Question

If $a > 2b > 0,$ then 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

• A. $\frac{2 b}{ \sqrt{a^2 - 4 b^2 }}$
• B. $\frac{ \sqrt{a^2 - 4 b^2 }}{2 b} $
• C. $\frac{ 2 b}{ a- 2 b}$
• D. $\frac{ b}{ a- 2 b}$

Answer 1

Answer: (A)

Given,$ y = mx -b \sqrt{1+m^2} $ touches both the circles, so
distance from centre = radius of both the circles.
$ \frac{|m a - 0 -b\sqrt{ 1+ m^2 |}}{ \sqrt{ m^2 +1}} = and \frac{ |- b \sqrt{ 1 + m^2}|}{ \sqrt{m^2 +1}} =b $
$ \Rightarrow |m a -b\sqrt{ 1+ m^2 } | \, $ and $ \, |- 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) $
$ \therefore m= \frac{ 2 b}{ \sqrt{ a^2 - 4 b^2}}$