Activity 1-7: The Overlapping Circles

www.carom-maths.co.uk

Draw three overlapping circles like so in Geogebra:

Add the straight lines AB, CD and EF, and then drag...

Task: can you PROVE that the three lines always meet?

The equation of the green circle AEB we can write as

x2 + y2 + a1x + b1y + c1= 0, or C1 = 0.

What happens if we subtract these two equations?

The equation of the blue circle ADB we can write as

x2 + y2 + a2x + b2y + c2 = 0, or C2 = 0.

We get (a1 - a2)x + (b1 - b2)y + (c1 - c2) = 0.

So we have a straight line – but which line is it?

A satisfies both C1 = 0 and C2 = 0, so must be on C1 - C2 = 0.

B satisfies both C1 = 0 and C2 = 0, so must be on C1 - C2 = 0.

So the line given by C1 - C2 = 0 must be the line AB.

By an exactly similar argument, the line given by C2 – C3 = 0 must be the line CD,

where C3 = 0 is the equation of the red circle CED,

x2 + y2 + a3x + b3y + c3 = 0.

Now think about the point where AB and CD meet.

Thus we have that the lines AB, CD and EF are concurrent.

This is on C1 – C2 = 0, and also on C2 – C3 = 0.

So this point must be on the sum of these equations,

which is C1– C3 = 0.

But what is this, but the equation of the line EF?

A new question now: what happens if we ADD the equations of our circles?

Clearly C1 + C2 + C3 = 0 is the equation of a circle.

Try the Autograph file below that shows what happens when you add three circles like this.

Task – What do you notice about the sum-circle?

can you prove this?

Take the green, blue and red circles. The purple circle is their sum – the ‘sum-circle’.

http://www.s253053503.websitehome.co.uk/carom/carom-files/carom-1-7-3.agg

Three Circles Autograph File

Point A is inside the green circle C1 < 0 for point A.

Point B is inside the blue circle C2 < 0 for point B.

So point D is inside the blue, red and green circles C1 + C2 + C3 < 0 for point D

point D is inside the sum-circle.

The purple circle, the sum-circle, seems to always enclose the intersection

of the red, green and blue circles.

Point C is inside the red circle C3 < 0 for point C.

Or else you might like to explore Friendly Circles...

The file below shows what happens as we vary a, b and c.

Friendly Circle Autograph file

http://www.s253053503.websitehome.co.uk/carom/carom-files/carom-1-7-4.agg

So we only need to consider the cases where a, b and c are all positive,

or where exactly one of a, b and c is negative.

Proof: clearly the theorem is true for (a, b, c) just if it is true for ( a, b, c).

Conjecture:some pair of these three circles always overlap,

even if only in a single point.In other words, the three circles are never disjoint.

Or we might say, the circles are Friendly.

Similarly we have a2 > ab + bc + ca, and c2 > ab + bc + ca.

Suppose a, b and c are all positive, and suppose the theorem is false.

So √((a b)2 + (b c)2) > c + a, and squaring gives b2 > ab + bc + ca.

Multiplying these gives (abc)2 > (ab + bc + ca)3, which is clearly absurd.

Suppose instead that just a is negative, with a = d. Then √(( d b)2 + (b c)2) > c + d,

which gives b2 + bd bc > cd.

Completing the square we have

Now taking the square root,

Similarly c > b, which is a contradiction. So the three circles are always friendly.

Carom is written by Jonny Griffiths, hello@jonny-griffiths.net

With thanks to:David Sharpe of Mathematical Spectrum

for publishing my article on Sumlines,and Shaun Stevens.