10. THE RADICAL AXIS

Exploring  

We have already briefly introduced the idea of the radical axis of two intersecting circles: it is the locus of points which have the same power with respect to the two circles. So in this diagram, we have

                    PQ.PR = PA.PB = PQ'.PR'

for all points on the axis.

For the original reference, see here ...

But what happens when the two circles do not intersect? Does a radical exis exist in this case? And if so, how can we locate it?

You might like to look at the different possible positions of the circles.

                 Case 1              Case 2                Case 3

                                                                                                                                                                                                                                                                                                    


































A Harmonic Problem

Suppose A, B, C, D are four points in any order on a straight line. We claim:

Lemma 10:1 There exist on the line two points X, Y which harmonically separate each of the pairs A, B and C, D.

To establish this we rather cheekily assume the existence of the points X, Y and the midpoint M of XY. We shall give each point the obvious coordinate with respect to some choice of origin O, and find x, y and m in terms of a, b, c and d.

We want (*). In terms of coordinates, the final equation of (*) becomes

(a – m)(b – m) = (c – m)(d – m)    ab – m(a + b) + m2 = cd – m(c + d) + m2    m(a + b + c + d) = ab – cd.

This gives m = (ab – cd) / (a + b + c + d).

So given a, b, c, d, we can calculate m, establishing the existence of a unique point M.
If we take M to be the origin, m = 0, and  ab = cd. From (*) we now get x2 = y2 = ab = cd, enabling us to calculate x and y. This places the points X, Y distant &Mac195;(ab) = &Mac195;(cd) on either side of M. Thus X and Y exist and this completes the proof.

Comments: Since ab may be negative, it is possible that for some arrangements of the points, x and y may be ‘imaginary’. Also, in calculating m there is the possibility that a + b + c + d = 0. You might investigate the placing of the points in this case.

                                                                                                                                                                                


































Radical Axis

Now suppose our two non-intersecting circles with centres O, Q meet the common diameter OQ in points A, B; C, D as in the diagram. As in Lemma 10.1, construct points X, Y which are harmonic conjugates to each of the pairs A, B; C, D. We prove

Theorem 10.1 The locus of a point P which has equal powers with respect to two non-intersecting circles is the perpendicular bisector of chord XY – the radical axis.

Proof Let P have equal powers with respect to the two circles. We must show that for all such points P, PX = PY. We do this by showing that PX = t where t is the common tangent length from P to the circles.

Let U, V be inverses of P with respect to the circles as shown. Then , so U, V, Q, O are concyclic.
Similarly, P, U, and X, Y are inverse pairs with respect to the circle centre O P, U, X, Y concyclic, and
                P, V, and X, Y are inverse pairs with respect to the circle centre Q P, V, X, Y concyclic.
We deduce that points P, U, V, X, Y are concyclic.

Now PXU = PVU (angles subtended by PU)
                      = POX (OQUV is a cyclic quadrangle with supplementary opposite angles).
Hence  PXU ~  POX, and so PU / PX = PX / PO from which .
Thus PX = t as required.
                                                                                                                                                                               






















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Summing Up

In the case of the intersecting circles, we looked at the common chord. We notice that for points lying on the portions of the common chord which lie outside both circles, the tangents to the circles have equal length. The radical axis is thus a sort of extension of the common chord for the case where the circles do not intersect.

In fact, the existence of the tangent is really quite irrelevant to the argument: the critical quantity is actually the power of P with respect to the two circles, . It just happens that when this value is non-negative, it equates to the t2 we have used in our argument.

It is common to use the term radical axis to cover all these cases.

The argument we used previously about the common chords of three intersecting circles meeting in a common point carries staraight over here, giving:

The radical axes of three circles taken in pairs are concurrent. They meet in a point called the radical cemtre of the three circles.

The points X, Y used in our argument above are called the limiting points of the two circles. We shall see later that there is good reason for this choice of name.
                                                                                                                                                      
                                                                                          





























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Extensions

1.  If two circles have two tangents in common, prove that these tangents are bisected by the radical axis.

2. If two circles have no common tangent (one circle completely contained in the other) how can we construct the radical axis?

3. Two circles are orthogonal if the tangent to each at a common point is a radius of the other. (This means that they intersect at right angles.) Two circles have limiting points X and Y. Show that every circle passing through X and Y cuts each of the two given circles orthogonally. [This is not hard, but if you have problems, we establish it in the next chapter.]

4. Two circles have limiting points X and Y. A common tangent meets the circles at P and Q. Prove that the circle with diameter PQ passes through X and Y.


                                            Hints and Solutions ...

For looking up ...

http://en.wikipedia.org/wiki/Radical_axis

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





























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Hints and Solutions

1.  If two circles have two tangents in common, prove that these tangents are bisected by the radical axis.

• Let a common tangent meet the circles in A, B. Let the circles meet in U, V, and let radical axis UV meet the common tangent in W. Then WA2 = WU.WV = WB2, so WA = WB as required.

2. If two circle have no common tangent (one circle completely contained in the other) how can we construct the radical axis?

• Draw a circle intercepting the two given circles. This gives two real radical axes meeting in a point P. Repeat this with another intercepting circle to obtain point Q. Then PQ is the required radical axis.

3. Two circles are orthogonal if the tangent to each at a common point is a radius of the other. (This means that they intersect at right angles.) Two circles have limiting points X and Y. Show that every circle passing through X and Y cuts each of the two given circles orthogonally. [This is not hard, but if you have problems, we establish it in the next chapter.]

• See next chapter.

4. Two circles have limiting points X and Y. A common tangent meets the circles at P and Q. Prove that the circle with diameter PQ passes through X and Y.

• The circle with diameter PQ cuts PQ orthogonally at P, Q, and hence is orthogonal to the two given circles. Hence it passes through the limiting points X, Y.