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3 circles of radii a,b,c (a<b<c) touch e...

3 circles of radii `a,b,c (a

A

a,b,c are in AP

B

`(1)/(sqrta)=(1)/(sqrtb)+(1)/(sqrtc)`

C

`sqrta, sqrtb,sqrtc " are in AP"`

D

`(1)/(sqrtb)=(1)/(sqrta)+(1)/(sqrtc)`

Text Solution

Verified by Experts

The correct Answer is:
B
According to given information, we have the following figure.

where, A, B, C are the centres of the circles
Clearly, AB = a + b (sum of radii) and BD = b - a
`thereforeAD=sqrt( (a+b)^(2)-(b-a)^(2))`
( usinf Pythagorus theorum in `DeltaABD`)
`=2sqrt(2ab)`
Similarly , AC = a + c and CE = c - a
`In DeltaACE, AE=sqrt((a+c)^(2)-(c-a)^(2))=2sqrtac`
Similarly, BC = b + c and CF = c - b
`therefore In DeltaBCF, BF = sqrt((B+c)^(2)-(c-b)^(2))=2sqrt(bc)`
`therefore AD+AE+BF`
`therefore 2sqrt(ab)+2sqrt(ac)+2sqrt(bc)`
`rArr" " (1)/(sqrtc)+(1)/(sqrtb)+(1)/(sqrta)`
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