Problem 944

Proof: Construct a complex plane as shown, let the complex numbers corresponding to points $P, Q$ be:
$$\begin{array}{l} p=r_{1} e^{i\left(b_{1} \mid w \tau\right)}, \\ q=a+r_{2} e^{j\left(\theta_{2}+w\right)} \end{array}$$
(Where $t$ is a parameter)
Assume that a point $M$ exists, whose corresponding complex number is $Q_{0}$, which is the point corresponding to $t=0$. Then,
$$\frac{m-p}{q-p}=\frac{m \quad p_{4}}{q_{n}-p_{4}} .$$
$$\begin{array}{l} \therefore m=p+\frac{\left(p-\frac{p_{0}}{(p)}(q-p)\right.}{\left(q-q_{0}\right)} \\ =r_{1} e^{v^{\left(i H_{1}\right.}+m e r}+\frac{r_{1} e^{j \theta_{1}}}{r_{1} e^{i \|_{1}}-r_{2} e^{i q_{2}}} \\ \text { - }\left[a \mid e^{j u c}\left(r_{2} e^{j w_{2}}-r_{1} e^{j /}\right)\right] \\ =\frac{r_{1} e^{r_{1}}}{r_{1} e^{j_{1}}-r_{2} e^{j \pi_{2}}} \cdot a \text {. } \\ \end{array}$$
(From this, we know that $r_{1} \neq r_{2}$, and the denominator is not zero)
It is independent of the parameter $t$, i.e., point $M$ is a fixed point.
Therefore, the fixed point $M$ exists.
Note: If the two circles have a common point $S$, and $P_{0}=Q_{0}=S$, then the conclusion does not hold. In some cases, similar triangles may degenerate into a set of collinear points.