解：如圖，過點(diǎn)?$A$?作?$AD⊥BC，$?垂足為?$D。$?
∵?$AB = AC = 5，$??$AD⊥BC，$??$BC = 6，$?
∴易得點(diǎn)?$O$?在直線?$AD$?上，?$BD = \frac {1}{2}BC = 3，$?
∴在?$Rt△ABD$?中，?$AD = \sqrt {AB^2-BD^2} = 4。$?
?$ $?當(dāng)點(diǎn)?$O?$?在?$AD$?的反向延長線上時(shí)，連接?$O?B。$?
∵?$O?D = AD + AO? = 4 + 3 = 7，$?
∴在?$Rt△O?BD$?中，?$O?B = \sqrt {O_{1}D^2+BD^2} = \sqrt {7^2+3^2} = \sqrt {58}。$?
?$ $?當(dāng)點(diǎn)?$O?$?在線段?$AD$?上時(shí)，連接?$O?B。$?
∵?$O?D = AD - AO? = 4 - 3 = 1，$?
∴在?$Rt△O?BD$?中，?$O?B = \sqrt {O_{2}D^2+BD^2} = \sqrt {1^2+3^2} = \sqrt {10}。$?
綜上所述，?$⊙O$?的半徑為?$\sqrt {58}$?或?$\sqrt {10}。$?
