1.

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 \[{{(x-a)}^{2}}+{{y}^{2}}={{b}^{2}}\], is            [IIT Screening 2002]

A. \[\frac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]
B. \[\frac{\sqrt{{{a}^{2}}-4{{b}^{2}}}}{2b}\]
C. \[\frac{2b}{a-2b}\]
D. \[\frac{b}{a-2b}\]
Answer» B. \[\frac{\sqrt{{{a}^{2}}-4{{b}^{2}}}}{2b}\]


Discussion

No Comment Found

Related MCQs