[Solution] Connect $O P, O Q, O T, O T^{\prime}$, then by $T P=T^{\prime \prime} P, O T=O T^{\prime \prime}=r$, and $\angle P T^{\prime \prime} O=\angle P T O=90^{\circ}$, thus $\triangle O T P \cong \triangle O T^{\prime \prime} \mathrm{P}$.
Similarly, we can prove $\triangle O T^{\prime} Q \cong \triangle O T^{\prime \prime} \mathrm{Q}$.
Let $x=\angle T O P=\angle P O T^{\prime \prime}$,
$$
y=\angle T^{\prime \prime} O Q=\angle Q O T^{\prime},
$$

then $2 x+2 y=180^{\circ}$.
Therefore, $\angle P O Q=x+y=90^{\circ}$.
Thus, $\triangle P O Q$ is a right triangle. And $O T^{\prime \prime}$ is the altitude on the hypotenuse $P Q$.
Therefore, $T^{\prime \prime} O^{2}=P T^{\prime \prime} \cdot T^{\prime \prime} Q$,
which means $r^{2}=4 \cdot 9$, yielding $r=6$.
Hence, the answer is $(B)$.
20.26 In the attached figure, $T P$ and $T^{\prime} Q$ are two parallel tangent lines of this circle with radius $r$, $T$ and $T^{\prime}$ are the points of tangency. $P T^{\prime \prime} Q$ is a third tangent line with the point of tangency $T^{\prime \prime}$.
If $T P=4, \quad T^{\prime} Q=9$, then $r$ is
(A) $\frac{25}{6}$.
(B) 6 .
(C) $\frac{25}{4}$.
(D) A number other than $\frac{25}{6}, 6, \frac{25}{4}$.
(E) No definite solution can be determined from the given conditions.