Question

Three circles of radii a, b, c ( a < b < c ) touch each other externally. If they have x – axis as a common tangent, then: (A) 1/√(a) = 1/√(b) + 1/√(c)
(B) 1/√(b) = 1/√(a) + 1/√(c)
(C) a, b, c are in A.P.
(D) √(a), √(b), √(c) are in A.P. [JEE Main 2019]

Answer 1

Answer:

what is the radii value

Answer 2

Three circles having radii a , b , c ( a < b < c ) touch each other externally and are having x – axis as in common, therefore  :  (B) 1 /√( b ) = 1 /√( a ) + 1 /√( c )

This can be found as follows:

• Since, x axis is common, therefore,

       √[ (a+b)² + (a-b)² ] + √[ (a+c)² + (a-c)² ]  =  √[ (b+ c)² - (b-c)² ]

• Upon solving the above equation, we get,

       √ab + √ac = √bc

       √ab + √ac -√bc = 0

• Thus,

        1/√(a) + 1/√(c) - 1/√(b) = 0

        1/√(a) + 1/√(c) = 1/√(b)