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設兩定點分別為A,B,以AB所在直線為x軸,AB的垂直平分線為y軸建立直角坐標系如圖:∵|AB|=6,則A(-3,0),B(3,0),設M(x,y),則|MA|2+|MB|2=26,即((x+3)2+y2)2+((x-3)2+y2)2=26.整理得:x2+y2=4.∴M的軌跡方程是x2+y2=4.
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