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Question
Two circle with centres 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.
Solution
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.
Therefore, AC = CB = `1/2` AB = 1 cm
In right ΔACP, we have AP2 = AC2 + CP2
`=>` 52 = 12 + CP2
`=>` CP2 = 25 – 1 = 24
`=> CP = sqrt(24)= 2 sqrt(6) cm`
Now, PQ = 2 CP
= `2 xx 2sqrt(6) cm`
= `4sqrt(6) cm`
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