If the circle x2 + y2 = 4 is known and the secant ABC of the circle is made through a (4,0), then the trajectory equation of the midpoint of the chord BC is () A. （x-2）2+y2=4B. （x-2）2+y2=4（0≤x＜1）C. （x-1）2+y2=4D. （x-1）2+y2=4（0≤x＜1）

If the circle x2 + y2 = 4 is known and the secant ABC of the circle is made through a (4,0), then the trajectory equation of the midpoint of the chord BC is () A. （x-2）2+y2=4B. （x-2）2+y2=4（0≤x＜1）C. （x-1）2+y2=4D. （x-1）2+y2=4（0≤x＜1）

Let the midpoint (x, y) of the chord BC, the slope of the straight line passing through a be K, and the equation of secant ABC be y = K (x-4); make the secant ABC of the circle, so the line between the midpoint and the center of the circle is perpendicular to the secant ABC, and the equation is x + KY = 0; because the intersection point is the midpoint of the chord and it is on these two straight lines, the trajectory equation of the point in the chord BC is x2 + y2-4x = 0, as shown in the figure, so select B