Three Identical Tangent Circles.

Let the distance of the test line tangent to the third circle from the centre of the middle circle be $x$ . Then

$\dfrac {15}{x}=\dfrac {25}{5}\implies x=3$ .

Therefore $|\overline {AB}|=2\sqrt {5^2-3^2}=\boxed 8$ .

Label the figure as above. We note that $\triangle O_3PQ$ is a right triangle and by Pythagorean theorem , we have $PQ = \sqrt{25^2-5^2} = 10\sqrt 6$ . Let $O_2M$ be perpendicular to $PQ$ . We note that $M$ is the midpoint of $AB$ . Let $AM=BM = d$ .

Now $\triangle O_2PM$ and $\triangle O_3PQ$ are similar. Then $\dfrac {PM}{PO_2} = \dfrac {PQ}{PO_3} \implies PM = \dfrac {10\sqrt 6 \times 15}{25} = 6\sqrt 6$ .

Since the two chords $AB$ and $CD$ of the middle circle intersect outside the circle at $P$ , we have:

$\begin{aligned} PA \cdot PB & = PC \cdot PD \\ (PM - d)(PM + d) & = 10 \cdot 20 \\ (6\sqrt 6 - d)(6\sqrt 6 + d) & = 200 \\ 216 - d^2 & = 200 \\ d^2 & = 16 \\ \implies d & = 4 \\ AB & = 2d = \boxed 8 \end{aligned}$