Question

数学问题见下图

Answer 1

设AB:y=x+m,①
代入x^2+y^2-2x+4y-4=0得2x^2+2mx+m^2-2x+4x+4m-4=0,
整理得2x^2+(2m+2)x+m^2+4m-4=0,
△/4=(m+1)^2-2(m^2+4m-4)=-m^2-6m+9>0,
m^2+6m-9<0,-3-3√2<m<-3+3√2.②
设A(x1,y1),B(x2,y2),则x1+x2=-m-1,x1x2=(m^2+4m-4)/2,
由①，y1y2=(x1+m)(x2+m)=x1x2+m(x1+x2)+m^2,
以AB为直径的圆过原点，
<==>向量OA*OB=x1x2+y1y2=0,
<==>2x1x2+m(x1+x2)+m^2=m^2+4m-4-m^2-m+m^2=0,
<==>m^2+3m-4=0,解得m=1或-4，满足②，
∴l:y=x+1或y=x-4.