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If a >2b >0, then find the positive valu...

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`

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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 \( x^2 + y^2 = b^2 \) and \( (x-a)^2 + y^2 = b^2 \), we will follow these steps: ### Step 1: Identify the equations of the circles The first circle is centered at the origin with the equation: \[ x^2 + y^2 = b^2 \] The second circle is centered at \( (a, 0) \) with the equation: ...
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