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設⊙O的半徑為R,連接OA、OB ∵OA=OB=R,N是AB的中點 ∴AN=AB/2=2√3/2=√3,ON⊥AB(垂徑分弦) ∴OA²;-ON²;=AN²; ∵MN=1 ∴ON=OM-MN=R-1 ∴R²;-(R-1)²;=3 ∴R=2 ∴⊙O的半徑為2
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