Six. (12 points) On the hypotenuse $AB$ of the right triangle $\triangle ABC$, color the points $P$ that satisfy $PC^{2} = PA \cdot PB$ in red. How many red points are there at least, and at most, on the hypotenuse?
Six, as shown, let $B P=x$, then $A P=c-x$. Draw the altitude $C H$ on the hypotenuse, then $C H=\frac{a b}{c}, B H$ $=\frac{a^{2}}{b}$. Therefore,
$$
\left.A P=\frac{a^{2}}{c}-x \right| \, \text { ( } P \text { can be between } A \text { and } Y \text {, also please consider } H \text { and }
$$
$B$ points.
In the right triangle $\triangle C$, we have
$$
\begin{aligned}
P C^{2} & =C H^{2}+H P^{2} \\
& =\frac{a^{2} b^{2}}{c^{2}}=\left(\frac{a^{2}}{c}-x\right)^{2} \\
& =x^{2}-\frac{2 a^{2}}{c} x+a^{2} .
\end{aligned}
$$

But from the given, we also have
$$
P C^{2}=P A \cdot P B=x(c-x) .
$$

From (1) and (2), we get
$$
x^{2}-\frac{2 a^{2}}{c} x+a^{2}=x(c-x) \text {, }
$$

which simplifies to $2 x^{2}-\frac{2 a^{2}+c^{2}}{c} x+a^{2}=0$.
Factoring gives $(2 x-c)\left(x-\frac{a^{2}}{c}\right)=0$.
Thus, $x_{1}=\frac{c}{2}$ (the midpoint of the hypotenuse),
$x_{2}=\frac{a^{2}}{c}$ (the foot of the altitude on the hypotenuse).
Since a quadratic equation has at most two real roots, the hypotenuse has at most two red points; when the right triangle $\triangle A B C$ is isosceles, the midpoint and the foot of the altitude coincide, so the hypotenuse has at least one red point.