20.54 In $\triangle A B C$, $A B=10, A C=8, B C=6$. Circle $P$ is the smallest circle passing through $C$ and tangent to $A B$. Let $Q, R$ be the points where circle $P$ intersects sides $A C, B C$ (not coinciding with $C$). The length of segment $Q R$ is
(A) 4.75 .
(B) 4.8 .
(C) 5 .
(D) $4 \sqrt{2}$.
(E) $3 \sqrt{3}$.
(29th American High School Mathematics Examination, 1978)
[Solution 1] Let $N$ be the center of circle $P$. If $T$ is the point of tangency of the tangent line $AB$, and $CH$ is the altitude from $C$ to side $AB$ of $\triangle ABC$, then
$$
C N + N T \geqslant C T \geqslant C H.
$$

Assuming circle $P$ is the smallest circle passing through $C$ and tangent to $AB$, the radius $CN$ should be minimal, meaning $T$ coincides with $H$ and $N$ is the midpoint of $CH$; that is, $CH$ is the diameter of circle $P$.
Since $\angle QCR = 90^{\circ}$, and $QR$ is the diameter, then
$$
\triangle CBH \sim \triangle ABC,
$$

and $\frac{CH}{BC} = \frac{AC}{AB}$, i.e., $\frac{CH}{6} = \frac{8}{10}$.
Thus, $CH = 4.8$, and therefore $QR = 4.8$.
Hence, the answer is $(B)$.
[Solution 2] From the given, $\triangle ABC$ is a right triangle with $\angle BCA = 90^{\circ}$, so $RQ$ is the diameter of the circle, and the circle is smallest when $RCQT$ is a rectangle.
At this point, $CT$ is the altitude from $C$ to $AB$, and $RQ = CT$.
$$
\therefore \quad CT = \frac{AC \cdot BC}{AB} = \frac{6 \cdot 8}{10} = 4.8.
$$

Hence, the answer is $(B)$.