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If $a > 2 b > 0$ then the positive value of m for which $y = m x − b 1 + m 2$ is a common tangent to $x 2 + y 2 = b 2$ and $x − a 2 + y 2 = b 2$ is

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a

2 b a 2 − 4 b 2

b

2 b a − 2 b

c

b a − 2 b

d

a 2 − 4 b 2 2 b

answer is A.

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Detailed Solution

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$y = m x − b 1 + m 2$ is a tangent to the circle $x − a 2 + y 2 = b 2$ apply d = r

$\Rightarrow b = \frac{m a - b \sqrt{1 + m^{2}}}{\sqrt{1 + m^{2}}} \Rightarrow 2 b = \frac{m a}{\sqrt{1 + m^{2}}} \Rightarrow \frac{a^{2}}{4 b^{2}} = \frac{1 + m^{2}}{m^{2}} \Rightarrow \frac{a^{2}}{4 b^{2}} - 1 = \frac{1}{m^{2}} \Rightarrow m = \frac{2 b}{\sqrt{a^{2} - 4 b^{2}}}$

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