<s>[INST] Limit when circumference shrinks 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., $r \rightarrow 0^+$) ?
 [/INST] 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
$$\frac{y-r}{x-0}=\frac{b-r}{a-0}=\frac{r\sqrt{1-(r/2)^2}-r}{r^2/2-0}.$$
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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