Four. (This question is worth 20 points) Given the circle $C: x^{2}+(y-d)^{2}=r^{2}(d>r)$. Take two points $M, N$ on the $x$-axis such that the circle with $MN$ as its diameter is externally tangent to circle $C$. Find the coordinates of point $A$ such that for all $M, N$ satisfying the condition, $\angle MAN$ is a constant angle.
Let the center of the circle with diameter $MN$ be $T(h, 0)$, and the radius be $s$. Then $M(h-s, 0), N(h+s, 0)$. Let the center of circle $C$ be $K(0, d)$.

Then $K T^{2}=(r+s)^{2}=h^{2}+d^{2}$.
By symmetry, the point $A$ we are looking for lies on the $y$-axis, so we can set $A(0, t)$, assuming $t>0$.
Let the slope of $A M$ be $k_{1}$, and the slope of $A N$ be $k_{2}$. Then
$$
\begin{array}{l}
k_{1}=\frac{-t}{h-s}, \quad k_{2}=\frac{-t}{h+s} . \\
\operatorname{tg} M A N=\frac{k_{1}-k_{2}}{1+k_{1} k_{2}}=\frac{2 s t}{h^{2}-s^{2}+t^{2}} .
\end{array}
$$

From $h^{2}+d^{2}=(r+s)^{2}$, we get
$$
\operatorname{tg} M A N=\frac{2 s t}{r^{2}+t^{2}-d^{2}+2 r s} .
$$

Thus, taking $r^{2}$
$$
+t^{2}-d^{2}=0,
$$

i.e., $t^{2}=d^{2}-r^{2}$,
we have
$$
\operatorname{tg} M A N=\frac{2 s t}{2 r s}
$$
$=\frac{t}{r}$ (a constant).
Therefore, the coordinates of $A$ are $(0$,
$$
\left.\pm \sqrt{d^{2}-r^{2}}\right) \text {. }
$$