[Pythagorean Theorem (direct and inverse)] [Theorem on the sum of the squares of the diagonals] Two circles with radii $\sqrt{5}$ and $\sqrt{2}$ intersect at point $A$. The distance between the centers of the circles is 3. A line through point $A$ intersects the circles at points $B$ and $C$ such that $A B=A C$ (point $B$ does not coincide with $C$). Find $A B$.
Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles, respectively. Drop perpendiculars $O_{1} M$ and $O_{2} N$ to the line $B C$ and denote

$A B=A C=2 x, O_{1} M=a, O_{2} N=b$. Then $b^{2}-a^{2}=\left(5-x^{2}\right)-\left(2-x^{2}\right)=3, O_{1} M^{2}=2-x^{2}, O_{2} N^{2}=5-x^{2}$.

The midline of the trapezoid $O_{1} M N O O_{2}$ is $1 / 2(b+a)$. On the other hand, it is the median of the triangle $O_{1} A O_{2}$, and therefore $(b+a)^{2}=2\left(O_{1} A^{2}+O_{2} A^{2}\right)-O_{1} O_{2}^{2}=5$ (see problem $\underline{55267}$ ).

Thus, $2 x^{2}=\left(2-a^{2}\right)+\left(5-b^{2}\right)=7-a^{2}-b^{2}, 9=(b-a)^{2}+4 x^{2}=(b-a)^{2}+14-2 a^{2}-2 b^{2}=14-(b+a)^{2}$, from which $(b+a)^{2}=5$,

$(b-a)^{2}=\left(b^{2}-a^{2}\right)^{2} /(b+a)^{2}=9 / 5, a^{2}+b^{2}=1 / 2\left((b+a)^{2}+(b-a)^{2}\right)=17 / 5, x^{2}=1 / 2\left(7-a^{2}-b^{2}\right)=9 / 5$. Therefore, $A B=$ $2 x=\frac{6}{\sqrt{5}}$

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-07.jpg?height=383&width=761&top_left_y=2207&top_left_x=657)

## Answer

Problem