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

Clearly, `y=ms-bsqrt(1+m^(2))` is a tangent to `x^(2)+y^(2)=b^(2)`. It will touch the circle `(x-a)^(2)+y^(2)=b^(2)`, if
`|(ma-0-bsqrt(1+m^(2)))/(sqrt(m^(2)+1))|=b`
`rArr |m a - b sqrt(1+m^(2))|=bsqrt(1+m^(2))`
`rArr ma-bsqrt(1+m^(2))=pm b sqrt(1+m^(2))`
`rArr ma=2b sqrt(1+m^(2)) or, ma=0`
`rArr m^(2)+a^(2)=4b^(2)+4b^(2)m^( 2)`
`rArr m = pm (2b)/(sqrt(a^(2)-4b6(2))rArr m = (2b)/(sqrt(a^(2)-4b^(2))) [:' m gt 0 ]`