Given: Radius of small circle (centre P), PA = PB = 5 cm.

Let M be the midpoint of the common chord AB. The line connecting the centers P and Q passes through M and is perpendicular to AB.
Given: Radius of small circle (centre P), $PA = PB = 5 \text{ cm}$.
Radius of big circle (centre Q), $QA = QB = 6 \text{ cm}$.
Length of common chord $AB = 8 \text{ cm}$.
Since M is the midpoint of AB, $AM = MB = \frac{1}{2} \times 8 = 4 \text{ cm}$.