$27 \cdot 8$ If the line $x=\frac{\pi}{4}$ is intercepted by the curve $C$:
$$
(x-\arcsin a)(x-\arccos a)+(y-\arcsin a)(y+\arccos a)=0
$$

to form a chord of length $d$, then the minimum value of $d$ as $a$ varies is
(A) $\frac{\pi}{4}$.
(B) $\frac{\pi}{3}$.
(C) $\frac{\pi}{2}$.
(D) $\pi$.
(China High School Mathematics League, 1993)
[Solution 1] From the problem, we know that the curve $C$ is a circle with the endpoints of its diameter at
$$
P_{1}(\arcsin a, \arcsin a), P_{2}(\arccos a, -\arccos a)
$$

The x-coordinate of the center of the circle is
$$
x_{\alpha}=\frac{\arcsin a+\arccos a}{2}=\frac{\pi}{4},
$$

so the line $x=\frac{\pi}{4}$ passes through the center of the circle, and $d$ is the diameter of the circle.
Since $d^{2}=2\left[(\arcsin a)^{2}+(\arccos a)^{2}\right] \geqslant(\arcsin a+\arccos a)^{2}=\frac{\pi^{2}}{4}$. Therefore,
$$
d \geqslant \frac{\pi}{2} .
$$

Furthermore, when $\arcsin a=\arccos a$, the corresponding $d_{\text {min }}=\frac{\pi}{2}$. Hence, the correct choice is $(C)$.
[Solution 2] Taking $a=1$ or 0, we get $d=\frac{\sqrt{2}}{2} \pi \frac{\pi}{2}$.
By symmetry, we know that $(C)$ is correct.
Therefore, the correct choice is $(C)$.