[ Pythagorean Theorem (direct and inverse). ] [ Mutual position of two circles ]

A segment connecting the centers of two intersecting circles is divided by their common chord into segments of 5 and 2.

Find the common chord, given that the radius of one circle is twice the radius of the other.
The segment connecting the points of intersection of the circles is perpendicular to the line of centers.

## Solution

Let the circles with centers $O_{1}$ and $O_{2}$ and radii $r$ and $2r$ respectively intersect at points $A$ and $B$, and the segments $O_{1} O_{2}$ and $A B$ intersect at point $K$. The line of centers of two intersecting circles is perpendicular to their common chord, so triangles $A O_{1} K$ and $A O_{2} K$ are right triangles. Since $A O_{1} < A O_{2}$, then

$K O_{1} < K O_{2}$. Therefore, $K O_{1} = 2$ and $K O_{2} = 5$.

By the Pythagorean theorem, $r^{2} - 4 = 4 r^{2} - 25$, from which $r^{2} = 7$, and $A K^{2} = 7 - 2^{2} = 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-38.jpg?height=511&width=717&top_left_y=-1&top_left_x=676)

## Answer

$2 \sqrt{3}$.