$AB$ and $CD$ are two fixed parallel chords of the circle $S$. $M$ is a variable point on the circle. $Q$ is the intersection of the lines $MD$ and $AB$. $X$ is the circumcenter of the triangle $MCQ$. 
Find the locus of $X$. 
What happens to $X$ as $M$ tends to 
(1) $D$, 
(2) $C$? 
Find a point $E$ outside the plane of $S$ such that the circumcenter of the tetrahedron $MCQE$ has the same locus as $X$.

1. **Define the circle and points:**
   - The circle \( S \) has the equation \( x^2 + y^2 = r^2 \).
   - Points \( A \) and \( B \) are on the x-axis, so \( A(-r, 0) \) and \( B(r, 0) \).
   - Points \( C \) and \( D \) are on a line parallel to the x-axis at a distance \( r \sin \beta \) from the origin, so \( C(-r \cos \beta, r \sin \beta) \) and \( D(r \cos \beta, r \sin \beta) \).
   - The variable point \( M \) on the circle can be parameterized as \( M(r \cos \alpha, r \sin \alpha) \).

2. **Find the intersection \( Q \) of \( MD \) and \( AB \):**
   - The line \( MD \) can be parameterized using the points \( M \) and \( D \):
     \[
     y - r \sin \alpha = \frac{r \sin \beta - r \sin \alpha}{r \cos \beta - r \cos \alpha} (x - r \cos \alpha)
     \]
     Simplifying, we get:
     \[
     y - r \sin \alpha = \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} (x - r \cos \alpha)
     \]
   - Since \( Q \) lies on \( AB \), \( y = 0 \). Substituting \( y = 0 \) in the equation of \( MD \):
     \[
     0 - r \sin \alpha = \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} (x_Q - r \cos \alpha)
     \]
     Solving for \( x_Q \):
     \[
     x_Q = r \cos \alpha - \frac{r \sin \alpha (\cos \beta - \cos \alpha)}{\sin \beta - \sin \alpha}
     \]

3. **Find the circumcenter \( X \) of \( \triangle MCQ \):**
   - The circumcenter \( X \) of a triangle is the intersection of the perpendicular bisectors of its sides.
   - The perpendicular bisector of \( MC \) is the line that is perpendicular to \( MC \) and passes through its midpoint.
   - The perpendicular bisector of \( CQ \) is the line that is perpendicular to \( CQ \) and passes through its midpoint.
   - The coordinates of \( X \) can be found by solving the system of equations formed by these perpendicular bisectors.

4. **Determine the locus of \( X \):**
   - By symmetry and the properties of the circle, the locus of \( X \) is a line.
   - The line can be derived from the geometric properties of the circle and the fixed points \( A, B, C, D \).
   - The equation of the locus is:
     \[
     2 \sin \beta \cdot x + 2 \cos \beta \cdot y + r \cdot \tan \beta = 0
     \]

5. **Behavior of \( X \) as \( M \) tends to \( D \) and \( C \):**
   - As \( M \) tends to \( D \), \( X \) will approach a specific point on the locus line.
   - As \( M \) tends to \( C \), \( X \) will approach another specific point on the locus line.

6. **Find a point \( E \) outside the plane of \( S \) such that the circumcenter of the tetrahedron \( MCQE \) has the same locus as \( X \):**
   - The point \( E \) should be chosen such that the circumcenter of \( MCQE \) lies on the same locus line.
   - One possible choice for \( E \) is a point on the z-axis, say \( E(0, 0, h) \), where \( h \) is a non-zero constant.

The final answer is \( \boxed{ 2 \sin \beta \cdot x + 2 \cos \beta \cdot y + r \cdot \tan \beta = 0 } \)