(5) In the Cartesian coordinate system, a fixed point $B$ is on the negative half of the $y$-axis. A moving line $l$ passing through $B$ intersects the ellipse $\frac{x^{2}}{2}+y^{2}=1$ at points $C$ and $D$. The circle with $CD$ as its diameter always passes through a fixed point $A$ above the $x$-axis. Then the coordinates of $A$ are $\qquad$.
(5) $A(0,1)$ Hint: Let $A(s, t), B(0,-m),(m>0, t>0)$, then the line $l$ passing through point $B$ with slope $k$ is $l: y=k x-m$. From $\left\{\begin{array}{l}y=k x-m, \\ \frac{x^{2}}{2}+y^{2}=1,\end{array}\right.$, we get
$$
\left(1+2 k^{2}\right) x^{2}-2 k m x+2 m^{2}-2=0 .
$$

Therefore,
$$
x_{C}+x_{D}=-\frac{2 k m}{1+2 k^{2}}, x_{C} x_{D}=\frac{2 m^{2}-2}{1+2 k^{2}} \text {. }
$$

Also, from
$$
\angle C A D=90^{\circ} \Rightarrow \overrightarrow{A C} \cdot \overrightarrow{A D}=0,
$$

we have
$$
\begin{array}{l}
\left(x_{C}-s\right)\left(x_{D}-s\right)+\left(k x_{C}-m-t\right)\left(k x_{D}-m-t\right)=0 \\
\Rightarrow\left(1+k^{2}\right) x_{C} x_{D}-(s+k m+k t)\left(x_{C}+x_{D}\right)+s^{2}+(m+t)^{2}=0 \\
\Rightarrow\left(-2+2 s^{2}+2 t^{2}\right) k^{2}-4 m s k+2 m^{2}-2+(m+t)^{2}=0 .
\end{array}
$$

Since $k \in \mathbf{R}$, we get
$$
\left\{\begin{array}{l}
-2+2 s^{2}+2 t^{2}=0, \\
4 m s=0, \\
2 m^{2}-2+s^{2}+(m+t)^{2}=0 .
\end{array} \Rightarrow A(s, t)=A(0,1) .\right.
$$

When the slope of $l$ does not exist, $C D$ is the minor axis of the ellipse, and point $A(0,1)$ also satisfies.