Tangent Circles Problem

Here is the problem: Given three lengths, construct a set of three circles that are mutually tangent and have those three lengths as their radii. In the figure above you can adjust the three lengths by moving the points labeled A₂, B₂ and C₂ at the top of the diagram. As you do so, the three circles change size appropriately to maintain their tangency.

The Solution

Thus the solution is pretty easy. First, we need to construct line segments that have the lengths d₁ = a+b, d₂ = b+c and d₃ = c+a. Then we need to construct a triangle having these three segments as lengths. The vertices of this triangle will serve as centers for the three circles.

It is easy to construct a line segment that is the sum of two segments—just draw a line and from a single point, draw circles having radii equal to the two lengths. The points on opposite sides of the single point where they intersect the line will be separated by a length that is the sum of the two lengths.

In this way the three lengths of the desired triangle could be constructed. Once we know the lengths d₁, d₂ and d₃, it is easy to construct the triangle. Draw circles of radii d₂ and d₃ about the two ends of a segment of length d₁. The point where those circles intersect is d₂ from one end and d₃ from the other, so we are done.

Being lazy, we would like to minimize the number of times we have to copy lengths, so here is a reasonable way to do the construction. Assume that you have the three lengths a, b and c laid out as in the figure above. Draw a new line with a point "V" on it. Using V as the center, draw circles of radius a and b. (This can be done with the Ctr PP=>C command. This command first takes the center of a circle and then uses the distance between the two points you select as the radius of the circle. In Geometer you simply need to select V as the center and then the endpoints A₁ and A₂ of the length-defining segment as your radius to obtain the first circle, for example.)

On one side of V find the intersection of one of the circles and on the other, the intersection with the other. The distance between these two intersection points will be a+b. See the figure above.

Now we have identified A and B on our triangle, and they are clearly separated by a+b. Now draw a third circle about V of radius c. It should be clear that the distance from B to the intersection of that line with the circle on the other side of V from B is b+c and that the distance from A to intersection of that circle on the other side of V from A is a+c. Thus we can draw two circles centered at A and B having edge points at the two intersections with the line of the circle of radius c centered at V. The intersection of these circles yields the location of the point C.

To complete the diagram, draw circles of radius a, b, and c about points A, B and C, respectively. To clear up the clutter, you probably will want to change the color of all your auxiliary lines to the invisible color.

At this point the problem is solved, but suppose you would like to convert your diagram to a beautiful one that would be more suitable for a classroom presentation. Load the diagram Demos/TangentCircs.T and play with it as you read these notes. Here are some things that you might do:

• You may want to change the names of the points on the controlling segments to A₁, A₂, ..., C₂ to make it clear to the viewer that the segment A₁A₂ controls the length of the radius of the circle centered at A. You can also name the segment A₁A₂ a, et cetera, to make the correspondence even more clear.
• If A₁ and A₂ are free points, as you drag them around, the figure will certainly work, but it is much more pleasing if you can just drag A₂ and it moves on a line parallel to B₁B₂ and C₁C₂. To do this in the diagram, A₁, B₁ and C₁ are pinned points. Pinned points cannot be moved. To create a pinned point, look in the Primitives pulldown menu under Points.

  Next, another point is chosen (pinned and invisible) that determines the three horizontal lines from A₁, B₁ and C₁. The points A₂, B₂ and C₂ are constrained to lie on those lines. Then visible line segments are drawn from A₁ to A₂, et cetera.

• Color-coordinate the drawing. All the lengths and points associated with A are one color, with B another, and so on.
• To make it completely clear that the radii of the final circles are equal to a, b and c, we need to draw those as well. AV and BV can be drawn easily, but we need to find the intersections of the lines AC with the circles and BC with the circles. Draw those lines, find the intersections using LC=>P, make the lines invisible, and then draw the shorter segments to those intersections in the appropriate colors.
• Add some text to the diagram to explain what is going on with the Text command under the Primitives pulldown menu.
• Finally, to assure that the lines A₁A₂, B₁B₂ and C₁C₂ are parallel and evenly spaced, use the text editor to modify the coordinates of the pinned points so that the lines are perfectly horizontal and evenly spaced.

Here is a complete listing of the Geometer code to produce the diagram:

.geometry "version 0.60";
v1 = .pinned(-0.78, 0.85, .red, "A\sub{1}");
v2 = .pinned(-0.78, 0.7, .green, "B\sub{1}");
v3 = .pinned(-0.78, 0.55, .cyan, "C\sub{1}");
v4 = .pinned(0.78, 0.85, .in);
v5 = .pinned(0.78, 0.7, .in);
v6 = .pinned(0.78, 0.55, .in);
l1 = .l.vv(v1, v4, .in);
l2 = .l.vv(v2, v5, .in);
l3 = .l.vv(v3, v6, .in);
v7 = .vonl(l1, -0.452096, 0.85, .red, "A\sub{2}");
v8 = .vonl(l2, -0.598802, 0.7, .green, "B\sub{2}");
v9 = .vonl(l3, -0.38024, 0.55, .cyan, "C\sub{2}");
l4 = .l.vv(v1, v7, .red);
l5 = .l.vv(v2, v8, .green);
l6 = .l.vv(v3, v9, .cyan);
v10 = .free(-0.517964, -0.0898204, .in);
v11 = .free(-0.0269461, -0.0898204, .in, "V");
l7 = .l.vv(v10, v11, .in, .longline);
c1 = .c.ctrvv(v11, v1, v7, .in);
c2 = .c.ctrvv(v11, v2, v8, .in);
v12 = .v.lc(l7, c1, 1, .red, "A");
v13 = .v.lc(l7, c2, 2, .green, "B");
c3 = .c.ctrvv(v11, v3, v9, .in);
v14 = .v.lc(l7, c3, 1, .in);
v15 = .v.lc(l7, c3, 2, .in);
c4 = .c.vv(v12, v15, .in);
c5 = .c.vv(v13, v14, .in);
v16 = .v.cc(c4, c5, 2, .cyan, "C");
c6 = .c.ctrvv(v16, v3, v9, .cyan);
c8 = .c.ctrvv(v13, v2, v8, .green);
c9 = .c.ctrvv(v12, v1, v7, .red);
l8 = .l.vv(v12, v16, .in);
l9 = .l.vv(v16, v13, .in);
v17 = .v.lc(l8, c6, 1, .in);
v18 = .v.lc(l9, c6, 2, .in);
l10 = .l.vv(v12, v11, .red);
l11 = .l.vv(v12, v17, .red);
l12 = .l.vv(v17, v16, .cyan);
l13 = .l.vv(v16, v18, .cyan);
l14 = .l.vv(v18, v13, .green);
l15 = .l.vv(v13, v11, .green);
.text("Given three lengths, construct three mutually
tangent circles having those three lengths
as their radii.");

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