过O点作OD⊥AC于D
∠B=90°,AC=13,AB=5,则BC=12
△AOD∽△ACB,则OD/BC=AO/AC,即:OD=BC*AO/AC=12*(5-r)/13
(1)当OD>r时,⊙O与AC相离.即:12*(5-r)/13>r,可解得: