$\left[\begin{array}{l}\text { [Intersecting Circles]} \\ {[\text { Chords and Secants (other) }]} \end{array}\right.$

Two equal circles intersect at point $C$. Through point $C$, two lines are drawn intersecting the given circles at points $A, B$ and $M, N$ respectively. Line $A B$ is parallel to the line of centers, and line $M N$ forms an angle $\alpha$ with the line of centers. It is known that $A B=a$. Find $N M$.
Drop perpendiculars from the centers of the circles to the lines $A B$ and $M N$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the circles, points $A$ and $M$ belong to the first circle, $B$ and $N$ belong to the second.

Drop perpendiculars $O_{1} X$ and $O_{2} Y$ to the line $A B$. Then $X$ and $Y$ are the midpoints of the chords $A C$ and $B C$. Therefore, $O_{1} O_{2} = X Y = A C / 2 + B C / 2 = A B / 2 = a / 2$.

Drop perpendiculars $O_{1} P$ and $O_{2} Q$ to the line $M N$. Then $P$ and $Q$ are the midpoints of the chords $M C$ and $N C$. Therefore, $P Q = M N / 2$.

Let $F$ be the projection of point $O_{1}$ onto $O_{2} Q$. Then $O_{1} F \| M N$. Consequently, $\angle F O_{1} O_{2} = \alpha$ and $O_{1} F = P Q$. Since $O_{1} F = O_{1} O_{2} \cos \alpha = a / 2 \cos \alpha$, then $M N = 2 P Q = a \cos \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-34.jpg?height=437&width=788&top_left_y=2308&top_left_x=635)

## Answer

## [ Pythagorean Theorem (direct and inverse). ]

Problem