consider rays OA OB OD where OD bisect the angle BOA. Draw a circle with center O and radius 1. Then inverse A B D with respect to the circle. We got A'B'D',and here A'B'D'O are cyclic. Also <A'OD'=<D'OB',by Ptolemy's Theorem. We have (A'O)+(B'O)=k(D'O) this leads to 1/OA+1/OB=k/OD So,if 1/OA+1/OB is a constant 1,line AB pass through a fixed point D on the angle bisector of <BOA. Take the midpoint C of OD as center,1 the radius. Because OC is the angle bisector, AC/BC=AO/BO=a/b (Let OA=a,OB=b) Let T U be the midpoint of OA,OB the vector <OC>=b/(a+b)<OT>+a/(a+b)<OU> and the radius 1=[b/(a+b)](a/2)+[a/(a+b)](b/2) (here we used 1/a+1/b=1) By now we can easily check that with the circle we draw, and the circles with diameter OA,OB. They have two common tangents all the time. Best regards, Chao