Construct the circle in the Cartesian plane with center and passing through A(0,1). By symmetry, the circle also passes through and .
Because the circle has center and passes through A(0,1), the equation of the circle is
This reduces to
x2 - sx + y2 - (p + 1)y + p = 0
So the x-intercepts of the circle are the solutions to x2 - sx + p = 0.
Alternate justification:
The segment joining the x-intercepts has a length , hence x1 + x2 = s.intercepts the arc AX1X2, and intercepts the opposite arc, hence the two angles are supplementary. But is also supplementary with , so is congruent to , which in turn impies that . Hence
,
which implies that , so x1 x2 = p.
Thus (x - x1)(x - x2) = x2 - sx + p, and the solutions to x2 - sx + p = 0 are x1 and x2 .
3 comments:
you might want to change the y coordinate of your circle center -- shows as \frac{p-1}{2} when you want \frac{p+1}{2}
you may want to change the y-coordinate of your circle center -- it reads \frac{p-1}{2} when you want \frac{p+1}{2}
Thanks.
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