Five. (20 points) Given an ellipse with semi-major axis $a$ and semi-minor axis $b$, one endpoint of the minor axis is $O$. $P$ and $Q$ are any two points on the ellipse different from point $O$, and $O P \perp O Q$. If the projection of point $O$ on line segment $P Q$ is $M$, find the locus of point $M$.

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Five, as shown in Figure 6, establish a rectangular coordinate system, then the equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{(y-b)^{2}}{b^{2}}=1
$$
$(a>b>0)$.
That is, $b^{2} x^{2}+a^{2} y^{2}-2 a^{2} b y=0$.
Let $P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right)$, and the equation of $P Q$ is $y=k x+m$.
Substitute $\frac{y-k x}{m}=1$ into equation (1), we get
$$
b^{2} x^{2}+a^{2} y^{2}-2 a^{2} b y \cdot \frac{y-k x}{m}=0 \text {. }
$$

Simplifying, we get the homogeneous equation
$$
m b^{2} x^{2}+2 a^{2} b k x y+\left(m a^{2}-2 a^{2} b\right) y^{2}=0,
$$

This equation can also be written as
$$
\left(m a^{2}-2 a^{2} b\right)\left(\frac{y}{x}\right)^{2}+2 a^{2} b k \frac{y}{x}+m b^{2}=0 \text {. }
$$

The roots of this equation are $\frac{y_{1}}{x_{1}} 、 \frac{y_{2}}{x_{2}}$.
Since $O P \perp O Q$, we have $\frac{y_{1}}{x_{1}} \cdot \frac{y_{2}}{x_{2}}=-1$,
Therefore, $\frac{m b^{2}}{m a^{2}-2 a^{2} b}=-1$, solving for $m$ gives $m=\frac{2 a^{2} b}{a^{2}+b^{2}}$.
Let $M(x, y)$. Since $O M \perp P Q$, then $k=-\frac{x}{y}$, substituting into the equation of line $P Q$, we have
$$
y=-\frac{x^{2}}{y}+\frac{2 a^{2} b}{a^{2}+b^{2}},
$$

That is,
$$
x^{2}+y^{2}-\frac{2 a^{2} b y}{a^{2}+b^{2}}=0 \text {. }
$$

Therefore, the locus of point $M$ is a circle with center $\left(0, \frac{a^{2} b}{a^{2}+b^{2}}\right)$ and radius $\frac{a^{2} b}{a^{2}+b^{2}}$, excluding the origin.