Given a circle $\gamma$ and two points $A$ and $B$ in its plane.  By $B$ passes a variable secant that intersects $\gamma$ at two points $M$ and $N$. Determine the locus of the centers of the circles circumscribed to the triangle $AMN$.
1. **Define the given elements:**
   - Let $\gamma$ be a circle with center $O(0,0)$ and radius $r$.
   - Let $A(a,0)$ and $B(b,c)$ be two points in the plane of the circle.
   - Consider a variable secant line passing through $B$ that intersects $\gamma$ at points $M$ and $N$.

2. **Equation of the secant line:**
   - The equation of the line passing through $B(b,c)$ can be written as:
     \[
     y - c = \lambda (x - b)
     \]
     where $\lambda$ is a parameter.

3. **Intersection of the secant line with the circle:**
   - The equation of the circle $\gamma$ is:
     \[
     x^2 + y^2 = r^2
     \]
   - Substitute $y = \lambda (x - b) + c$ into the circle's equation:
     \[
     x^2 + (\lambda (x - b) + c)^2 = r^2
     \]
   - Expand and simplify:
     \[
     x^2 + \lambda^2 (x - b)^2 + 2\lambda c (x - b) + c^2 = r^2
     \]
     \[
     x^2 + \lambda^2 (x^2 - 2bx + b^2) + 2\lambda c x - 2\lambda bc + c^2 = r^2
     \]
     \[
     (1 + \lambda^2) x^2 + (2\lambda c - 2\lambda^2 b) x + (\lambda^2 b^2 - 2\lambda bc + c^2 - r^2) = 0
     \]
   - This is a quadratic equation in $x$, and solving it gives the $x$-coordinates of $M$ and $N$.

4. **Midpoint of $M$ and $N$:**
   - Let $M(x_1, y_1)$ and $N(x_2, y_2)$ be the points of intersection.
   - The midpoint $P$ of $M$ and $N$ is:
     \[
     P\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
     \]
   - Since $M$ and $N$ lie on the circle, their coordinates satisfy the circle's equation.

5. **Circumcenter of $\triangle AMN$:**
   - The circumcenter $L$ of $\triangle AMN$ is the point equidistant from $A$, $M$, and $N$.
   - Using the formula for the circumcenter of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, we get:
     \[
     L\left(\frac{\lambda(a^2 - r^2)}{2(a\lambda - b\lambda + c)}, -\frac{a^2 - r^2}{2(a\lambda - b\lambda + c)}\right)
     \]

6. **Locus of the circumcenter $L$:**
   - To find the locus of $L$, we need to eliminate the parameter $\lambda$.
   - The coordinates of $L$ suggest a linear relationship between $x$ and $y$.
   - By analyzing the expressions, we derive the equation of the locus:
     \[
     2(a - b)x - 2cy + r^2 - a^2 = 0
     \]

The final answer is $\boxed{2(a - b)x - 2cy + r^2 - a^2 = 0}$