6. Given that $A B C D$ is a cyclic quadrilateral inscribed in a circle with radius $R$, $A B=12, C D=6$, and the extensions of $A B$ and $D C$ intersect at $P$, with $B P=8$ and $\angle A P D=60^{\circ}$. Then $R$ equals ( . .
(A) 10
(B) $2 \sqrt{21}$
(C) $12 \sqrt{2}$
(D) 14
6. (R).

By the secant theorem, we have
$$
P E \cdot P A = P C \cdot P D,
$$

so $8 \times 20$
$$
= P C (P C + 6).
$$

Solving this, we get $P C = 10$.
As shown in Figure 3, connect AC. In
$\triangle P A C$, since $P A =$
$2 P C$ and $\angle A P C = 60^{\circ}$, we have
$\angle P C A = 90^{\circ}$. Therefore, $A D$ is the diameter of the circle. By the Pythagorean theorem, we get
$$
\begin{aligned}
A D^{2} & = A C^{2} + C D^{2} = \left(P A^{2} - P C^{2}\right) + C D^{2} \\
& = 20^{2} - 10^{2} + 6^{2} = 336. \\
\therefore R & = \frac{1}{2} A D = 2 \sqrt{21}.
\end{aligned}
$$