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可設弦BC的中點為M. 則由垂徑定理可知,OM⊥BC 且BM=CM=3 又OB=OC=R=5, ∴由勾股定理可知,恆有OM=4 即動點M到原點O的距離恆為4. ∴中點M的軌跡是以原點為圓心,半徑為4的圓, ∴軌跡方程為x²+y²=16
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