An acute angle with the vertex $O$ and the rays $Op_1$ and $Op_2$ is given in a plane. Let $k_1$ be a circle with the center on $Op_1$ which is tangent to $Op_2$. Let $k_2$ be the circle that is tangent to both rays $Op_1$ and $Op_2$ and to the circle $k_1$ from outside. Find the locus of tangency points of $k_1$ and $k_2$ when center of $k_1$ moves along the ray $Op_1$.

1. **Define the given elements and setup the problem:**
   - Let \( O \) be the origin \((0,0)\).
   - Let the angle between the rays \( Op_1 \) and \( Op_2 \) be \( 2\alpha \).
   - Let \( k_1 \) be a circle with center \( O_1 \) on \( Op_1 \) and radius \( r \), tangent to \( Op_2 \).
   - Let \( k_2 \) be a circle tangent to both rays \( Op_1 \) and \( Op_2 \) and to the circle \( k_1 \) from outside.

2. **Determine the radius of \( k_1 \):**
   - Since \( k_1 \) is tangent to \( Op_2 \), the radius \( r \) of \( k_1 \) is the perpendicular distance from \( O_1 \) to \( Op_2 \).
   - If \( O_1 \) is at \((\lambda, 0)\) on \( Op_1 \), then the distance from \( O_1 \) to \( Op_2 \) is \( \lambda \sin 2\alpha \).
   - Therefore, \( r = \lambda \sin 2\alpha \).

3. **Determine the center and radius of \( k_2 \):**
   - Let the center of \( k_2 \) be \( O_2 \) at \((t, t \cdot \tan \alpha)\) and radius \( t \cdot \tan \alpha \).
   - The distance between \( O_1 \) and \( O_2 \) is given by:
     \[
     (O_1O_2)^2 = t^2 \tan^2 \alpha + (\lambda - t)^2
     \]
   - Since \( k_2 \) is tangent to \( k_1 \) from outside, the distance between their centers is equal to the sum of their radii:
     \[
     O_1O_2 = r + t \cdot \tan \alpha = \lambda \sin 2\alpha + t \cdot \tan \alpha
     \]

4. **Solve for \( t \):**
   - Equate the two expressions for \( O_1O_2 \):
     \[
     \sqrt{t^2 \tan^2 \alpha + (\lambda - t)^2} = \lambda \sin 2\alpha + t \cdot \tan \alpha
     \]
   - Square both sides to eliminate the square root:
     \[
     t^2 \tan^2 \alpha + (\lambda - t)^2 = (\lambda \sin 2\alpha + t \cdot \tan \alpha)^2
     \]
   - Expand and simplify to solve for \( t \):
     \[
     t^2 \tan^2 \alpha + \lambda^2 - 2\lambda t + t^2 = \lambda^2 \sin^2 2\alpha + 2\lambda t \sin 2\alpha \cdot \tan \alpha + t^2 \tan^2 \alpha
     \]
     \[
     \lambda^2 - 2\lambda t + t^2 = \lambda^2 \sin^2 2\alpha + 2\lambda t \sin 2\alpha \cdot \tan \alpha
     \]
     \[
     \lambda^2 - 2\lambda t = \lambda^2 \sin^2 2\alpha + 2\lambda t \sin 2\alpha \cdot \tan \alpha
     \]
     \[
     t = \lambda (1 - \sqrt{2} \sin \alpha)^2
     \]

5. **Find the locus of the tangency points:**
   - The line \( O_1O_2 \) has the equation:
     \[
     y = \frac{t \cdot \tan \alpha}{t - \lambda}(x - \lambda)
     \]
   - Substitute \( t = \lambda (1 - \sqrt{2} \sin \alpha)^2 \):
     \[
     y = \frac{\lambda (1 - \sqrt{2} \sin \alpha)^2 \cdot \tan \alpha}{\lambda (1 - \sqrt{2} \sin \alpha)^2 - \lambda}(x - \lambda)
     \]
     \[
     y = \frac{\lambda (1 - \sqrt{2} \sin \alpha)^2 \cdot \tan \alpha}{\lambda ((1 - \sqrt{2} \sin \alpha)^2 - 1)}(x - \lambda)
     \]
     \[
     y = \frac{\lambda (1 - \sqrt{2} \sin \alpha)^2 \cdot \tan \alpha}{\lambda (1 - 2\sqrt{2} \sin \alpha + 2 \sin^2 \alpha - 1)}(x - \lambda)
     \]
     \[
     y = \frac{\lambda (1 - \sqrt{2} \sin \alpha)^2 \cdot \tan \alpha}{\lambda (-2\sqrt{2} \sin \alpha + 2 \sin^2 \alpha)}(x - \lambda)
     \]
     \[
     y = \frac{(1 - \sqrt{2} \sin \alpha)^2 \cdot \tan \alpha}{-2\sqrt{2} \sin \alpha + 2 \sin^2 \alpha}(x - \lambda)
     \]
     \[
     y = \frac{\sin 2\alpha}{\cos 2\alpha + 2} \cdot x
     \]

The final answer is \( \boxed{ y = \frac{\sin 2\alpha}{\cos 2\alpha + 2} \cdot x } \).