Problem 6

For each parabola y = x2 + px + q meeting the coordinate axes in three distinct points, a circle through these points is drawn. Show that all of the circles pass through a single point which is (0,1).
Solution. Let x1,  x2 be the real roots of the equation x2 + px + q = 0. Because all three points are distinct, it follows that x1 ≠ x2 and q ≠ 0. Therefore, the three meeting points are (x1, 0), (x2, 0), (0, q). Consider the circle
x2 + y2 + ax + by + c = 0
where a,  b,  c are real numbers and contains the above three points.
Then
x21 + ax1 + c = 0 x22 + ax2 + c = 0 q2 + bq + c = 0
Since x1, x2 are the roots of the equationx2 + px + q = 0, it follows that a = p, c = q and b =  − q − 1. Then the circle has the equation
x2 + y2 + px − (q + 1)y + q = 0
and it pass through the point (0,1).