Answer
The point $R(x_R,0)$ tends to $(-8,0)$ as $r$ tends to $0$.
Work Step by Step
The equation of circle $C_2$ is : $x^2+y^2 = r^2$ ...(1)
And the equation of circle $C_1$ is : $(x-1)^2+y^2 = 1$ ... (2)
To find the intersection point $Q$ we solve the quations (1) and (2) for $x$ and $y$.
From (2) we have $x^2+y^2-2x+1 = 1$, or $x^2+y^2-2x = 0$.
Now using (1) we get $r^2-2x =0$ or $x = \frac{r^2}{2}$.
Substituting this value in (1) we get $y^{2} = r^2- \frac{r^4}{4} \Rightarrow y = \pm\sqrt{r^2- \frac{r^4}{4} }$.
Since the $y$ coordinates of $Q$ is positive, we can write the coordinates of $Q$ as $(x,y) = ( \frac{r^2}{2},\sqrt{r^2- \frac{r^4}{4}})$.
So the equation of line passing through $P(0,r)$ and $Q(\frac{r^2}{2},\sqrt{r^2- \frac{r^4}{4}})$ is $$\frac{y-r}{x-0} = \frac{\sqrt{r^2- \frac{r^4}{4}}-r}{\frac{r^2}{2}-0}.$$
To locate $R(x_R,0)$ we put $x = x_R, y =0$ in the equation of line $PQ$ to get, $$\frac{-r}{x_R} = \frac{\sqrt{r^2- \frac{r^4}{4}}-r}{\frac{r^2}{2}}$$ or $$x_R = -\frac{r^3}{2(\sqrt{r^2- \frac{r^4}{4}}-r)}.$$
Hence $$\lim\limits_{r \to 0}x_R = -\lim\limits_{r \to 0}\frac{r^3}{2(\sqrt{r^2- \frac{r^4}{4}}-r)}= -\lim\limits_{r \to 0}\frac{r^2}{2(\sqrt{1- \frac{r^2}{4}}-1)} ..(3)$$
Using binomial theorem for $|r|<1$we have, $\sqrt{1- \frac{r^2}{4}} = \Sigma_{k=0}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}$
$$\Rightarrow \sqrt{1- \frac{r^2}{4}} -1= \Sigma_{k=1}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}$$
$$\Rightarrow 2(\sqrt{1- \frac{r^2}{4}} -1)= 2\Sigma_{k=1}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}$$
$$\Rightarrow \frac{r^2}{2(\sqrt{1- \frac{r^2}{4}} -1)}=\frac{r^2}{ 2\Sigma_{k=1}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}}$$
$$=\frac{r^2}{ 2(\frac{1}{2})(\frac{1}{2}-1)(- \frac{r^2}{4})+2\Sigma_{k=2}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k}}$$
$$=\frac{1}{ \frac{1}{8}+2\Sigma_{k=2}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k-1}}$$
$$\Rightarrow \lim\limits_{r \to 0}\frac{r^2}{2(\sqrt{1- \frac{r^2}{4}}-1)}= \lim\limits_{r \to 0} \frac{1}{ \frac{1}{8}+2\Sigma_{k=2}^{\infty}\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k)}{k!}(- \frac{r^2}{4})^{k-1}} = \frac{1}{\frac{1}{8}}=8$$
Hence from (3) $$\lim\limits_{r \to 0}x_R = -\lim\limits_{r \to 0}\frac{r^2}{2(\sqrt{1- \frac{r^2}{4}}-1)} =-8$$
Hence the point $R(x_R,0)$ tends to $(-8,0)$ as $r$ tends to $0$.
