# The lines of intersection and metallic numbers

We are looking at an infinite family of circles. The first circle is centred on the point and has radius . Every other circle is centred on the South pole of the previous circle and has radius times that of the previous circle. (In the first example in the main article we have , but we give a sketch of the proof for general .) The family of circles with the two lines of intersection. The y-intercept is the silver ratio.

We want to show that all the intersection points that lie on the same side of the -axis lie on a straight line, as shown in the diagram above.

We will also show that the two lines of intersection meet the -axis in the point If we have as claimed in the main article.

First notice that the radius of the circle is . This is because the first circle has radius and any other circle has radius times the radius of the previous circle.

Also notice that for the circle has centre That’s because the centre of the first circle is and the -coordinate of the centre of any other circle is the -coordinate of the centre of the previous circle minus the radius of the previous circle. (The first circle has centre .) The centres and radii of the first three circles.

Our proof rests on the fact that there is quite a simple transformation of the plane which sends each circle in the family to the next one along. This transformation, call it , consists of a scaling by a factor of , followed by a translation by 1 in the direction of the negative -axis. In other words, sends a point with coordinates to the point with coordinates We leave the following two facts for you to prove:

1. The transformation sends a circle with centre and radius to circle with centre and radius 2. The transformation sends a straight line with equation to the straight line with equation .

By point (1), the circle in our family gets sent by to a circle with radius and centre And that’s exactly the circle in our family, as required.

To show that the intersection points of circles lie on a straight line as claimed, let’s first think about the line segment which connects the intersection points and as shown below. Points of intersection and the line segments connecting them.

By point (2) above, sends to another straight line segment (rather than some curvy shape). Because sends each circle to the next one along, that line segment, call it is the one connecting the intersection points and Point (2) also says that and have the same slope — and from this it follows that , and all lie on a straight line. You can apply the same argument to other line segments connecting consecutive intersection points, to show that all intersection points lie on the same straight line, call it .

The argument above also shows that sends to itself. This implies that Rearranging gives as required.

Back to the main article.