[Two tangents drawn from one point] [Auxiliary similar triangles]

Circle $C_{1}$ with radius $2 \sqrt{6}$ and center $O_{1}$ and circle $C_{2}$ with radius $\sqrt{6}$ and center $O_{2}$ are positioned such that $O_{1} O_{2}=\sqrt{70}$. Line $l_{1}$ is tangent to the circles at points $A_{1}$ and $A_{2}$, and line $l_{2}$ is tangent at points $B_{1}$ and $B_{2}$. Circles $C_{1}$ and $C_{2}$ lie on the same side of line $l_{1}$ and on opposite sides of line $l_{2}$, with $A_{1}, B_{1} \in C_{1}$ and $A_{2}, B_{2} \in C_{2}$. Points $A_{1}$ and $B_{1}$ lie on opposite sides of line $O_{1} O_{2}$. A line $l_{3}$ is drawn through point $B_{2}$, perpendicular to line $l_{2}$. Line $l_{1}$ intersects line $l_{2}$ at point $A$, and line $l_{3}$ at point $B$. Find $A_{1} A_{2}$, $B_{1} B_{2}$, and the sides of triangle $A B B_{2}$.
Let $E$ be the base of the perpendicular dropped from point $O_{2}$ to the radius $O_{1} A_{1}$. Then $E A_{1} A_{2} O_{2}$ is a rectangle, and $E O_{1} = O_{1} A_{1} - A_{1} E = \sqrt{6}$.

From the right triangle $O_{1} E O_{2}$, we find that $O_{2} E = \sqrt{O_{1} O_{2}^{2} - O_{1} E^{2}} = 8$.

Therefore, $A_{1} A_{2} = O_{2} E = 8$. Let $F$ be the base of the perpendicular dropped from point $O_{2}$ to the extension of the radius $O_{1} B_{1}$. Then $F B_{1} B_{2} O_{2}$ is a rectangle, and $F O_{1} = O_{1} B_{1} + B_{1} F = 3 \sqrt{6}, O_{2} F = \sqrt{O_{1} O_{2}^{2} - O_{1} F^{2}} = 4$.

Therefore, $B_{1} B_{2} = O_{2} F = 4$. Since $O_{2} B_{2} \perp l_{2}$, and the line $l_{3}$ is perpendicular to $l_{2}$ and passes through point $B_{2}$, points $O_{2}, B_{2}$, and $B$ lie on the same line. Let $B B_{2} = a, A B_{2} = b$ (the legs of the right triangle $A B_{2} B$), and $A B = c$ (the hypotenuse). From the equality of the segments of tangents drawn from a point to a circle, it follows that $A A_{2} = A B_{2} = b, A A_{1} = A B_{1} = A B_{2} = B_{1} B_{2} = b + 4, A A_{1} = A B_{1} = A_{1} A_{2} - A A_{2} = 8 - b$.

From the equation $b + 4 = 8 - b$, we find that $b = 2$. Right triangles $A B B_{2}$ and $O_{2} B A_{2}$ are similar by two angles (the angle at vertex $B$ is common), and the similarity ratio is $\frac{A B_{2}}{O_{2} A_{2}} = \frac{b}{O_{2} A_{2}} = \frac{\sqrt{2}}{\sqrt{3}}$. Therefore, $a = B B_{2} = \frac{\sqrt{2}}{\sqrt{3}} A_{2} B = \frac{\sqrt{2}}{\sqrt{3}} (c + 2)$. From the system $a^{2} + 2^{2} = c^{2}, a = \frac{\sqrt{2}}{\sqrt{3}} (c + 2)$, we find that $c = 10, a = 4 \sqrt{6}$.

## Answer

$A_{1} A_{2} = 8, B_{1} B_{2} = 4, A B_{2} = 2, A B = 10, B B_{2} = 4 \sqrt{6}$.