A) \[\frac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]
B) \[\frac{\sqrt{{{a}^{2}}-4{{b}^{2}}}}{2b}\]
C) \[\frac{2b}{a-2b}\]
D) \[\frac{b}{a-2b}\]
Correct Answer: A
Solution :
Given, \[y=mx-b\sqrt{1+{{m}^{2}}}\]touches both die circles. So, distance from centre is equal to radius of both the circles. \[\frac{|-b\sqrt{1+{{m}^{2}}}|}{\sqrt{{{m}^{2}}+1}}=b\]and\[\frac{|ma-0-b\sqrt{1+{{m}^{2}}}|}{\sqrt{{{m}^{2}}+1}}=b\] \[\Rightarrow \]\[|ma-b\sqrt{1+{{m}^{2}}}|=|-b\sqrt{1+{{m}^{2}}}|\] \[\Rightarrow \]\[{{m}^{2}}{{a}^{2}}-2abm\sqrt{1+{{m}^{2}}}+{{b}^{2}}(1+{{m}^{2}})\] \[={{b}^{2}}(1+{{m}^{2}})\] \[\Rightarrow \]\[ma-2b\sqrt{1+{{m}^{2}}}=0\]\[\Rightarrow \]\[{{m}^{2}}{{a}^{2}}=4{{b}^{2}}(1+{{m}^{2}})\]\[\Rightarrow \]\[m=-\frac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]
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