Activity 1-7: The Overlapping Circles

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 x 2 + y 2 + a 1 x + b 1 y + c 1 = 0, or C 1 = 0. What happens if we subtract these two equations? The equation of the blue circle ADB we can write as x 2 + y 2 + a 2 x + b 2 y + c 2 = 0, or C 2 = 0. We get (a 1 - a 2 )x + (b 1 - b 2 )y + (c 1 - c 2 ) = 0. So we have a straight line – but which line is it?

A satisfies both C 1 = 0 and C 2 = 0, so must be on C 1 - C 2 = 0. B satisfies both C 1 = 0 and C 2 = 0, so must be on C 1 - C 2 = 0. So the line given by C 1 - C 2 = 0 must be the line AB. By an exactly similar argument, the line given by C 2 – C 3 = 0 must be the line CD, where C 3 = 0 is the equation of the red circle CED, x 2 + y 2 + a 3 x + b 3 y + c 3 = 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 C 1 – C 2 = 0, and also on C 2 – C 3 = 0. So this point must be on the sum of these equations, which is C 1 – C 3 = 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 C 1 + C 2 + C 3 = 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’. carom/carom-files/carom agg Three Circles Autograph File

Point A is inside the green circle  C 1 < 0 for point A. Point B is inside the blue circle  C 2 < 0 for point B. So point D is inside the blue, red and green circles  C 1 + C 2 + C 3 < 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  C 3 < 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 carom/carom-files/carom 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 a 2 > ab + bc + ca, and c 2 > 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 b 2 > 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 b 2 + 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, With thanks to: David Sharpe of Mathematical Spectrum for publishing my article on Sumlines, and Shaun Stevens.