Answer
As $C_2$ shrinks, $R$ approaches the point $(4,0)$
Work Step by Step
We can find an expression for the point $Q$.
The equation of the shrinking circle is $x^2+y^2 = r^2$
Then $y^2 = r^2-x^2$
The equation of the fixed circle is $(x-1)^2+y^2 = 1$
Then $y^2 = 1-(x-1)^2 = 2x-x^2$
$Q$ is a point on both circles.
We can equate the expressions for $y^2$ to find the x-coordinate of $Q$:
$r^2-x^2 = 2x-x^2$
$x = \frac{r^2}{2}$
We can find the y-coordinate of $Q$:
$y^2 = r^2-x^2$
$y^2 = r^2-(\frac{r^2}{2})^2$
$y^2 = r^2-\frac{r^4}{4}$
$y^2 = r^2(1-\frac{r^2}{4})$
$y = \sqrt{r^2(1-\frac{r^2}{4})}$
$y = r\sqrt{1-\frac{r^2}{4}}$
$P$ is the point $(0,r)$
$Q$ is the point $( \frac{r^2}{2}, r\sqrt{1-\frac{r^2}{4}}~)$
We can find the slope of the line connecting $P$ and $Q$:
$m = \frac{r-r\sqrt{1-\frac{r^2}{4}}}{0- \frac{r^2}{2}}$
$m = \frac{2r(\sqrt{1-\frac{r^2}{4}}-1)}{r^2}$
$m = \frac{2(\sqrt{1-\frac{r^2}{4}}-1)}{r}$
We can let the point $R$ be $(R_x,0)$
Since the point $(R_x,0)$ is on this line, the slope of the line connecting $P$ and the point $(R_x,0)$ has the same slope.
$m = \frac{r-0}{0-R_x}$
$\frac{2(\sqrt{1-\frac{r^2}{4}}-1)}{r} = \frac{r}{-R_x}$
$R_x = \frac{r^2}{2(1-\sqrt{1-\frac{r^2}{4}})}$
We can find the value of $R_x$ as $r$ approaches 0:
$\lim\limits_{r \to 0}R_x$
$= \lim\limits_{r \to 0}\frac{r^2}{2(1-\sqrt{1-\frac{r^2}{4}})}$
$= \lim\limits_{r \to 0}\frac{r^2}{2(1-\sqrt{1-\frac{r^2}{4}})}\cdot \frac{1+\sqrt{1-\frac{r^2}{4}}}{1+\sqrt{1-\frac{r^2}{4}}}$
$= \lim\limits_{r \to 0} \frac{r^2(1+\sqrt{1-\frac{r^2}{4}})}{2[1-(1-\frac{r^2}{4})]}$
$= \lim\limits_{r \to 0} 2(1+\sqrt{1-\frac{r^2}{4}})$
$= 2(1+\sqrt{1-\frac{0}{4}})$
$= 2(1+\sqrt{1})$
$= 2(1+1)$
$= 4$
As $C_2$ shrinks, the point $R$ approaches the point $(4,0)$
