In a Cartesian coordinate system, the points

$$
A(0,9), \quad B(6,3), \quad C\left(-\frac{1}{3},-1\right) \quad \text { and } D\left(3,-\frac{7}{2}\right)
$$

are given. Determine the point such that $P A=P B$ and $P C: P D=2: 3$.
Let the coordinates of the sought point be denoted by $x$ and $y$, then according to the problem,

$$
x^{2}+(y-9)^{2}=(x-6)^{2}+(y-3)^{2}
$$

and

$$
\left[\left(x+\frac{1}{3}\right)^{2}+(y+1)^{2}\right]:\left[(x-3)^{2}+\left(y+\frac{7}{2}\right)^{2}\right]=4: 9
$$

From (1), after expanding, rearranging, and simplifying, we get

$$
y=x+3
$$

This geometric locus is the perpendicular bisector of the distance $AB$. From (2),

$$
9\left(x^{2}+\frac{2}{3} x+\frac{1}{9}+y^{2}+2 y+1\right)=4\left(x^{2}-6 x+9+y^{2}+7 y+\frac{49}{4}\right)
$$

which simplifies to

$$
x^{2}+y^{2}+6 x-2 y-15=0
$$

This geometric locus is therefore a circle, the well-known Apollonian circle.

The abscissas of the common points of the two geometric loci are obtained by substituting the value of $y$ from (3) into (4).

$$
x^{2}+(x+3)^{2}+6 x-2 x-6-15=0
$$

or

$$
x^{2}+5 x-6=0
$$

from which

$$
\begin{array}{lll}
x_{1}=1, & \text { and thus } & y_{1}=4 \\
x_{2}=-6, & \text { and thus } & y_{2}=-3 .
\end{array}
$$

Therefore, there are two solutions: $P_{1}(1,4)$ and $P_{2}(-6,-3)$.