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Let $C_1$ be a fixed circumference with equation $(x-1)^2 + y^2 = 1$ and $C_2$ a circumference to be shrinked, with center at $(0, 0)$ and radius $r$.

Let $P$ be the point $(0, r)$, $Q$ the upper intersection between $C_1$ and $C_2$ and $R$ the intersection between the line $PQ$ with the $x$ axis.

What happens with $R$ when $C_2$ shrinks (i.e., $r \rightarrow 0^+$) ?

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up vote 3 down vote accepted

Since I cannot think of anything clever, I will compute. The answer turns out to be different from what geometric intuition might suggest.

The point $P=P(r)$ has coordinates $(0,r)$. We find the coordinates of $Q$. So we are solving the system $$(x-1)^2+y^2=1,\qquad x^2+y^2=r^2.$$ Subtract. We get $2x=r^2$, so $x=r^2/2$. The $y$-coordinate of the upper point of intersection is therefore $\sqrt{r^2-r^4/4}$. Now we find the equation of the line $PQ$. In the usual way we obtain the equation $$\frac{y-r}{x}=\frac{\sqrt{r^2-r^4/4} -r}{r^2/2}.$$ The $x$-intercept is obtained by putting $y=0$. We get $$x=\frac{-r^3/2}{\sqrt{r^2-r^4/4}-r}.$$ Now it is time to take the limit. We can simplify our expression slightly to $$x= \frac{r^2/2}{1-\sqrt{1-r^2/4}}.$$ The taking of the limit is routine. We can for example multiply top and bottom by $1+\sqrt{1-r^2/4}$. Or recall more informally that if $\epsilon$ is small, $\sqrt{1-\epsilon}\approx 1-\epsilon/2$. Either way, the limit turns out to be $4$. Surprise!

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Yeah, this problem is intriguing. It would be nice to see a zoomed image showing that it's really 4, but i don't know if that is possible. – Fernando Apr 7 '12 at 20:11

The upper intersection point $T=(a,b)$ is on both $C_1$ and $C_2$ and has positive $y$-component. Thus

$$\begin{cases}a^2+b^2=r^2 \\ (a-1)^2+b^2=1. \end{cases}$$

Subtracting the first from the second, we obtain $1-2a=1-r^2$, and thus $a=r^2/2$. We can then solve for $b$ as $b=\sqrt{r^2-(r^2/2)^2}=r\sqrt{1-(r/2)^2}$. Therefore $T=\left(r^2/2,r\sqrt{1-(r/2)^2}\right)$. The line going between $T$ and $(0,r)$ is given by


The $x$-intercept occurs when we set $y=0$; solving for $x$ gives us

$$x=\frac{-r(r^2/2)}{r\sqrt{1-(r/2)^2}-r} \cdot \frac{1+\sqrt{1-(r/2)^2}}{1+\sqrt{1-(r/2)^2}}=2\left(1+\sqrt{1-(r/2)^2}\right).$$

Clearly as $r\to0$, $x\to 4$.

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