# Two Circles and a Limit Analytic Proof

### Konhauser, Velleman, and Wagon

We are solving the Two Circles and a Limit problem:

A stationary circle of radius 3 is centered at (3, 0). Another circle of variable radius r is centered at the origin and meets the positive y-axis in point A. Let B be the common point of the two circles in the upper half-plane. Let E be the intersection of AB extended with the x-axis. What happens to E as r grows smaller and smaller?

The problem has been included in a wonderful book Which Way Did the Bicycle Go? by J. Konhauser, D. Velleman, S. Wagon as Problem #5. This is where the following solution comes from. The solution is analytic, but in an elegant way. It entirely avoids the common drudgery associated with repeated applications of the Pythagorean theorem.

Let C(r) be the circle of radius r centered at the origin; S, the other circle centered at (3, 0) with radius 3.

The former has the equation x² + y² = r², the latter, (x - 3)² + y² = 3², which can be rewritten as

x² + y² = r² and
x² + y² = 6x.

Since point B(x, y) lies on C(r), x² + y² = r². And, since it lies on S, x² + y² = 6x. Combining the two gives, x = r²/6. Then from the first equation, y = r² - (r²/6)². Simplifying, we get the coordinates of B:

B(r²/6, r/6·36 - r²).

Point A has coordinates (), r), so that the slope K of line AB is K = (36 - r² - 6)/r. The equation of AB is, say,

y = Kx + r.

The line intersects x-axis at (0 - r)/K = r²/(6 - 36 - r²) = (36 - r² + 6), which when r→∞ tends to 12.

### References

1. E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #396
2. J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #5 ### Limits in Geometry 