Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

We know that,

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 - 3) cm = 2 cm.

Also, the common chord PQ is the perpendicular bisector of AB.

Thus, AC = CB = $\dfrac{1}{2}AB = \dfrac{1}{2} \times 2$ = 1 cm.

In right ∆ACP, we have

⇒ AP² = AC² + CP² [By pythagoras theorem]

⇒ 5² = 1² + CP²

⇒ CP² = 25 - 1 = 24

⇒ CP = $\sqrt{24} = 2\sqrt{6}$ cm

Now,

PQ = 2 CP = 2 x $2\sqrt{6} = 4\sqrt{6}$ cm ≈ 9.8 cm.

Hence, the length of PQ is $4\sqrt{6}$ cm ≈ 9.8 cm.