Problem 1482

1. Identify the given points and their relationships:
- Let $A$ and $B$ be two points on a plane.
- Let $M$ be the midpoint of $AB$.
- Let $r$ be a line, and let $R$ and $S$ be the projections of $A$ and $B$ onto $r$, respectively.

2. Understand the projections:
- The projections $R$ and $S$ are such that $AR \perp r$ and $BS \perp r$.

3. **Circumcircle of $\triangle AMR$:**
- We need to show that the circumcircle of $\triangle AMR$ has the same radius as the circumcircle of $\triangle BSM$.

4. **Consider the circumcircle of $\triangle AMR$:**
- Let $(AMR)$ denote the circumcircle of $\triangle AMR$.
- Let $O_1$ be the center of $(AMR)$ and $R_1$ be its radius.

5. **Consider the circumcircle of $\triangle BSM$:**
- Let $(BSM)$ denote the circumcircle of $\triangle BSM$.
- Let $O_2$ be the center of $(BSM)$ and $R_2$ be its radius.

6. **Use the fact that $M$ is the midpoint of $AB$:**
- Since $M$ is the midpoint of $AB$, we have $AM = MB$.

7. Analyze the angles:
- Let $(RSM)$ meet $RS$ at $R$ and $P$.
- We have $\angle SPM = \angle RAM = 180^\circ - \angle SBM$.
- This implies that $SPMB$ is cyclic.

8. Use the property of cyclic quadrilaterals:
- Since $SPMB$ is cyclic, $\angle PMA = \angle PMB = 90^\circ$.
- This means that $P$ is on the perpendicular bisector of $AB$.

9. **Conclude that $PA = PB$:**
- Since $P$ is on the perpendicular bisector of $AB$, we have $PA = PB$.

10. Equal radii of the circumcircles:
- Since $PA = PB$, the circles $(AMR)$ and $(BSM)$ have equal diameters.
- Therefore, the radii of the circumcircles $(AMR)$ and $(BSM)$ are equal.

$$\boxed{R_1 = R_2}$$