Today I finished the textbook exercises from the section on limit laws. There were a few interesting ones among the higher numbers. I may discuss more of them in coming days, but today I want to give a sketch of the last problem in the section:
“The figure shows a fixed circle $C_1$ with equation $(x-1)^2+y^2=1$ and a shrinking circle $C_2$ with radius $r$ and center the origin. $P$ 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 $PQ$ with the $x$-axis. What happens to $R$ as $C_2$ shrinks, that is, as $r\rightarrow 0^+$?”

What do you think? Make a guess.
It turns out the answer is that it approaches the point $(4,0)$. This is not at all in line with my intuition that, since at the limit point $P$ and $Q$ have the same $y$-coordinate, the $x$-coordinate of $R$ would grow without bound as $\overline{PQ}$ approached horizontal.
You can see the actual path of $R$ in this manipulable model of the problem that I made in Desmos after I found the solution. I did that by crunching the algebra arising from the equations of the circles and line in the diagram. I’m not going to give a full account of that here, but the basic procedure was a follows:
- Find the coordinates of the intersections of the two circles.
$(\frac{r^2}{2}, \pm\frac{r}{2}\sqrt{4-r^2})$ - The one with a positive $y$-coordinate is $Q$.
- Use the coordinates of $P$ and $Q$ to find the slope of $\overline{PQ}$.
$\frac{\sqrt{4-r^2}-2}{r}$ - Plug this and $y$-coordinate of $P$ into the slope-intercept form of a line.
$y=\frac{\sqrt{4-r^2}-2}{r}x+r$ - Find the $x$-intercept of this line.
$(\sqrt{4-r^2}+2,0)$ - This is $R$. Find it’s value as $r\rightarrow 0^+$.
It turns out that there is also a trigonometric solution, which you can read about in this thread on Math Stack Exchange. I stumbled on this when I did a Google search for the problem, hoping to check my (to me) surprising answer.
I also did a lot of thinking today about my review project and its ultimate goal of allowing me to do some online tutoring. I may say more about that in the coming days, as well.
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