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.
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=AB2=1cm
Now, in right angled Δ PLA, we have:
AP2=AL2+PL2
⇒ PL2=AP2−AL2=24
⇒ PL=2√6
Thus PQ=2×PL=4√6