解：如圖，過點(diǎn)?$A $?作?$AD⊥BC，$?垂足為?$D$?
?$ ∵AB=AC=5，$??$AD⊥BC，$??$BC=6$?
?$ ∴BD=\frac 1 2BC=3，$??$AD$?垂直平分?$BC$?
∴點(diǎn)?$O$?在直線?$AD$?上
∴在?$Rt△ABD$?中，?$AD=\sqrt {{AB}^2-{BD}^2}=4$?
當(dāng)點(diǎn)?${O}_1$?在?$AD$?的反向延長(zhǎng)線上時(shí)，連接?${O}_1B$?
?$ {O}_1D=AD+A{O}_1=4+3=7$?
在?$Rt△{O}_1BD$?中，?${O}_1B=\sqrt {{{O}_1D}^2+{BD}^2}=\sqrt {58}$?
當(dāng)點(diǎn)?${O}_2$?在線段?$AD$?上時(shí)，連接?${O}_2B$?
?$ {O}_2D=AD-A{O}_2=4-3=1$?
在?$Rt△{O}_2BD$?中，?${O}_2B=\sqrt {{{O}_2D}^2+{BD}^2}=\sqrt {10}$?
綜上所述，?$⊙O$?的半徑為?$\sqrt {58}$?或?$\sqrt {10}$
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