Osculating Circles and Hyperbola



The circle C0 described by the equation x2+y2=1 and the hyperbola H described by the equation x2-y2=1 are drawn in the plane. Now, a sequence C1, C2, C3, C4 , ..., of circles is drawn with the following property: Each Ck+1 is tangent to and above Ck and tangent to both branches of H. Prove that all of these circles have integral radii.
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By symmetry, the center of each circle is on the y-axis. Hence, the circle Ck is described by the equation x2+(y-ck)2 = rk2, for a suitable choice of ck and rk.

The circle Ck meets the y-axis at two points, (0,Sk-1) and (0,Sk), where
Sk-1 = ck-rk and Sk = ck+rk = ck+1-rk+1.


The circle Ck meets H at the points (xk,yk) and (-xk,yk).

Since (xk,yk) is on Ck, xk2+(yk-ck)2 = rk2.

Since (xk,yk) is on H, xk2-yk2 = 1.

We next calculate the slope of the curves Ck and H by implicit differentiation.

Ck: 2x+2(y-ck)y' = 0, y' = -x / (y-ck). (Note: y-ck = 0 forces x = 0 which is clearly impossible as a choice for xk.

H: 2x-2yy' = 0, y'= x/y. (Here, y = 0, is a point of intersection, but only for C0.)

Since these two slopes must be the same at the point (xk,yk), we have -xk / (yk-ck) = xk / yk, which simplifies to ck = 2yk. (Dividing by xk is permissible, since our point must be on H, and H does not cross the y-axis.)

At this point we have five families ofequations in the five unknowns ck, rk, Sk, xk, yk:
Sk  =  ck+rk  (1)
ck+1-rk+1  =  ck+rk (2)
ck  =  2yk (3)
xk2-yk2  =  1 (4)
xk2+(yk-ck)2  =  rk2 (5)


Subtracting (4) from (5), and then substituting (3) into the resultant yields:
yk2+(yk-ck)2  =  rk2-1 
4yk2+(2yk-2ck)2  =  4rk2-4 
ck2+(ck-2ck)2  =  4rk2-4 
2ck2  =  4rk2-4 
4ck2  =  8rk2-8 (6)


We can use (1) and (2) to write 2ck = Sk-1+Sk and 2rk = Sk-Sk-1, which we then substitute into (6):
(Sk-1+Sk)2  =  2(Sk-Sk-1)2-8 
Sk-12+2Sk-1Sk+Sk2  =  2Sk-12-4Sk-1Sk+2Sk2-8  
0  =  Sk-12-6Sk-1Sk+Sk2-8  (7k)


This is an non-homogeneous difference equation, and most of us find it a bit easier to think in terms of homogeneous difference equations. However, there is an easy trick here. Subtract the equations (7k-1) from the equation (7k).
(Sk-12-6Sk-1Sk+Sk2-8) - (Sk-22-6Sk-2Sk-1+Sk-12-8)  =  0  
Sk2 - 6Sk-1Sk + 6Sk-2Sk-1 - Sk-22  =  0  (8k)


We will also need to calculate S1 as a second initial value. From (71), S02-6S0S1+S12-8 = 0, and we already know that S0 = 1.
Hence, S12-6S1-7 = 0, (S1+1)(S1-7) = 0. Since S1 > S0, we deduce that S1 = 7.

We have traded a first-order non-linear non-homogeneous difference equation (7k) for a second-order non-linear homogeneous difference equation (8k). There are nice techniques for solving a second-order linear homogeneous difference equations, but ours is non-linear. Still, one should never give up. Note that Sk = Sk-2 is a solution of (8k), but that Sk > Sk-2 is required by the statement of the problem. We can factor (8k) and remove the factor of Sk - Sk-2.

0 = Sk2 - 6Sk-1Sk + 6Sk-2Sk-1 - Sk-22 = (Sk - Sk-2)(Sk-6Sk-1+Sk-2).


We know know that Sk = 6Sk-1-Sk-2, and hence Sk is always an odd integer. Since 2rk = Sk-Sk-1 , rk is always an integer.

Further techniques learned in Discrete Mathematics class can be used to provide a non-recursive formula for Sk, rk, ck, yk, xk.

The characteristic equation for Sk = 6Sk-1-Sk-2 is t2 - 6t +1 = 0, which has solutions t = 3 + sqrt(8) and t = 3 - sqrt(8). Thus, for suitable choices of A and B,
Sk = A(3 + sqrt(8))k + B(3 - sqrt(8))k.
Substituting k = 0 and k = 1 yields 1 = A + B and 7 = A(3 + sqrt(8)) + B(3 - sqrt(8)).
Substituting B = 1 - A into the second equation gives
7 = A(3 + sqrt(8)) + (1-A)(3 - sqrt(8)) = 3A + Asqrt(8) + 3 - sqrt(8) -3A + Asqrt(8) = 2Asqrt(8) + 3 - sqrt(8)
and A = 1 + sqrt(2), B = 1 - sqrt(2).
A further simplification is reached by noticing that (1 + sqrt(2))2 = 3 + sqrt(8). Hence, Sk = (1 + sqrt(2))2k+1 + (1 - sqrt(2))2k+1.

We leave the determination of explicit formula for the other variables as an exercise. Simplifying the formula for xk is the only slightly challenging part of that exercise.

This problem appeared in the M.A.A. Mathematical Monthly.

URL: http://www.math.fau.edu/locke/courses/Rec-Math/osculate.htm