Question

In the adjoining figure, two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Answer 1

Two circles with centres A and B of respective radii 5 cm and 3 cm touch each other internally.
The perpendicular bisector of AB meets the bigger circle at P and Q.
Join AP.
Let PQ intersect AB at point L.
Here, AP = 5 cm
Then AB = (5 - 3) cm = 2 cm
Since PQ is the perpendicular bisector of AB, we have:
AL=$\frac{AB}{2}=1cm$
Now, in right angled Δ PLA, we have:
$A{P}^2=A{L}^2+P{L}^2$
⇒ $P{L}^2=A{P}^2-A{L}^2=24$
⇒ $PL=2\sqrt{6}$
Thus $PQ=2\times PL=4\sqrt{6}$