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設線段P1P2的中點為M,∵P1(4,9)和P2(6,3),∴圓心M(5,6),又|P1P2|=(4−6)2+(9−3)2=210,∴圓的半徑為12|P1P2|=10,則所求圓的方程為:(x-5)2+(y-6)2=10.故答案為:(x-5)2+(y-6)2=10
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