7. APOLLONIUS CIRCLE


Apollonius of Perga (ca. 262 BC – ca. 190 BC) was a Greek geometer and astronomer, of the Alexandrian school, noted for his writings on conic sections. It was Apollonius who gave the ellipse, the parabola, and the hyperbola the names by which we know them. The Apollonius crater on the moon was named in his honor.

Exploring

Let A , B be two fixed points on a line. Our problem is to find the locus of a point P which moves so that the ratio of its distances from A and B is constant.

     (a) What is the locus if = 1?

           Check your answer ...

     (b)  Let’s now change the diagram. Now think about the case = 2.            Try to plot a few points of this locus.
           What do you think the locus might be?

           Check your answer ...   
                                                                                                                                                                                                                                                                                                                       


































Harmonic Set

To fully understand the Apollonius circle, we return briefly to cross ratios.

If you have been doing your Extensions exercises (!) you will have
discovered that four points on a line determine 24 cross ratios, generally with six distinct values:
m, 1 – m, 1/m, 1/(1 – m), (m–1)/m, m/(m – 1). When m = –1, this reduces to three values: –1, 2 and 1/2. You can check that when the cross ratio (A, B; C, D) = – 1, it remains constant when a single pair (A, B) or (C, D) is interchanged (the Single Interchange Property). We see that (A, B; C, D) = –1 is equivalent to (*). This means that points C and D divide segment AB internally and externally in the same ratio, and we say that C and D separate A and B harmonically. Points C and D are called harmonic conjugates.


For later reference we prove:

Lemma 7.1    Let O be the midpoint of AB. Then (A, B; C, D) = –1      .

Proof   Starting with (*):      

, (noting that )

(expanding and simplifying, and noting that ).
                                                                                                                                                                                   


































The Circle

In the diagram, is assumed constant, and we wish to show that the locus of P is the indicated circle. Let the internal and external bisectors of APB meet the line AB in points X and Y. Since the bisectors of the vertical angle of a triangle divide the base in the ratio of the sides, we obtain

                  

Now the ratio AP : PB is given, so points X and Y are fixed. We also know that the angle between two bisectors of a given angle is 90°, and so XPY is a right angle. If follows that the locus of P is a fixed circle on XY as diameter.

Finally we observe that , remembering to incorporate the signs. Hence X, Y separate A, B harmonically.

We next work towards finding a converse of this result, but to do this we need another results about the four collinear points A, B, X and Y.

                                 















































Harmonic Pencil

          (D, B; Z, )  =  (A, B; X, Y)  =  –1 (our given assumption).

But (D, B; Z, )  =   =  –1     DZ = ZB.
Since PZ PY and PY // DB, so PZ DB. Now triangles PZD, PZB are congruent (S.A.S.), and PZ is an angle bisector of APB. We deduce immediately that PY is the other angle bisector. This completes the proof of the lemma.

We have defined a collinear set of points A, B; C, D to be harmonic if (A, B; C, D) = –1. We define a harmonic pencil to be a pencil for which any transversal cuts the lines of the pencil in a harmonic set.

Lemma 7.2 If pencil P(A, B, X, Y) is harmonic, and XPY = 90°, then PX, PY are angle bisectors of APB.

Referring to the adjacent diagram, we draw line DZB parallel to YP. We are going to follow a useful convention and define a ‘point at infinity’, , on the line DZB. We think of this point as being where DZB ‘meets’ the parallel line PY. To justify this we might think of line DZB being slightly angled to PY with the lines intersecting in point U, and then think of as being a limiting case as we move towards the parallel position.

The advantage of this convention is that we can now regard the line
DZB as a genuine transversal of the pencil P(A,B, X,Y). It follows that

                                   
A lemma is a ‘helping theorem’. In German the word is Hilf-satz.
































A Converse Result

Suppose now that A, B are two fixed points determining line AB, and that X, Y separate them harmonically. Then if P is any point on the circle with diameter XY, the ratio PA/PB is constant (in fact = XA/XB = YA/YB).

Proof  We are given that the pencil P(A, B, X, Y) is harmonic. Also, P lies on the circle on XY as diameter, so PX PY. Therefore, by Lemma 7.2, PX and PY are angle bisectors of APB.

It follows that for all positions of P

              '

as required

•••••••••••••••••••••••••••••••••••••••••••••••••••••
Apollonius wrote about another circle problem: Given three non-intersecting circles in the plane, is it possible to draw a fourth circle which is tangent to them all, in how many ways, and how are such a circles constructed?
                                                                                                                  






























P

Extensions

1. If A, B are two points on a line, then the midpoint of segment AB is the harmonic conjugate of ‘the point at infinity’ of the line AB.

2. Two circles meet in U and V. A common tangent to the circles touches them at A and B respectively. Another circle is drawn through U and V, cutting the common tangent in points C and D. Show that (A, B; C, D) = – 1.

3. Prove that if (A, B; C, D) = (A, B; D, C), then the set of four points is harmonic, with the common value of the cross ratios equal –1.

4. Let A, B be two fixed points. By plotting some points, make a conjecture about the locus of a point P for which PA + PB is constant.

5. Given two intersecting circles, why do there not exist two points A and B such that each circle is a circle of Apollonius with respect to these points?
                                                                                 Hints and Solutions ...

For looking up ...

http://www.jimloy.com/cindy/apoll.htm

Geometry for Advanced Pupils, Maxwell, E. A. (Oxford 1963)
                                                                                                                                           
                                                                                          
































P

Hints and Solutions

1. If A, B are two points on a line, then the midpoint of segment AB is the harmonic conjugate of ‘the point at infinity’ of the line AB.

• Without loss of generality, take A(0), B(2), C(1) and D(). Then treating like an ordinary number,
(A, B; C, D) = (0, 2; 1, ) = Alternatively you might think of as approximated by a large number L, and then take the limit.

2. Two circles meet in U and V. A common tangent to the circles touches them at A and B respectively. Another circle is drawn through U and V, cutting the common tangent in points C and D. Show that (A, B; C, D) = – 1.

• Let UV meetAB in O. Then OA2 = OB2 = OC.OD, and we can apply Lemma 7.1.

3. Prove that if (A, B; C, D) = (A, B; D, C), then the set of four points is harmonic, with the common value of the cross ratios equal –1.

• The cross ratios have values m and 1/m for some value of m. Equating gives m = 1, and m  1 if the four points are distinct.

4. Let A, B be two fixed points. By plotting some points, make a conjecture about the locus of a point P for which PA + PB is constant.

• The locus of P in this case is an ellipse. You might also investigate the case when PA – PB is constant.

5. Given two intersecting circles, why do there not exist two points A and B such that each circle is a circle of Apollonius with respect to these points?

• Let M be midpoint of chord AB, and consider the circle described by P with AP/BP = k. If k > 1, the circle cuts AB on the B side of M. If k < 1, the circle cuts AB on the A side of M. Hence the proposed circles cannot intersect.