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)+4y1y...