122 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. $\pi$
122 C. From the problem, we know that the curve $C$ is a circle with the line segment connecting points $P_{1}(\operatorname{arc} \sin \alpha, \operatorname{arc} \sin \alpha)$ and $P_{2}(\arccos \alpha, -\operatorname{arc} \cos \alpha)$ as its diameter. The coordinates of the center of the circle are $\left(\frac{\arcsin \alpha + \operatorname{arc} \cos \alpha}{2}, \frac{\arcsin \alpha - \operatorname{arc} \cos \alpha}{2}\right)$.
Since $\operatorname{arc} \sin \alpha + \operatorname{arc} \cos \alpha = \frac{\pi}{4}$, $d$ is the diameter of the circle.
$$
d = P_{1} P_{2} = \sqrt{\left(2 \arcsin \alpha - \frac{\pi}{2}\right)^{2} + \left(\frac{\pi}{2}\right)^{2}}
$$

When $\operatorname{arc} \sin \alpha = \frac{\pi}{4}$, i.e., $\alpha = \frac{\sqrt{2}}{2}$, the minimum value is $\frac{\pi}{2}$.