Eight, prove as shown in the figure, connect $0 R, P R$, draw RC $\perp \mathrm{RQ}$ at C,
$$\Rightarrow \underset{O \text { is the midpoint of } P Q}{\mathrm{PA} / / \mathrm{QB}}\} \Rightarrow \mathrm{AR}=\mathrm{BR},$$
$$\left.\begin{array}{l} \angle \mathrm{ARP}=\angle \mathrm{RQP} \\ \mathrm{PQ} \text { is the diameter of } \odot \mathrm{O} \end{array}\right\} \Rightarrow\left\{\begin{array}{l} \angle \mathrm{PRC} \text { and } \angle \mathrm{CRQ} \\ \angle \mathrm{CRQ} \text { and } \angle \mathrm{RQC} \end{array}\right.$$

are complementary $\}$
$$\Rightarrow \angle \mathrm{PRC}=\angle \mathrm{RQC} \Rightarrow \angle \mathrm{ARP}=\angle \mathrm{PRC}$$
$$\Rightarrow R t \triangle R P A \cong R t \triangle R P C \Rightarrow A R=C R=B R \Rightarrow$$ CR is the radius of the circle with center $R$ and $R A$ as the radius, passing through point $B$, and tangent to $P Q$ at $C$.