On the attraction "Merry parking", the auto has only two position* of a steering wheel: "right", and "strongly right". So the auto can move along an arc with radius $r_1$ or $r_2$. The auto started from a point $A$ to the Nord, it covered the distance $\ell$ and rotated to the angle $a < 2\pi$. Find the locus of its possible endpoints.
1. **Initial Setup and Problem Reformulation:**
   - We start with a point \( A \) and a ray \( \ell \) originating from \( A \) on the plane.
   - Given two radii \( r_1 \) and \( r_2 \) with \( r_1 > r_2 \), and two angles \( \alpha_1 \) and \( \alpha_2 \) such that \( \alpha_1 + \alpha_2 < \pi \).
   - The trajectory \( \Gamma \) consists of arcs with radii \( r_1 \) and \( r_2 \), with the sum of the angle measures of these arcs being \( 2\alpha_1 \) and \( 2\alpha_2 \) respectively.
   - The arcs are tangent to each other at their endpoints and lie on the same side of the tangent.

2. **Finding the Locus of Endpoints:**
   - Let \( O_1 \) and \( O_2 \) be the endpoints of arcs satisfying the first condition with radii \( r_1 \) and \( r_2 \) respectively, and angle measures \( 2(\alpha_1 + \alpha_2) \).
   - Consider two discs \( W_1 \) and \( W_2 \) centered at \( O_1 \) and \( O_2 \) with radii \( 2(r_1 - r_2) \sin \alpha_2 \) and \( 2(r_1 - r_2) \sin \alpha_1 \) respectively.
   - The required locus is the intersection of these discs.

3. **Proof of the Locus:**
   - Let \( PQ \) be an arbitrary arc of \( \Gamma \) with radius \( r_1 \), and \( R \) be a point on segment \( PQ \) such that \( PR : PQ = r_2 : r_1 \).
   - An arc \( PR \) touching \( \Gamma \) at \( P \) has radius \( r_2 \) and the tangent to it at \( R \) is parallel to the tangent to \( \Gamma \) at \( Q \).
   - Replace the arc \( PQ \) with \( PR \) and translate the part \( QB \) of \( \Gamma \) to \( \overrightarrow{QR} \).
   - Repeating this operation for each arc with radius \( r_1 \), we translate \( B \) to \( O_2 \), and \( \overrightarrow{O_2 B} = \frac{r_1 - r_2}{r_1} \sum \overrightarrow{P_i Q_i} \).
   - Similarly, \( O_1 B = \frac{r_2 - r_1}{r_2} \sum \overrightarrow{Q_i P_{i+1}} \) (assuming \( Q_0 = A \) and \( P_{n+1} = B \)).

4. **Homothety and Vector Mapping:**
   - Put all arcs of \( \Gamma \) on the unit circle starting from point \( X \). The endpoint of the last arc is a point \( Y \) such that \( \smile XY = 2(\alpha_1 + \alpha_2) \).
   - The homothety \( H \) with coefficient \( \frac{1}{r_1 - r_2} \) maps vector \( O_1 O_2 \) to \( XY \).
   - Color all arcs obtained from the arcs with radius \( r_1 \) red, and color blue the remaining arcs.
   - The homothety \( H \) maps vector \( O_2 B \) to the sum of red vectors and maps \( BO_1 \) to the sum of blue vectors.
   - Let \( Z \) be a point on arc \( XY \) such that \( \smile XZ = 2\alpha_1 \). The homothety \( H \) maps the boundary circles of \( W_2 \) and \( W_1 \) to the circles centered at \( X \) and \( Y \) respectively and passing through \( Z \).

5. **Maximizing the Lengths:**
   - Prove that the length of \( O_2 B \) is maximal if the red vector is unique.
   - Let the tangent \( m \) to the unit circle parallel to \( O_2 B \) touch the circle at \( E \).
   - Consider an arc \( T \) with length \( 2\alpha_1 \) and midpoint \( E \). If the endpoints of \( T \) lie on red arcs, divide these arcs into two parts.
   - The projection to \( m \) of a red vector with length \( \varphi \) lying outside \( T \) is less than \( \varphi \cos \alpha_1 \), therefore the sum of projections to \( m \) of all red vectors is less than the projection of \( T \).
   - Similarly, the length of \( O_1 B \) is maximal if the blue vector is unique. Thus \( B \) lies inside the intersection of two discs.

6. **Uniqueness of the Trajectory:**
   - Clearly, the common points of boundary circles of \( W_1 \) and \( W_2 \) correspond to the trajectories containing exactly one arc of each radius, and the points of these circles correspond to the trajectories containing one arc of some radius and two arcs of the remaining one.
   - Prove that each point \( B \) inside both discs corresponds to the unique trajectory containing four arcs and starting from the arc with radius \( r_1 \).

7. **Constructing the Parallelogram:**
   - Let \( D \) be the image of \( B \) in homothety \( H \). Find on arc \( XY \) such points \( E, F, G \) that \( \overrightarrow{XE} + \overrightarrow{FG} = \overrightarrow{XD} \).
   - Let \( E, G \) be the common points of the perpendicular bisector to \( DZ \) with arcs \( XZ, YZ \), and \( F \) be the second common point of arc \( XY \) with the line passing through \( Z \) and parallel to \( EG \).
   - Then \( DEFG \) is a parallelogram which is equivalent to the required equality. It is easy to see that \( E, F, G \) are uniquely defined by \( D \).

\(\blacksquare\)

The final answer is the intersection of the two discs centered at \( \boxed{ O_1 } \) and \( O_2 \) with radii \( 2(r_1 - r_2) \sin \alpha_2 \) and \( 2(r_1 - r_2) \sin \alpha_1 \) respectively.