设抛物线方程为y^2=2kx,p（1,－2）代入求得：k=2,则抛物线方程为y^2=4x
设原点为O,A（x1,y1）,B（x2,y2）,AB直线方程为y=m(x-4),代入抛物线方程得：m^2*(x-4)^2=4x,m^2*x^2-(8m^2+4)*x+16*m^2=0,
则x1+x2=(8m^2+4)/m^2,
x1*x2=16,
y1+y2=m(x1+x2-8)=4/m,
y1*y2=m^2*(x1-4)(x2-4)=-16
AB^2=(x1-x2)^2+（y1-y2）^2=(x1+x2)^2-4x1*x2+(y1+y2)^2-4y1*y2
=(8m^2+4)^2/m^4-64+16-m^2+64
=(80m^4+64m^2+16)/m^4
OA^2+OB^2=x1^2+y1^2+x2^2+y2^2=(x1+x2)^2-2x1*x2+(y1+y2)^2-2y1*y2
=(8m^2+4)^2/m^4-32+16-m^2+32=(80m^4+64m^2+16)/m^4=AB^2
所以,以线段AB为直径的圆恒过坐标原点