解：?$(1)$?全等三角形有?$\triangle OAP≌\triangle OBP$?
?$\triangle OAC≌\triangle OBC，$??$\triangle AP C≌\triangle BP C$?
?$ (2)$?設(shè)?$\odot O$?半徑為?$r$?
在?$Rt\triangle OAP {中}，$??$OA = r，$??$OP=r + 2，$??$P A = 4$?
根據(jù)勾股定理?$OA^2+P A^2=OP^2$?
即?$r^2+4^2=(r + 2)^2，$?解得?$r = 3$?
∴?$\odot O$?的半徑為?$3$?
