(A. Myakishev, 8--9) In the plane, given two concentric circles with the center $ A$. Let $ B$ be an arbitrary point on some of these circles, and $ C$ on the other one. For every triangle $ ABC$, consider two equal circles mutually tangent at the point $ K$, such that one of these circles is tangent to the line $ AB$ at point $ B$ and the other one is tangent to the line $ AC$ at point $ C$. Determine the locus of points $ K$.
1. **Define the problem setup:**
   - Let \( A(0,0) \) be the center of the two concentric circles.
   - Let \( B(r,0) \) be a point on one of the circles with radius \( r \).
   - Let \( C(a,b) \) be a point on the other circle with radius \( R \), so \( a^2 + b^2 = R^2 \).

2. **Define the circles tangent to \( AB \) and \( AC \):**
   - The first circle, tangent to \( AB \) at \( B \), has center \( M_1 \) and radius \( n \). Since it is tangent to \( AB \) at \( B \), \( M_1 \) lies on the perpendicular bisector of \( AB \), which is the y-axis. Thus, \( M_1 = (r, n) \).
   - The second circle, tangent to \( AC \) at \( C \), has center \( M_2 \) and radius \( n \). The line \( CM_2 \) is perpendicular to \( AC \).

3. **Find the coordinates of \( M_2 \):**
   - The equation of the line \( CM_2 \) is \( y - b = -\frac{a}{b}(x - a) \).
   - The equation of the circle centered at \( C \) with radius \( n \) is \( (x - a)^2 + (y - b)^2 = n^2 \).
   - Solving these equations, we find \( M_2 = \left(a + \frac{bn}{R}, b - \frac{an}{R}\right) \).

4. **Calculate \( n \) using the distance between \( M_1 \) and \( M_2 \):**
   - The distance \( M_1M_2 \) must be \( 2n \), so:
     \[
     (a + \frac{bn}{R} - r)^2 + (b - \frac{an}{R} - n)^2 = 4n^2
     \]

5. **Find the coordinates of \( K \):**
   - Point \( K(x,y) \) is the midpoint of \( M_1 \) and \( M_2 \):
     \[
     x = \frac{r + a + \frac{bn}{R}}{2}, \quad y = \frac{n + b - \frac{an}{R}}{2}
     \]

6. **Simplify the equations:**
   - From the midpoint equations:
     \[
     a + \frac{bn}{R} = 2x - r, \quad b - \frac{an}{R} = 2y - n
     \]
   - Substitute these into the distance equation:
     \[
     (2x - r - r)^2 + (2y - n - n)^2 = 4n^2
     \]
   - Simplify to find \( n \):
     \[
     n = \frac{(x - r)^2 + y^2}{2y}
     \]

7. **Solve for \( a \) and \( b \):**
   - Using the midpoint equations:
     \[
     a = \frac{R(2Rx - 2ny + n^2 - rR)}{n^2 + R^2}, \quad b = \frac{R(2nx + 2Ry - nr - nR)}{n^2 + R^2}
     \]
   - Substitute into the circle equation \( a^2 + b^2 = R^2 \):
     \[
     4x^2 - 4rx + 4y^2 - 4ny + r^2 - R^2 = 0
     \]
   - Using the expression for \( n \):
     \[
     2x^2 + 2y^2 - r^2 - R^2 = 0
     \]

8. **Conclusion:**
   - The locus of points \( K \) satisfies the equation:
     \[
     2x^2 + 2y^2 = r^2 + R^2
     \]
   - This is the equation of a circle centered at the origin with radius \( \sqrt{\frac{r^2 + R^2}{2}} \).

The final answer is \(\boxed{2x^2 + 2y^2 = r^2 + R^2}\)