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

To find the positive value of \( m \) for which the line \( y = mx - b\sqrt{1 + m^2} \) is a common tangent to the circles defined by \( x^2 + y^2 = b^2 \) and \( (x - a)^2 + y^2 = b^2 \), we can follow these steps:

### Step 1: Identify the centers and radii of the circles
- The first circle \( x^2 + y^2 = b^2 \) has:
- Center: \( (0, 0) \)
- Radius: \( b \)

- The second circle \( (x - a)^2 + y^2 = b^2 \) has:
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