Problem source

\noindent
\begin{center}
\begin{tikzpicture}[x=1.0cm, y=1.0cm, line width=0.3pt]
    % The asterisk in pspicture* means the drawing is clipped to the bounding box
    \clip (-2.94,-1.87) rectangle (7.07,3.86);
    % Circles
    \draw (0,1) circle (1.25);
    \draw (3,0) circle (0.55);
    % Lines
    \draw (-2.44,-0.03) -- (6.18,-0.85);
    \draw (-2.04,3.71) -- (6.55,-1.48);
    % Labels
    % \rput[tl] aligns the text to the Top-Left (North-West) of the node coordinate
    \node[anchor=north west, inner sep=0.1em] at (5.33,-0.41) {$P$};
    \node[anchor=north west, inner sep=0.1em] at (-0.18,1.1) {$C_1$};
    \node[anchor=north west, inner sep=0.1em] at (2.85,0.15) {$C_2$};
    \node[anchor=north west, inner sep=0.1em] at (-0.65,3.29) {$L'$};
    \node[anchor=north west, inner sep=0.1em] at (-1.5,-0.34) {$L$};
\end{tikzpicture}
\end{center}
The diagram above shows two non-overlapping circles $C_1$ and $C_2$ of different sizes. The lines $L$ and $L'$ are the two common tangents to $C_1$ and $C_2$ such that the two circles lie on the same side of each of the tangents. The lines $L$ and $L'$ intersect at the point $P$ which is called the \textit{focus} of $C_1$ and $C_2$.
\begin{questionparts}
\item Let $\mathbf{x}_1$ and $\mathbf{x}_2$ be the position vectors of the centres of $C_1$ and $C_2$, respectively. Show that the position vector of $P$ is 
\[
\frac{r_1 \mathbf{x}_2- r_2 \mathbf{x}_1}{r_1-r_2} \,,
\]
where $r_1$ and $r_2$ are the radii of $C_1$ and $C_2$, respectively.
\item The circle $C_3$  does not overlap either $C_1$ or $C_2$ and its   radius, $r_3$, satisfies  $r_1 \ne r_3 \ne r_2$. The focus of $C_1$ and $C_3$ is $Q$, and the focus of $C_2$ and $C_3$ is $R$. Show that $P$, $Q$ and $R$ lie on the same straight line.
\item Find a condition on $r_1$, $r_2$ and $r_3$ for $Q$ to lie half-way between $P$ and $R$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Notice that $P$ lies on $C_1C_2$, and that the triangles formed from $C_iPT_i$ where $T_i$ are the tangent points are similar, with ratios $\frac{r_1}{r_2}$. Therefore $\frac{C_1P}{r_1} = \frac{C_2P}{r_2}$, and hence $\frac{C_1P}{C_1C_2} = \frac{C_1P}{C_1P-C_2P} = \frac{1}{1-\frac{r_2}{r_1}} = \frac{r_1}{r_1-r_2}$

So we have $\mathbf{p}  = \mathbf{x_1} + (\mathbf{x}_2 - \mathbf{x}_1)\cdot\frac{r_1}{r_1-r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1-r_2}$

\item Suppose $\mathbf{x}_3 = \binom{\alpha}{\beta}$ in the basis of $\{ \mathbf{x}_1, \mathbf{x}_2 \}$, then we can see that

\begin{align*}
&& \mathbf{p} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} \\
&& \mathbf{q} &= \frac{r_1(\alpha \mathbf{x}_1 +\beta \mathbf{x}_2) - r_3\mathbf{x}_1}{r_1-r_3} \\
&&&= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\
&& \mathbf{r} &=\frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\
&& \mathbf{p}-\mathbf{q} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1}  - \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\
&&&= \frac{1}{(r_1-r_2)(r_1-r_3)} \binom{(r_1-r_3)(-r_2)-(r_1-r_2)(r_1\alpha-r_3)}{(r_1-r_3)r_1 - (r_1-r_2)r_1\beta} \\
&&&= \frac{r_1}{(r_1-r_2)(r_1-r_3)} \binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \\
&& \mathbf{q} - \mathbf{r} &= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta}  - \frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3}  \\
&&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(r_2-r_3)(r_1\alpha-r_3) - (r_1-r_3)r_2\alpha)}{(r_2-r_3)r_1\beta - (r_1-r_3)(r_2\beta - r_3)} \\
&&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(-r_2r_3+r_3^2) - \alpha(r_1r_3-r_3r_2)}{r_3(r_1-r_3)-\beta(r_1-r_2)} \\
&&&= \frac{r_3}{(r_1-r_3)(r_2-r_3)}\binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} 
\end{align*}

Therefore they are clearly parallel, and hence lie on a line.

\item $Q$ is halfway between $P$ and $R$ if \begin{align*}
&& \frac{r_1}{(r_1-r_2)(r_1-r_3)} &= \frac{r_3}{(r_1-r_3)(r_2-r_3)} \\
\Leftrightarrow && r_1(r_2-r_3) &= r_3(r_1-r_2) \\
\Leftrightarrow && r_1r_2 - r_1r_3 &= r_1r_3 - r_2r_3 \\
\Leftrightarrow && r_2 &= \frac{2r_1r_3}{r_1+r_3}
\end{align*}
\end{questionparts}