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`

Explanation

`y=mx-b sqrt(1+m^(2))` is a tangnet to the circle `x^(2)+y^(2)=b^(2)` for all vales of m.
If it also touches the circle `(x-a)^(2)+y^(2)=b^(2)` , then the length of the perpendicular from tis center (a,0) on this line is equal to the radiu b of the cirle, which give
`(ma-bsqrt(1+m^(2)))/(sqrt(1+m^(2)))=+-b`
Taking the negative value of the RHS, we get `m=0` . So, we neglect it.
Taking the positive value of the RHS, we get
`ma=2bsqrt(1+m^(2))`
or `m^(2)(a^(2)-4b^(2))=4b^(2)`
or `m=(2b)/(sqrt(a^(2)-4b^(2)))`