Four. (20 points) Let $k$ be a given non-zero real number, in the Cartesian coordinate system $x O y$, the equation of curve $C_{1}$ is $|x| \neq|y|$ and $\frac{x^{2}}{k-x}=\frac{y^{2}}{k+y}$, and point $A(k, k)$.
(1) Let $P$ be any point on $C_{1}$, try to find the equation of the locus $C_{2}$ of the midpoint $Q$ of segment $A P$, and identify the type and position of curve $C_{2}$;
(2) Find the angle $\theta$ (acute angle) between the tangents of $C_{1}$ and $C_{2}$ at their intersection point $B$ (expressed using an inverse trigonometric function).
$$
\begin{array}{l}
\text { (1) } \frac{x^{2}}{k-x}=\frac{y^{2}}{k+y} \\
\Leftrightarrow x^{2}(k+y)=y^{2}(k-x)(x \neq k \text { and } y \neq-k) \\
\Leftrightarrow(x+y)(y-x) k=(x+y) x y \\
\Leftrightarrow(y-x) k=x y(\text { by }|x| \neq|y| \text {, we get } x+y \neq 0) \\
\Leftrightarrow y(k-x)=k x \\
\Leftrightarrow y=\frac{k x}{k-x}=-k+\frac{k^{2}}{k-x} \\
\Leftrightarrow(x-k)(y+k)=-k^{2}(x \neq 0 \text { and } x \neq 2 k) .
\end{array}
$$

Therefore, the curve $C_{1}$ is the curve obtained by removing the points $(0,0)$ and $(2 k,-2 k)$ from the hyperbola
$$
(x-k)(y+k)=-k^{2}
$$

Let the midpoint of $P A$ be $Q(X, Y)$, then
$$
X=\frac{x+k}{2}, Y=\frac{y+k}{2} \text {. }
$$

Thus, $x=2 X-k, y=2 Y-k$.
Substituting into the equation $(x-k)(y+k)=-k^{2}$, we get
$$
(X-k) Y=-\left(\frac{k}{2}\right)^{2}\left(X \neq \frac{k}{2} \text { and } X \neq \frac{3}{2} k\right) \text {. }
$$

Therefore, the equation of the trajectory $C_{2}$ of $Q$ is
$$
(x-k) y=-\left(\frac{k}{2}\right)^{2}\left(x \neq \frac{k}{2} \text { and } x \neq \frac{3}{2} k\right),
$$

Its center is $(k, 0)$, and the asymptotes are $x=k$ and $y=0$, i.e., $C_{2}$ is the curve obtained by removing the points $\left(\frac{k}{2}, \frac{k}{2}\right)$ and $\left(\frac{3}{2} k,-\frac{k}{2}\right)$ from the hyperbola
$$
(x-k) y=-\left(\frac{k}{2}\right)^{2}
$$

(2) Solving the system of equations
$$
\left\{\begin{array}{l}
(x-k) y=-\left(\frac{k}{2}\right)^{2}, \\
(x-k)(y+k)=-k^{2} .
\end{array}\right.
$$
(2) $\div$ (1) gives $\frac{y+k}{y}=4$.

Solving for $y$ yields $y=\frac{1}{3} k$, substituting into equation (2) gives $x=\frac{k}{4}$.
Thus, the intersection point of $C_{1}$ and $C_{2}$ is $B\left(\frac{k}{4}, \frac{k}{3}\right)$.
Differentiating $(x-k) y=-\left(\frac{k}{2}\right)^{2}$ with respect to $x$ gives $\frac{\mathrm{d} y}{\mathrm{~d} x}(x-k)+y=0$, i.e.,
$$
\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{B}=-\left.\frac{y}{x-k}\right|_{B}=-\frac{\frac{k}{3}}{\frac{k}{4}-k}=\frac{4}{9} .
$$

Differentiating $(x-k)(y+k)=-k^{2}$ with respect to $x$ gives $\frac{\mathrm{d} y}{\mathrm{~d} x}(x-k)+y+k=0$, i.e.,
$$
\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{B}=-\left.\frac{y+k}{x-k}\right|_{B}=-\frac{\frac{k}{3}+k}{\frac{k}{4}-k}=\frac{16}{9} .
$$
The tangent of the angle (acute angle) between the tangents of $C_{1}$ and $C_{2}$ at the intersection point $B$ is
$$
\tan \theta=\frac{\frac{16}{9}-\frac{4}{9}}{1+\frac{16}{9} \times \frac{4}{9}}=\frac{9 \times 12}{81+64}=\frac{108}{145} .
$$

Thus, $\theta=\arctan \frac{108}{145}$.