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 the circles. So, distance from centre = Radius of both the circles \[\frac{|-b\sqrt{1+{{m}^{2}}}|}{\sqrt{{{m}^{2}}+1}}=b\] and \[\frac{|ma-0-b\sqrt{1+{{m}^{2}}}|}{\sqrt{{{m}^{2}}+1}}\] \[\Rightarrow \] \[|ma-b\sqrt{1+{{m}^{2}}}|=|-b\sqrt{1+{{m}^{2}}}|\] \[\Rightarrow \] \[{{m}^{2}}{{a}^{2}}-2abm\sqrt{1+{{m}^{2}}}+{{b}^{2}}(1+{{m}^{2}})\] \[={{b}^{2}}(1+{{m}^{2}})\] \[\Rightarrow \] \[ma-2b\sqrt{1+{{m}^{2}}}=0\] \[\Rightarrow \] \[{{m}^{2}}{{a}^{2}}=4{{b}^{2}}(1+{{m}^{2}})\] \[\Rightarrow \] \[{{m}^{2}}({{a}^{2}}-4{{b}^{2}})=4{{b}^{2}}\] \[\Rightarrow \] \[m=\frac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]
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