JEE Advance - Mathematics (2002 - No. 14)
If $$a > 2b > 0$$ then the positive value of $$m$$ for which $$y = mx - b\sqrt {1 + {m^2}} $$ is a common tangent to $${x^2} + {y^2} = {b^2}$$ and $${\left( {x - a} \right)^2} + {y^2} = {b^2}$$ is
$${{2b} \over {\sqrt {{a^2} - 4{b^2}} }}$$
$${{\sqrt {{a^2} - 4{b^2}} } \over {2b}}$$
$${{2b} \over {a - 2b}}$$
$${{b} \over {a - 2b}}$$
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