It's rather x = 2 + (4 - r^2)^(1/2).

Curious, without the calculations, I would vouch that in the limit the x-intersept goes to infinity.

The circles intersect at (r^2/2, r(1-(r/2)^2)^(1/2), which leads to a slope of (1/r)((4-r^2)^(1/2) - 2) for the desired line. I think this will give an x-intercept of x = 2 + (4-r^2)^(1/2), as you said.
In this case, the limit is 4, not 2.

I also think it is counterintuitive that the limit exists.

Now, how about an ellipse centered at (a,o) with axes lengths 2a and 2b and and another centered at the origin with corresponding axes lengths ka and kb? What happens to the x-intercept of the corresponding line as k approaches 0? (I haven't worked on it.)

The limit will be 4a. The easy way to see this is to start with the original problem with circles of radius r and 1. Now zoom the picture by a factor of a to get circles of radius ra and a. This is just the same picture, but larger, so the intersection point is obviously 4a. Next rescale just the y coordinate by a factor of b/a. This produces ellipses just as you described them, but only affects the y coordinates of all the points of intersection in the picture. Therefore the limiting value of the intersection's x coordinate remains 4a.

--Stuart Anderson

As k approaches 0, the intercept approaches 4a, which agrees with the solution to Stuart's problem (for a = b = 1).

Owen lost a factor of 2 in the y coordinate of the intersection point of the circles.

>Curious, without the calculations, I would vouch that in the
>limit the x-intersept goes to infinity.

Yes, that was what I expected too when I first glanced at the problem. The reason this problem was in a calculus class is that the students were not expected to rationalize the denominator in the formula for x, but rather to use L'Hopital's rule to take the limit as r--> 0 in its original form. Of course rationalizing the denominator is easier, but in my experience, university calculus courses do not emphasize insight, but rather applications of rules from a standard "calculus toolkit."

Now to the geometric approach:

Let A be the point (0,r), B be the intersection of circles C1 and C2, C be the center of C1 at (1,0), O be the origin, and X be the intersection of line AB with the x axis. Draw the chord OB of C1 and consider the triangle OCB. This is of course isoceles, so the angle bisector at C cuts the chord OB at its midpoint D, and meets OB at a right angle. Therefore angle AOB = angle DCO = 1/2 angle BCO. Next, look at triangle AOB, which is also isoceles, and repeat the same process, so that there is an angle bisector at O meeting chord AB perpendicularly at its midpoint E. Then since OE is perpendicular to BX, angle BXO = angle AOE = 1/2 angle AOB.

Therefore angle BXO = 1/4 angle BCO. This is true regardless of the size of r, so in the limit as r --> 0, B moves down to the origin and in the small angle limit we get OX = 4*OC. But OC = 1, so OX --> 4. One could put more details into taking this limit, but it is pretty clear that the ratio approaches 4. Although the other methods also give the limit as 4, this appears somewhat mysteriors. The fact that the angles BCO and BXO are in 4:1 ratio gives the clearest most satisfying way to see why the limit is 4, at least in my opinion.

--Stuart Anderson

Such a disrespect towards students. I hope the professor would not disqualify against rationalizing the denominator.

>The
>fact that the angles BCO and BXO are in 4:1 ratio gives the
>clearest most satisfying way to see why the limit is 4, at
>least in my opinion.
>

Yes, it carries a solid grain of explanation of the phenomenon. Thank you.