Let A,B,C be three collinear points and a circle T(A,r).
If  M and N are two diametrical opposite variable points on T,
Find locus geometrical of the intersection BM and CN.
1. **Define the problem setup:**
   Let \( A, B, C \) be three collinear points and a circle \( T \) with center \( A \) and radius \( r \). Let \( M \) and \( N \) be two diametrically opposite points on \( T \).

2. **Identify the intersection points:**
   We need to find the locus of the intersection point \( U \) of lines \( BM \) and \( CN \).

3. **Use projective geometry properties:**
   Take a point \( X \) on line \( AB \) such that the cross-ratio \((B, C; A, X) = -1\). This means that \( X \) is the harmonic conjugate of \( A \) with respect to \( B \) and \( C \).

4. **Define additional intersection points:**
   Let \( U = BM \cap CN \) and \( V = BN \cap CM \). By the properties of the complete quadrilateral, the line \( UV \) passes through \( X \).

5. **Analyze the line \( UV \):**
   Let \( S = UV \cap MN \). Since \( (BM, BN; BA, BS) = -1 \) and \( A \) is the midpoint of \( MN \), it follows that \( BS \parallel MN \). Therefore, \( UV \parallel MN \).

6. **Determine the midpoint relationship:**
   Since \( UV \parallel MN \) and \( X \) is the midpoint of \( UV \), we have that \( XU \) is constant.

7. **Use homothety:**
   The ratio \(\frac{XU}{AN} = \frac{CX}{CA}\) is fixed, implying that \( XU \) is constant. Therefore, \( U \) lies on a circle obtained from \( T \) by a homothety with centers \( B \) and \( C \). One of these centers is the external center, and the other is the internal center.

8. **Conclusion:**
   The locus of the intersection point \( U \) is a circle obtained from \( T \) by a homothety centered at \( B \) and \( C \).

The final answer is the locus of the intersection point \( \boxed{ U } \) is a circle obtained from \( T \) by a homothety centered at \( B \) and \( C \).