$k_1$ denotes one of the arcs formed by intersection of the circumference $k$ and the chord $AB$. $C$ is the middle point of $k_1$. On the half line (ray) $PC$ is drawn the segment $PM$. Find the locus formed from the point $M$ when $P$ is moving on $k_1$.

[i]G. Ganchev[/i]
1. **Define the Circle and Points:**
   - Let the circle \( k \) have center \( O(0,0) \) and radius \( r \).
   - Let \( AB \) be a chord of the circle, and \( k_1 \) be one of the arcs formed by the intersection of the circle and the chord \( AB \).
   - Let \( C \) be the midpoint of the arc \( k_1 \), so \( C \) is directly above the center \( O \) at \( (0, r) \).

2. **Parameterize Point \( P \):**
   - Let \( P \) be a point on the arc \( k_1 \). We can parameterize \( P \) as \( P(r \cos \alpha, r \sin \alpha) \) where \( \alpha \) is the angle subtended by \( P \) at the center \( O \).

3. **Define the Segment \( PM \):**
   - Let \( PM = c \) be a constant length, and \( M \) be a point on the ray \( PC \). Since \( M \) lies on the ray \( PC \), we can express \( M \) as a point on the line extending from \( P \) through \( C \).

4. **Find the Equation of the Line \( PC \):**
   - The line \( PC \) can be parameterized as:
     \[
     \text{Line } PC: \left( x, y \right) = \left( r \cos \alpha, r \sin \alpha \right) + t \left( 0 - r \cos \alpha, r - r \sin \alpha \right)
     \]
     Simplifying, we get:
     \[
     \left( x, y \right) = \left( r \cos \alpha, r \sin \alpha \right) + t \left( -r \cos \alpha, r - r \sin \alpha \right)
     \]
     \[
     \left( x, y \right) = \left( r \cos \alpha (1 - t), r \sin \alpha + t (r - r \sin \alpha) \right)
     \]
     \[
     \left( x, y \right) = \left( r \cos \alpha (1 - t), r (t + \sin \alpha (1 - t)) \right)
     \]

5. **Determine the Coordinates of \( M \):**
   - Since \( PM = c \), we need to find \( t \) such that the distance from \( P \) to \( M \) is \( c \):
     \[
     \sqrt{(r \cos \alpha (1 - t) - r \cos \alpha)^2 + (r (t + \sin \alpha (1 - t)) - r \sin \alpha)^2} = c
     \]
     Simplifying the distance formula:
     \[
     \sqrt{(r \cos \alpha (-t))^2 + (r (t + \sin \alpha - t \sin \alpha) - r \sin \alpha)^2} = c
     \]
     \[
     \sqrt{(r \cos \alpha t)^2 + (r (t + \sin \alpha - t \sin \alpha - \sin \alpha))^2} = c
     \]
     \[
     \sqrt{(r \cos \alpha t)^2 + (r t (1 - \sin \alpha))^2} = c
     \]
     \[
     \sqrt{r^2 t^2 (\cos^2 \alpha + (1 - \sin \alpha)^2)} = c
     \]
     \[
     r t \sqrt{\cos^2 \alpha + (1 - \sin \alpha)^2} = c
     \]
     \[
     t = \frac{c}{r \sqrt{\cos^2 \alpha + (1 - \sin \alpha)^2}}
     \]

6. **Substitute \( t \) Back into the Coordinates of \( M \):**
   - Using the value of \( t \):
     \[
     M = \left( r \cos \alpha (1 - \frac{c}{r \sqrt{\cos^2 \alpha + (1 - \sin \alpha)^2}}), r (\frac{c}{r \sqrt{\cos^2 \alpha + (1 - \sin \alpha)^2}} + \sin \alpha (1 - \frac{c}{r \sqrt{\cos^2 \alpha + (1 - \sin \alpha)^2}})) \right)
     \]

7. **Simplify the Coordinates of \( M \):**
   - Let \( k = \sqrt{\cos^2 \alpha + (1 - \sin \alpha)^2} \):
     \[
     M = \left( r \cos \alpha (1 - \frac{c}{r k}), r (\frac{c}{r k} + \sin \alpha (1 - \frac{c}{r k})) \right)
     \]
     \[
     M = \left( r \cos \alpha - \frac{c \cos \alpha}{k}, \frac{c}{k} + r \sin \alpha - \frac{c \sin \alpha}{k} \right)
     \]
     \[
     M = \left( r \cos \alpha - \frac{c \cos \alpha}{k}, r \sin \alpha + \frac{c (1 - \sin \alpha)}{k} \right)
     \]

8. **Form the Locus Equation:**
   - The locus of \( M \) is given by the equation:
     \[
     (x^2 + y^2)^2 - 2r^2(x^2 + y^2) - c^2[x^2 + (y - r)^2] + r^4 = 0
     \]

The final answer is \( \boxed{ (x^2 + y^2)^2 - 2r^2(x^2 + y^2) - c^2[x^2 + (y - r)^2] + r^4 = 0 } \).