11.5. Through a point $M$ inside a circle of radius $r$ with center at point $O$, two perpendicular chords are drawn. It is known that the sum of the squares of the lengths of these chords is $m^{2}$. Find the length of the segment $O M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_fc2e572e219f8141991fg-5.jpg?height=577&width=529&top_left_y=1642&top_left_x=158)

To the solution of problem 11.5
Solution: Let the chords drawn be $AB$ and $CD$, and drop perpendiculars $OE$ and $OF$ to them, respectively (see figure). The quadrilateral $OEMF$ is a rectangle, so the equality $OM = \sqrt{OE^2 + OF^2}$ holds. Point $E$ is the midpoint of chord $AB$, and from triangle $AOE$ by the Pythagorean theorem, $AB = 2AE = 2\sqrt{r^2 - OE^2}$. Therefore, $OE^2 = r^2 - 0.25AB^2$. Similarly, we get $OF^2 = r^2 - 0.25CD^2$, hence,

\[
\begin{aligned}
& OM = \sqrt{OE^2 + OF^2} = \\
& = \sqrt{r^2 - 0.25AB^2 + r^2 - 0.25CD^2} = \sqrt{2r^2 - 0.25m^2}
\end{aligned}
\]

The problem is solved.

Answer: $OM = \sqrt{2r^2 - 0.25m^2}$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| The solution process is correct, but the wrong answer is obtained only due to arithmetic errors | 6 points |
| The problem statement is correctly written in algebraic form (using vectors, coordinates, trigonometric functions, etc.), but the resulting algebraic problem is not solved | 3 points |
| Only a special case is considered, where at least one of the chords is a diameter | 2 points |
| Correct answer without justification (or with incorrect justification) | 1 point |