A fixed point $A$ inside a circle is given. Consider all chords $XY$ of the circle such that $\angle XAY$ is a right angle, and for all such chords construct the point $M$ symmetric to $A$ with respect to $XY$ . Find the locus of points $M$.
1. **Given Setup**: Let $(O, R)$ be the given circle with center $O$ and radius $R$. A fixed point $A$ is inside the circle. We consider all chords $XY$ of the circle such that $\angle XAY$ is a right angle. For each such chord, we construct the point $M$ symmetric to $A$ with respect to $XY$.

2. **Intersection Points**: Let lines $AX$ and $AY$ intersect the circle again at points $Z$ and $W$, respectively. Let $N$ be the midpoint of the chord $XY$.

3. **Angles and Perpendiculars**: Since $\angle XAY = 90^\circ$, we have $\angle XAN = \angle AXN = \angle AWZ$. This implies that $NA \perp ZW$. Similarly, if $L$ denotes the midpoint of $ZW$, then $LA \perp XY$.

4. **Parallelogram Formation**: As a result, quadrilateral $OLAN$ is a parallelogram. This implies that segments $AO$ and $NL$ bisect each other at point $U$. The orthogonal projection of $A$ on $XY$ lies on the circle with diameter $\overline{NL}$.

5. **Median Theorem Application**: By the median theorem in $\triangle ALN$, we have:
   \[
   UN^2 = \frac{1}{2}(AN^2 + AL^2) - \frac{1}{4}AO^2
   \]

6. **Pythagorean Theorem Application**: Using the Pythagorean theorem for right triangles $\triangle OZL$ and $\triangle OYN$, we get:
   \[
   AN^2 = R^2 - \frac{1}{4}ZW^2
   \]
   \[
   AL^2 = R^2 - \frac{1}{4}XY^2
   \]

7. **Summing the Squares**: Adding these equations, we obtain:
   \[
   AN^2 + AL^2 = 2R^2 - \left(\frac{1}{4}ZW^2 + \frac{1}{4}XY^2\right)
   \]
   Since $ZW = XY$ (as $Z$ and $W$ are reflections of $X$ and $Y$ across $A$), we have:
   \[
   AN^2 + AL^2 = 2R^2 - \frac{1}{2}XY^2
   \]

8. **Simplifying the Expression**: Therefore, we get:
   \[
   AN^2 + AL^2 = R^2
   \]

9. **Locus of $M$**: The locus of $M$ is a circle with center $O$ and radius $OM$. Using the derived expressions, we find:
   \[
   OM = 2 \sqrt{\frac{1}{2}R^2 - \frac{1}{4}AO^2}
   \]

The final answer is $\boxed{2 \sqrt{\frac{1}{2}R^2 - \frac{1}{4}AO^2}}$