Let m be perpendicular to OQ and OP, MB 1=MA 1, angle BMB 1= angle AMA 1, triangle BMB 1 and triangle AMA 1 congruence, so AM=BM.

It is easy to know from the figure that OB+OA=OP=4 (fixed value), AB square =OB square +OA square =(OB+OA) square -2oa * ob = 16-2oa * ob.

When OB (or OA)=0, the square of AB is the smallest, that is, AB=4 is also the smallest, and OB+OA+AB=8 is also the smallest.

However, because OB=0, O and B coincide, A and P coincide, and OAB is not a triangle, so the circumference of triangle AOB has no minimum.