题目如图.已知⊙O\odot O⊙O的半径为333,OA=8OA=8OA=8,点PPP为⊙O\odot O⊙O上一动点.以PAPAPA为边作等边△PAM\triangle PAM△PAM,则线段OMOMOM的长的最大值为( )A.141414B.999C.121212D.111111知识点:切线长定理章节:图形的性质 / 圆