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 solve the problem step by step, we need to find the positive value of \( m \) such that the line \( y = mx - b\sqrt{1 + m^2} \) is a common tangent to the circles defined by the equations \( x^2 + y^2 = b^2 \) and \( (x - a)^2 + y^2 = b^2 \).

### Step 1: Understand the Tangent Condition
A line \( y = mx + c \) is a tangent to a circle if the perpendicular distance from the center of the circle to the line equals the radius of the circle.

### Step 2: Find the Distance from the Center of the First Circle
The first circle \( x^2 + y^2 = b^2 \) has its center at \( (0, 0) \) and radius \( b \). The distance \( d \) from the center \( (0, 0) \) to the line \( y = mx - b\sqrt{1 + m^2} \) can be calculated using the formula:
\[
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