Two circles with radii 1 and $\sqrt{2}$ intersect at point A. The distance between the centers of the circles is 2.

Chord $A C$ of the larger circle intersects the smaller circle at point $B$ and is bisected by this point.

Find this chord.
Draw lines through the centers of the circles, perpendicular to the chord $A C$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles, respectively, $K$ be the projection of point $O_{1}$ onto the line $O_{2} B$, $P$ be the projection of point $O_{1}$ onto the line $A C$, and $F$ be the intersection point of the ray $O_{2} K$ with the smaller circle. Then $P$ is the midpoint of $A B$, and $K$ is the midpoint of $B F$. Let $A P = x$. Then $A C = 4 x$. We need to find $x$.

First method. $O_{1} K = P B = x$, $K B^{2} = O_{1} B^{2} - O_{1} K^{2} = 1 - x^{2}$, $O_{2} B^{2} = O_{2} A^{2} - A B^{2} = 2 - 4 x^{2}$. By the Pythagorean theorem, $\left(\sqrt{2 - 4 x^{2}} + \sqrt{1 - x^{2}}\right)^{2} + x^{2} = 4$, from which $x^{2} = 7 / 32$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-08.jpg?height=658&width=740&top_left_y=1020&top_left_x=676)

Second method. From the equality of triangles $A P O_{1}$ and $O_{1} K F$ (by leg and hypotenuse), it follows that point $O_{1}$ lies on the line $A F$. In triangle $A O_{2} F$, we know two sides and the median $O_{2} O_{1}$. By the formula from problem $\underline{55267}$

$O_{2} F^{2} = 2 O_{2} O_{1}^{2} + \frac{1}{2} A F^{2} - O_{2} A^{2} = 8$. By Heron's formula

$S_{O O F F} = \sqrt{\left(\frac{3}{2} + \sqrt{2}\right)\left(\sqrt{2} - \frac{1}{2}\right)\left(\sqrt{2} + \frac{1}{2}\right)\left(\frac{3}{2} - \sqrt{2}\right)} = \sqrt{\left(\frac{9}{4} - 2\right)\left(2 - \frac{1}{4}\right)} = \frac{\sqrt{7}}{4}$, from which $\quad x = \frac{2 S_{O_{1} O_{2} F}}{O_{2} F} = \frac{\sqrt{7}}{4 \sqrt{2}}$.

## Answer

$\sqrt{\frac{7}{2}}$