theorem 8.1 above, and this is true in the case where the line PR intersects the circle. But what if PR does not intersect the circle?

In the diagram, let S be the inverse of R. Then .
It follows that points P, Q, R, S are concyclic.

Since PQR is a right angle, this circle has diameter PR. Hence PSR is a right angle.

In other words, PS is the line through S perpendicular to line ORS – the polar of R.

It follows that P lies on the polar r of R.

                                   




































P

Extensions

1. Show that if the polar of P with respect to a circle meets the polar of Q in R, then the polar of R is PQ.

2. Tangents at points P, Q of a circle meet in point R. What can you say about the polar of R?

3. Two circles meet in points U and V. Show that the poles of UV with respect to these two circles lie on the line of centres of the circles.

4. Chords PQ, RS of a circle meet in point O. Tangents to P and Q meet in U, and tangents to R and S meet in V.
   (a) show that UV is the polar of O. (b) Show that PR meets QS on UV.

5. Chords PQ, RS of a circle meet in point O. PR meets QS in L, PS meets QR in M. Show that LM is the polar of O.

                                                                          Hints and Solutions ...
For looking up ...

http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml
 
Geometry for Advanced Pupils, Maxwell, E. A. (Oxford 1963)
                                                                                                                                          
                                                                                          






































P

Hints and Solutions

1. Show that if the polar of P with respect to a circle meets the polar of Q in R, then the polar of R is PQ.

• By the Reciprocal Property, R on p Pon r, and R on q Q on r. That is r = PQ.

2. Tangents at points P, Q of a circle meet in point R. What can you say about the polar of R?

• Polar r of R is PQ. For tangent at P is polar p, tangent at Q is polar q. R on p, q r = PQ.

3. Two circles meet in points U and V. Show that the poles of UV with respect to these two circles lie on the line of centres of the circles.

• Polar of U is the tangent u at U, polar of V is the tangent v at V. Hence the pole of UV is the intersection point of these tangents which lies on the line of centres,

4. Chords PQ, RS of a circle meet in point O. Tangents to P and Q meet in U, and tangents to R and S meet in V.
   (a) show that UV is the polar of O. (b) Show that PR meets QS on UV.

• (a) O lies on polars of U, V U, V lie on polar of O. Thus PQ is polar of O. (b) Let PQ.UV = J, RS.UV = K. Then (P, Q; J, O) = (R, S; K, O) = –1, showing that PR and QS meet on JK = UV.

5. Chords PQ, RS of a circle meet in point O. PR meets QS in L, PS meets QR in M. Show that LM is the polar of O.

• Let LM meet PQ, RS in J, K. Then (P, Q; J, O) = (R, S; K, O) (from M) = (Q, P; J, O) (from L) = –1 (single interchange) and the result follows.