EMAT 6690

Ken Montgomery

The Three Tangent Circles Problem

Problem Statement: Given two line segments, construct externally tangent circles with these segments and construct a common external tangent line. Then construct a circle tangent to the two given circles having the same external tangent line. Find the radius, c, of the new circle in terms of the radii a, and b, of the first two circles.

Part 1: Construction of externally tangent circles and tangent line

We begin with arbitrary line segments a, and b and construct the green circle with center B and radius, b. We construct and then we construct the red, dashed circle, with center C, but with radius, a. Notice that the intersection of Circle C and is A, where we construct the red circle A, which is tangent to circle B (Figure 1).

Figure 1: Arbitrary line segments, a and b, with tangent circles A and B

If we consider the segment, to be the hypotenuse of a right triangle, then the length of the base would correspond to the horizontal distance between A and B. The problem now becomes the construction of the vertex F, corresponding to the right angle of . We know that the height of the triangle should be equal to b – a, and that the length of the hypotenuse is equal to b + a. Assume and, then represents the base of the triangle and we have Equation 1 from the Pythagorean theorem.

Equation 1:

Solving Equation 1 foryields Equation 2.

Equation 2:

Multiplying out the squared terms yields Equation 3.

Equation 3:

Distributing the minus sign gives Equation 4.

Equation 4:

Combining like terms provides Equation 5.

Equation 5:

Taking the square root of both sides gives Equation 6.

Equation 6:

Simplification yields Equation 7, which gives the horizontal distance in terms of a and b, between Circles A and B.

Equation 7:

To construct the result of Equation 7, we first find the midpoint ofand construct a circle, with this segment as its diameter. A line, perpendicular toand its intersections with the circle are then constructed. By similar triangles, the orange segment,is a construction of the desired length, (Figure 2).

Figure 2: Construction of , given by the orange segment,

Since the base ofcorresponds to segmentand the height of the triangle should have the value b – a, we construct a circle, with center A and radius congruent to. We also construct a circle, centered at B, but with radius b – a. The intersection, F, of these two circles is the third vertex of our right triangle (Figure 3).

Figure 3: Construction of

At this point, we wish to construct a line, tangent to Circles A and B. Since we know that the line will be tangent to each circle and also a distance, a (the radius of Circle A), from the respective vertices, A and B, we know that the line will be parallel to. To simplify figure four, we hide part of our previous construction and proceed to construct a line, parallel to, through point G (Figure 4).

Figure 4: is tangent to and

Part 2: Derivations of radius, c and distance, as a function of radii a, and b

We will next wish to construct a third circle, externally tangent toandand tangent to. If we now consider the segment, to be the hypotenuse of a right triangle, then the length of the base would correspond to the horizontal distance between B and C. We know that the height of the triangle should be equal to b – c, and that the length of the hypotenuse is equal to b + c. Assume and, then represents the base of the triangle and from the Pythagorean theorem we have Equation 8.

Equation 8:

Solving Equation 8 for, we have Equation 9.

Equation 9:

Squaring the binomial terms yields Equation 10.

Equation 10:

Simplifying Equation 10, provides Equation 11.

Equation 11:

Taking the square root of both sides, gives Equation 12.

Equation 12:

Simplification allows us to obtain Equation 13.

Equation 13:

We have an expression representing the horizontal (x-axis) distance between B and C, but it is in terms of the radius, c, which we still do not know and now seek to determine. Again, using the Pythagorean theorem, we obtain Equation 14.

Equation 14:

Solving Equation 14 for, we obtain Equation 15.

Equation 15:

Squaring the binomial terms yields Equation 16.

Equation 16:

Distributing the minus sign and simplifying gives Equation 17.

Equation 17:

Taking the square root of both sides, gives Equation 18.

Equation 18:

Further simplifying and solving Equation 18 for yields Equation 19.

Equation 19:

We restate a previous results: Equation 7 here as Equation 20 and Equation 13 as Equation 21.

Equation 20:

Equation 21:

Substitution of Equations 20 and 21 into Equation 19 results in Equation 22.

Equation 22:

Division of both sides by 2 yields Equation 23.

Equation 23:

Rewriting the left side of the equation, we have Equation 24.

Equation 24:

Factoring the left side, we obtain Equation 25.

Equation 25:

Solving Equation 25 for, provides Equation 26.

Equation 26:

Taking the square root of both sides, we have Equation 27.

Equation 27:

Substitution of this result into Equation 13 yields the expression for, in Equation 28, below.

Equation 28:

Simplifying this equation yields Equation 29.

Equation 29:

Now that we have obtained expressions, in terms of a and b, for (the horizontal distance between B and C), and c, the radius of , we return to our problem of constructing a third circle, tangent toand externally tangent toand .

Part 3: Construction of a third circle externally tangent to circles A, B, and

In our construction, we will make use of the relations provided in Equation 13 and Equation 27. Since Equation 13 depends upon the value of c, we begin with a construction of this value. Equation 27 is restated below as Equation 30.

Equation 30:

Squaring the binomial in the denominator and applying the commutative property of addition, we obtain Equation 31.

Equation 31:

Dividing both sides of the equation by a, yields the ratio given in Equation 32.

Equation 32:

Each of the terms in this ratio is easily constructible, via similar triangles. First, we construct, with radius a + b. Then we construct, with length, (Figure 5).

Figure 5: Construction of

We next hideand construct, with radius. We wish to construct an isosceles triangle, with base, b. So we next construct, with radius, b. We then construct the intersection ofandat J (Figure 6).

Figure 6: Construction of

We next hideandand construct another circle with center, G and radius, a. We construct and connect the intersections of this circle and sidesand, to obtain, which, by Equation 32 and similar triangles, has length, c (Figure 7).

Figure 7: Construction of, with length equal to c

We next make use of Equation 13, restated below in Equation 33.

Equation 33:

This is a relatively simple construction that we have already done for , previously. We construct a right triangle, inscribed in a circle, with hypotenuse equal to b + c. Since the height of the right triangle (with the hypotenuse as the base) is equal to, by similar, right triangles, we construct the intersections of the circle and the line perpendicular at the point joining the lengths, b and c (Figure 8).

Figure 8: Construction of

In Figure 8, we have constructed the circle with center, K and radius, b. We then constructed and, which has length, b + c. We then construct the midpoint, P, of and. A line, perpendicular tois constructed at K, the point separating the segments equal to b and c, respectively. Then, we construct, which has length equal to . Next, we hide some of this construction and construct a circle, with center, G and radius(Figure 9).

Figure 9: Construction of, with radius equal to

In Figure 9, we also constructed a line, perpendicular toat Q. We next (Figure 10), construct a circle, with center, Q and radius equal to c (congruent to) and its intersection (R) with the perpendicular line.

Figure 10: Construction of, with radius, c

At this stage, we hide unnecessary parts of the construction and construct the third, externally tangent circle (), with radius, c, which is a horizontal distance,from the point, B (Figure 11).