P(x1,y1)、Q(x2,y2)
联立直线与椭圆,(3+a^2)x^2+2ax-1=0.
韦达定理,x1+x2=-2a/(3+a^2),x1x2=-1/(3+a^2).----(1)
并且y1=ax1+1,y2=ax2+1.----(2)
圆心(x0,y0),半径r:(x-x0)^2+(y-y0)^2=r^2.
以PQ为直径,x1+x2=2x0,y1+y2=2y0,(2r)^2=(x2-x1)^2+(y2-y1)^2.----(3)
圆过原点,x0^2+y0^2=r^2.----(4)
联立(1)(2)(3)(4),得a=1或-1.