设原点为O,A(x1,y1),B(x2,y2)因为线段ab为直径的圆过原点,所以OA⊥OB所以x1x2+y1y2=0联立y=x+m与y2=4x消去y得x^2+(2m-4)x+m^2=0△=(2m-4)^2-4m^2=16-16m＞0所以m＜1由韦达定理得x1+x2=4-2m,x1x2=m^2所以y1y2=(x1+m)(...