解：連接?$OA.$?
∵?$CD = 10，$?
∴?$OA=OC = OD = 5$?
.∵?$OM：OC = 3：5，$?
∴?$OM = 3，$?
∴在?$Rt\triangle AMO$?中，?$AM=\sqrt {OA^2-OM^2}=\sqrt {5^2-3^2} = 4.$?
如圖①，當點?$M$?在半徑?$OD$?上時，?$CM=OC + OM = 5 + 3 = 8.$?
∴在?$Rt\triangle AMC$?中，?$AC=\sqrt {AM^2+CM^2}=\sqrt {4^2+8^2} = 4\sqrt {5}.$?
如圖②，當點?$M$?在半徑?$OC$?上時，?$CM=OC - OM = 5 - 3 = 2.$?
∴在?$Rt\triangle AMC$?中，?$AC=\sqrt {AM^2+CM^2}=\sqrt {4^2+2^2} = 2\sqrt {5}.$?
綜上所述，弦?$AC$?的長為?$4\sqrt {5}$?或?$2\sqrt {5}$?
