I posted this problem a few weeks ago:

The figure shows a fixed circle with equation and a shrinking circle with radius and center the origin (in red). is the point , is the upper point of intersection of the two circles, and is the point of intersection of the line and the -axis. What happens to as shrinks, that is, as ?

And I also posted this applet, so you could investigate it yourself.

Now, I post the solution. Initially, I thought that the -coordinate of would go off to infinity or zero, I wasn’t sure which. The equation for is . Solving for the intersection of the two circles, , we find it has coordinates

Remembering that , we now find the equation of line in point-slope form.

Now, we seek to find the coordinates of , the -intercept of the line. Letting in the above equation, we solve for and find

We now take the limit of as .

If we try to evaluate this limit by plugging in 0, we get an indeterminant form 0/0. We can either use L’hopital’s Rule or evaluate it numerically. Either way, we find.

I have to say, this result surprised me. Like I said, I expected this limit to evaluate to 0 or infinity–but 4?? I had to get a bit more of an intuitive understanding, so I built that Geogebra applet.

Also, I was told afterward, by the student who brought this to me, that there’s a straight-forward geometric way of deriving the answer, based on the observation that point *Q* and point *R* and the point (2,0) (0,2) form an isosceles triangle for all values of *r*. The observation isn’t trivial. Can you prove it’s true? Once you do realize this fact, though, the above result is clear.

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You have bright students if they can think along the lines that you say about isosceles triangles.

John, I think you mean that the right triangle is formed by the points Q, R, and (2,0). Otherwise, I really like this problem and its non-intuitive limit! Once you see the isosceles triangle, the answer is clear!

Right! The isosceles triangle formed by Q, R, and (2,0). Thanks, Matthew.

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