4. THREE INTERSECTING
CIRCLES

The intersection points of two intersecting circles determine a straight line. What happens if we have three intersecting circles?

Exploring

We need first to examine an interesting property of the circle.

Let P be a point external to a given circle, and let a line through PQR meet the circle in points Q, R. Also, let PT be tangent to the circle at T as in the diagram. Reproduce a copy of this diagram with a larger scale, and for different lines PQR, measure PQ, PR. Tabulate your results as below. What appears to be true?

 
PQ
PR
PT
PT 2
Case 1
Case 2
...
 
 
 
 


Check ...

                                 



































Power of a Point

Let P be a point external to a given circle, and let a line through PQR meet the circle in points Q, R. Also, let PT be tangent to the circle at T as in the diagram. Reproduce a copy of this diagram with a larger scale, and for different lines PQR, measure PQ, PR. Tabulate your results as below. We asked what appears to be true?

The evidence points to the result: PQ.PR = PT 2.

It is fairly easy to prove this by adding a couple of lines to the diagram.

Now the triangles PTQ, PRT are similar, since they have P in common,
and PTQ = PRT.

It follows that the sides of the triangles are proportional, so PT / PQ = PR / PT.
That is, PQ . PR = PT 2. Notice that this result holds for all chords passing through the point P.

The number PT 2 is called the power of the point P with respect to the circle.

You might like to show that if P lies inside the circle, then the quantity PQ . PR remains constant for various chords through P. In this case this common constant number is again the power of the point P with respect to the circle. Thinking of the signed lengths being in opposite directions, this power is assigned a negative number. Of course, in this case there is no PT 2.

                                 

It is easy to see why these two angles must be equal.

Let TD be the diameter at T .
This means that PTD = 90° (angle between tangent and diameter), and TQD = 90° (angle in a semicircle).

Hence PTQ = 90° – DTQ = TDQ

                        = R (and all other angles subtended by                                         the chord QT ).
































Two Intersecting Circles

Consider two circles intersecting in points A, B. The line AB is called the radical axis of the two circles. It has a special property.

   Let P be an arbitrary point on AB.
   What can you say about PQ . PR?
   What can you say about PQ' . PR' ?
   Why are these two quantities equal?

We see that
PQ . PR  = PA . PB (power property of the blue circle)
=  PQ'. PR' (power property of the yellow circle).

So point P has the same power with respect to both circles. Notice that the argument still holds if P lies at an intersection of the circles, or on the line segment AB. Thus

The radical axis of two circles is the locus of points which have the same power with respect to both the circles.

Now what happens if we have three intersecting circles?

                                   




































P

Three Intersecting Circles

In this figure we have three intersecting circles. Taking them in pairs, they give rise to three radical axes. You will notice that the diagram shows them all passing through a common point O.

    Can you explain why this is so?
    There is a very simple argument.

We show

The radical axes of three circles pass through a common point. This point is known as the radical centre of the circles.

Proof To prove this assertion, suppose that axes BB' and CC' meet in O. Does AA' also pass through O?

Since O lies on BB', O has the same power with respect to the green circle and the magenta circle.
Since O lies on CC', O has the same power with respect to the blue circle and the magenta circle.
Hence, O has the same power with respect to the green circle and the blue circle.
Hence O lies on the radical axis AA' of these two circles.

Thus the three radical axes meet in the common point O.

                                 


































Extensions

1. Let C be a circle of radius r and centre O. Sometimes the power of P with respect to C is defined as OP 2 – r 2. How does this compare with the definition we have given? What happens when P lies inside the circle? on the circle?

2. In the diagram, look at OQ . OQ', and OR . OR'. Can you deduce that the points Q, Q', R, R' lie on a circle? Find two similar sets of concyclic points.

3. In the diagram, is it possible for B'AC', C'BA', A'CB' to be straight lines?

4. Show that if the points Q, R, B', C' are concyclic (lie on a circle), then we have B'C' // Q'R'.

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

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

http://mathworld.wolfram.com/RadicalLine.html

                                 






































Hints and Solutions

1. Let C be a circle of radius r and centre O. Sometimes the power of P with respect to C is defined as OP 2 – r 2. How does this compare with the definition we have given? What happens when P lies inside the circle? on the circle?

• It is better in that it remains valid if P lies inside the circle.

2. In the diagram, look at OQ . OQ', and OR . OR'. Can you deduce that the points Q, Q', R, R' lie on a circle? Find two similar sets of concyclic points.

• Both equal OA . OA'. Other such sets are R, R', P, P'; P, P', Q, Q'.

3. In the diagram, is it possible for B'AC', C'BA', A'CB' to be straight lines?

• If A'B'C' is equilateral and O the orthocntre, the circles have the sides of the triangle as diameters.

4. Show that if the points Q, R, B', C' are concyclic (lie on a circle), then we have B'C' // Q'R'.

• We are given OQ.OB' = OR.OC'. Also, OQ.OQ' = OR.OR'. Hence OQ' / OB' = OR' / OC', and so R'Q' // C'B'.