This is actually pretty easy once you figure it out, D2 instead of D3 hard.
Since we know y has 2 real roots then we can write y=(x-p)*(x-q) where p+q=a and p*q=b.
Then the two x-intercepts are at (p,0) and (q,0) and the y-intercept is at (0,pq). The circle through these points intersects the y-axis at one more place, call that point (0,z).
For any circle if chords AB and CD are perpendicular and intersect at O then AO*OB = CO*OD.
From our four points then we can have AO=p, OB=q, CO=pq and OD=z. Then p*q=pq*z, which makes z=1.
Therefore the circle must also pass through (0,1).
Solution
