If `a >2b >0,` then find the positive value of `m` for which `y=m x-bsqrt(1+m^2)` is a common tangent to `x^2+y^2=b^2` and `(x-a)^2+y^2=b^2dot`

Options

A. `(2b)/(sqrt(a^(2)-4b^(2)))`
B. `(sqrt(a^(2)-4b^(2)))/(2b)`
C. `(2b)/(a-2b)`
D. `(b)/(a- 2b)`

Answer

A

Explanation

given ,y= mx -` b sqrt( 1+m^(2))` touches both the circles , so distance from caentre = radius of both the circles.
`(|ma-0 -b sqrt(1+m^(2))|)/(sqrt(m^(2))+1)=b and (|-bsqrt(+m^(2))|)/(sqrt(m^(2)+1))=b`
` implies |ma-b sqrt(1m^(2)|=|-bsqrt(1+m^(2))|`
`implies m^(2) a^(2) - 2abm sqrt(1+m^(2))+b^(2)(1+m^(2))=b^(2)(1+m^(2))`
`implies ma - 2bsqrt(1+m^(2))=0`
`implies m^(2) a^(2) = 4b^(2)(1+m^(2))`
`therefore m= (2b)/(sqrt(a^(2)- 4b^(2)))`