You just know that your original figure is a circle of the radius 4 with the center at the origin 
of the coordinate system of the coordinate plane.

You also know that two yours basic points are the endpoints of the diameter of the circle, 
which lies on the horizontal axis y = 0. These points are A1 = (-4,0) and A2 = (4,0).

Now, the answer is:

    The locus of the midpoints of all chords of this circle that contain A1 is the circle of the radius 2 
    with the center at the point C1 = (-2,0). 

        It is the circle of the radius half of the radius of the original circle and touching 
        the original circle from the interior at the point A1 (shown in red in the Figure 1)

    The locus of the midpoints of all chords of this circle that contain A2 is the circle of the radius 2 
    with the center at the point C2 = (2,0). 

        It is the circle of the radius half of the radius of the original circle and touching 
        the original circle from the interior at the point A2 (shown in green in the Figure 1).







             Figure 1


   


                 Figure 2


Proof

Let A be an arbitrary, but fixed point of the circle O (Figure 2).
Let AB be some chord in the given circle with the points A and B on the circle (Figure 2). 
Let D be the middle point of the chord AB. Let the point C bisects the radius OA.
We need and we are going to prove that the length of the interval CD is half of the length of the radius OA.

Connect the center of the circle O with the point D.
Then OD is the median in the triangle AOB.
This triangle is isosceles (OA and OB are congruent as the radii of the circle).
In the isosceles triangle the median to the base coincides with the altitude (see the lesson 
  An altitude a median and an angle bisector in the isosceles triangle in this site).
So, OD is perpendicular to AB and the triangle ADO is right-angled triangle with the right angle at the vertex D.

Now, CD is the median in the right-angled triangle ADO drawn to its hypotenuse.
In the right-angled triangle the median drawn to hypotenuse has the measure half of the hypotenuse
  (see the lesson Median drawn to the hypotenuse of a right triangle in this site).

Thus we proved our statement: the distance from the point C to the middle of the chord AB is half the radius of the 
original circle. In other words, middle points of all chords passing through the point A are equidistant from the point C.