4. If the line $x=\frac{\pi}{4}$ is intercepted by the curve $C$:
$$
(x-\arcsin \alpha)(x-\arccos \alpha)+(y-\arcsin \alpha)(y+\arccos \alpha)=0
$$

to form a chord of length $d$, the minimum value of $d$ as $\alpha$ varies is ( ).
(A) $\frac{\pi}{4}$;
(B) $\frac{\pi}{3}$;
(C) $\frac{\pi}{2}$;
(D) $\pi$.
4. (C)

From the problem, we know that the curve $C$ is a circle with the points $P_{1}(\arcsin a, \arcsin a)$ and $P_{2}(\arccos a, -\arccos a)$ as the endpoints of its diameter. The x-coordinate of the center of the circle is
$$
x_{0}=\frac{\arcsin a+\arccos a}{2}=\frac{\pi}{4}
$$

Therefore, the line $x=\frac{\pi}{4}$ passes through the center of the circle, and $d$ is the diameter of the circle. And
$$
\begin{array}{c}
d^{2}=2\left[(\arcsin a)^{2}+(\arccos a)^{2}\right] \geqslant(\arcsin a+\arccos a)^{2}=\frac{\pi^{2}}{4} . \\
\therefore d \geqslant \frac{\pi}{2}, \text { and when } \arcsin a=\arccos a \text {, the corresponding } d=\frac{\pi}{2},
\end{array}
$$

Thus, the answer is (C).