The limit about the line connecting the intersection of a circle and the $y$-axis and the intersection of the shrinking circle and a fixed circle

Question

There is 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 ray $PQ$ and the $x$-axis. What happens to $R$ as $C_2$ shrinks, that is, as $r \to 0^+$? (The figure is made with GeoGebra)

In order to solve this problem, I made a script using GeoGebra in which the circle $C_2$ is a dynamic one whose radius $r$ can be adjusted with a slider. As I set $r \to 0^+$, the figure seems to suggest that $R \to (4,0)$. In particular, this is the state with $r = 0.001$, in which $R$ is reported to be $(3.9999997523053,0)$:

However, I would like to find out a way to prove (or disprove, though unlikely) my guess that $$\lim_{r \to 0} R = (4,0).$$ But I have little idea. Any help would be appreciated.

Answer 1

First let's find the equation for $Q$. It's the intersection of the circles $(x-1)^2 + y^2 = 1$ and $x^2 + y^2 = r^2$. So:

$$1 = x^2 -2x + 1 + y^2 \implies 0 = r^2 - 2x \implies x = \frac{r^2}{2}$$

From here we conclude that $y = \sqrt{\frac{4r^2 -r^4}{4}}$. Now the line joining $P(0,r)$ and $Q\left(\frac{r^2}{2},\sqrt{\frac{4r^2 -r^4}{4}}\right)$ is given by.

$$y= \frac{\sqrt{\frac{4r^2 -r^4}{4}}-r}{\frac{r^2}{2}} \cdot x + r = \frac{\sqrt{4r^2 - r^4} - 2r}{r^2}\cdot x + r$$

Hence the $x$-intercept is given by setting $y=0$. Then:

$$- r^3 = \left(\sqrt{4r^2 -r^4}-2r\right)\cdot x \implies x = \frac{-r^3}{\sqrt{4r^2 -r^4}-2r}$$

Finally:

$$\lim_{r \to 0^{+}} x = \lim_{r \to 0^{+}} \frac{-r^3}{\sqrt{4r^2 -r^4}-2r} = \lim_{r \to 0^{+}}\frac{-r^3(\sqrt{4r^2 -r^4}+2r)}{4r^2 -r^4-4r^2} = \lim_{r \to 0^{+}} \frac{\sqrt{4r^2 -r^4}+2r}{r} $$ $$= \lim_{r \to 0^{+}} \sqrt{4 - r^2} + 2 = 4$$

Answer 2

An alternative solution using geometry.

Let $\angle POQ = 2t$, so $\angle QOR = \frac{\pi}{2} - 2t$.

Let $O = (0, 0)$ and $A = (1, 0)$. Since $OP = r$ and $OA = AQ = 1$, we have $$ OP = r = 2 \sin(2t). $$

Since $OP = OQ = r$, we have $\angle P = \frac{\pi}{2} - t$, so $\angle R = t$. Thus, $$ OR = \frac{OP}{\tan \angle R} = \frac{r}{\tan t} = \frac{2 \sin(2t)}{\tan t} $$

Note that $\angle POQ \to 0^+$ as $r \to 0^+$. Therefore, $$ \lim_{r \to 0^+} OR = \lim_{t \to 0^+} \frac{2 \sin(2t)}{\tan t} = \lim_{t \to 0^+} \frac{4 \sin t \cos t}{\frac{\sin t}{\cos t}} = 4 (\cos 0)^2 = 4. $$