In Figure, two circles with centres `A\ a n d\ B` and of radii `5\ c m\ a n d\ 3\ c m` touch each other internally. If the perpendicular bisector of segment `A B` meets the bigger circle in `P\ a n d\ Q` , find the length of `P Q`
`A \and\ B` are the centres of the circles with radii 5cm and 3cm respectively.
`C` is the mid-point of AB.
Extend `AB` upto `O` point on circumference of outer circle.
`AB=AO−BO=5−3=2cm` (since AO and BO are radii of larger and smaller circles)
`AC=AB/2=2/2=1cm`
now in right angled triangle AMP,`AC=1cm,AP=5cm`
by pythagoras therom , `AP^2=PC^2+AC^2`
`PC^2=sqrt(AP^2−AC^2)`
`PC^2=sqrt(5^2−1^2)`
Therefore,`PQ=2PC=2sqrt24`
`=4sqrt6cm (CP=CQ)`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the length of segment PQ.

### Step-by-Step Solution:

1. **Identify the Radii and Centers**:
- Let the radius of the larger circle (center A) be \( r_A = 5 \, \text{cm} \).
- Let the radius of the smaller circle (center B) be \( r_B = 3 \, \text{cm} \).

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