Consider a circle $k$ with center $S$ and radius $r$. Denote $\mathsf M$ the set of all triangles with incircle $k$ such that the largest inner angle is twice bigger than the smallest one. For a triangle $\mathcal T\in\mathsf M$ denote its vertices $A,B,C$ in way that $SA\ge SB\ge SC$. Find the locus of points $\{B\mid\mathcal T\in\mathsf M\}$.
1. **Understanding the problem and setting up the conditions:**
   - We are given a circle \( k \) with center \( S \) and radius \( r \).
   - We need to consider triangles with an incircle \( k \) such that the largest inner angle is twice the smallest one.
   - For a triangle \( \mathcal{T} \in \mathsf{M} \), we denote its vertices \( A, B, C \) such that \( SA \geq SB \geq SC \).

2. **Rewriting the condition using angle notation:**
   - Let the angles of the triangle be \( \alpha, \beta, \gamma \) with \( \gamma \geq \beta \geq \alpha \).
   - Given that the largest angle is twice the smallest, we have \( \gamma = 2\alpha \).
   - The sum of the angles in a triangle is \( 180^\circ \), so \( \alpha + \beta + \gamma = 180^\circ \).
   - Substituting \( \gamma = 2\alpha \), we get \( \alpha + \beta + 2\alpha = 180^\circ \) which simplifies to \( \beta = 180^\circ - 3\alpha \).

3. **Finding the range of \( \alpha \):**
   - From the inequality \( \alpha \leq 180^\circ - 3\alpha \leq 2\alpha \):
     \[
     \alpha \leq 180^\circ - 3\alpha \implies 4\alpha \leq 180^\circ \implies \alpha \leq 45^\circ
     \]
     \[
     180^\circ - 3\alpha \leq 2\alpha \implies 180^\circ \leq 5\alpha \implies \alpha \geq 36^\circ
     \]
   - Therefore, \( 36^\circ \leq \alpha \leq 45^\circ \).

4. **Setting up the Cartesian coordinate system:**
   - Assume \( AC = 1 \) and place \( A \) at \( (0,0) \) and \( C \) at \( (1,0) \).
   - Let \( t = \tan{\alpha} \).

5. **Finding the equation of lines \( AB \) and \( CB \):**
   - The equation of line \( AB \) is:
     \[
     y = xt
     \]
   - The equation of line \( CB \) is:
     \[
     y = -\frac{2t}{1-t^2}(x-1)
     \]

6. **Solving for the \( x \)-coordinate of \( B \):**
   - Equating the two equations:
     \[
     xt = -\frac{2t}{1-t^2}(x-1)
     \]
     \[
     xt(1-t^2) = -2t(x-1)
     \]
     \[
     x(1-t^2) = -2(x-1)
     \]
     \[
     x - x t^2 = -2x + 2
     \]
     \[
     x + 2x = x t^2 + 2
     \]
     \[
     3x = x t^2 + 2
     \]
     \[
     x(3 - t^2) = 2
     \]
     \[
     x = \frac{2}{3 - t^2}
     \]

7. **Substituting \( y = xt \) into the equation:**
   - We have \( y = \frac{2t}{3 - t^2} \).
   - Substituting into the hyperbola equation:
     \[
     x = \frac{2}{3 - \frac{y^2}{x^2}}
     \]
     \[
     3x^2 - y^2 - 2x = 0
     \]
     \[
     \frac{\left(x - \frac{1}{3}\right)^2}{\left(\frac{1}{3}\right)^2} - \frac{y^2}{\left(\frac{1}{\sqrt{3}}\right)^2} = 1
     \]

8. **Conclusion:**
   - The locus of points \( \{B \mid \mathcal{T} \in \mathsf{M}\} \) is a hyperbola with center \( \left(\frac{1}{3}, 0\right) \) and foci \( \left(-\frac{1}{3}, 0\right) \) and \( (1, 0) \).

The final answer is \(\boxed{\text{hyperbola with center } \left(\frac{1}{3}, 0\right) \text{ and foci } \left(-\frac{1}{3}, 0\right) \text{ and } (1, 0)}\).