已知圆x²+y²+x-6m和直线x+2y-3=0交与PQ两点,且以PQ为直径的圆恰过坐标原点,求m
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已知圆x²+y²+x-6m和直线x+2y-3=0交与PQ两点,且以PQ为直径的圆恰过坐标原点,求m
已知圆x²+y²+x-6m和直线x+2y-3=0交与PQ两点,且以PQ为直径的圆恰过坐标原点,求m
已知圆x²+y²+x-6m和直线x+2y-3=0交与PQ两点,且以PQ为直径的圆恰过坐标原点,求m
x+2y-3=0,即x=3-2y,
将x=3-2y代入圆x²+y²+x-6m=0,得(3-2y)²+y²+3-2y-6m=0,
整理得5y²-14y+12-6m=0,
设P(x1,y1),Q(x2,y2),
则y1+y2=14/5,y1y2=(12-6m)/5,
∴x1x2=(3-2y1)(3-2y2)=9-6(y1+y2)+4y1y2=9-6×14/5+4(12-6m)/5=(9-24m)/5,
∵以PQ为直径的圆恰过坐标原点O,
∴OP⊥OQ,故x1x2+y1y2=(9-24m)/5+(12-6m)/5=0,
解得m=7/10,
将m=7/10代入方程5y²-14y+12-6m=0,得5y²-14y+39/5=0,
∵△=(-14)²-4×5×(39/5)=40>0,
∴m=7/10符合题意.