∠AOB=45°,OP= 4根号2,OQ=7
根据余弦定理：
PQ^2=OP^2+OQ^2-2OP*OQcos45°
=(4根号2)^2+7^2-2*4根号2*7*根号2/2=25
PQ=5
PD/DQ=2/3
PD/(PD+DQ)=2/(2+3)
PD/PQ=2/5
半径PD=PQ*2/5=5*2/5=2
OP=4根号2,PD=2,QD=OQ=r,根据余弦定理：
PQ^2=OP^2+OQ^2-2OP*OQcos45°
如果外切：PQ=PD+QD=2+OQ
(PD+QD)^2=(4根号2)^2+OQ^2-2*4根号2*OQ*根号2/2
(2+OQ)^2=32-8OQ
OQ^2+12OQ-28=0
(OQ+14)(OQ-2)=0
OQ+14＞14
∴OQ=2
如果内切：PQ=QD-PD=OQ-2
(QD-PD)^2=(4根号2)^2+OQ^2-2*4根号2*OQ*根号2/2
(OQ-2)^2=32-8OQ
(OQ+2)^2=32
(OQ+2+4根号2)(OQ+2-4根号2)=0
OQ+2+4根号2＞0
∴OQ+2-4根号2=0
∴OQ=4根号2-2