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∵兩圓相交於點A(1,3)、B(m,-1),兩圓的圓心均在直線l:x+y+c=0上,∴直線l垂直平分線段AB.∴kAB•kl=−11+m2+3−12+c=0,解得.m=−3c=0∴m+c=-3.故選C.
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