Really Fun Limit Problem (revisited)

I posted this problem a few weeks ago:

The figure shows a fixed circle $C_1$ with equation $\left(x-1\right)^2+y^2=1$ and a shrinking circle $C_2$ with radius $r$ and center the origin (in red). $P$ is the point $(0,r)$, $Q$ is the upper point of intersection of the two circles, and $R$ is the point of intersection of the line $PQ$ and the $x$-axis. What happens to $R$ as $C_2$ shrinks, that is, as $r\rightarrow 0^{+}$? And I also posted this applet, so you could investigate it yourself.

Now, I post the solution. Initially, I thought that the $x$-coordinate of $R$ would go off to infinity or zero, I wasn’t sure which. The equation for $C_2$ is $x^2+y^2=r^2$. Solving for the intersection of the two circles, $Q$, we find it has coordinates $Q=\left(\frac{r^2}{2},\sqrt{r^2-\frac{r^4}{4}}\right)$

Remembering that $P=(0,r)$, we now find the equation of line $PQ$ in point-slope form. $y-r=\left(\frac{r-\sqrt{r^2-\frac{r^4}{4}}}{0-\frac{r^2}{2}}\right)(x-0)$

Now, we seek to find the coordinates of $R$, the $x$-intercept of the line. Letting $y=0$ in the above equation, we solve for $x$ and find $x=\frac{\frac{r^2}{2}}{1-\sqrt{1-\frac{r^2}{4}}}$

We now take the limit of $x$ as $r\rightarrow 0$. $\displaystyle\lim_{r\rightarrow 0}\frac{\frac{r^2}{2}}{1-\sqrt{1-\frac{r^2}{4}}}$

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. $\displaystyle\lim_{r\rightarrow 0}x=4$

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.

4 thoughts on “Really Fun Limit Problem (revisited)”

1. Gene Chase on said:

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

2. Matthew on said:

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!

3. Mr. Chase on said:

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

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