Q82 / 200 · Manipal (MET) 2012 · Mathematics
If $a\gt2 b\gt0$, then positive value of $m$ for which $y=m x-b \sqrt{1+m^2}$ is a common tangent to $x^2+y^2=b^2$ and $(x-a)^2+y^2=b^2$ is
- Correct answerA$\frac{2 b}{\sqrt{a^2-4 b^2}}$
- B$\frac{\sqrt{a^2-4 b^2}}{2 b}$
- C$\frac{2 b}{a-2 b}$
- D$\frac{b}{a-2 b}$
Solution
Given, $y=m x-b \sqrt{1+m^2}$ touches both the circles.
So, distance from centre = Radius of both the circles
$\frac{\left|-b \sqrt{1+m^2}\right|}{\sqrt{m^2+1}}=b$
$\begin{array}{cc}\text { and } & \frac{\left|m a-0-b \sqrt{1+m^2}\right|}{\sqrt{m^2+1}}=b \\ \Rightarrow & \left|m a-b \sqrt{1+m^2}\right|=\left|-b \sqrt{1+m^2}\right| \\ \Rightarrow & m^2 a^2-2 a b m \sqrt{1+m^2}+b^2\left(1+m^2\right) \\ \Rightarrow & =b^2\left(1+m^2\right) \\ \Rightarrow & m a-2 b \sqrt{1+m^2}=0 \\ \Rightarrow & m^2 a^2=4 b^2\left(1+m^2\right) \\ \Rightarrow & m^2\left(a^2-4 b^2\right)=4 b^2 \\ & m=\frac{2 b}{\sqrt{a^2-4 b^2}}\end{array}$
Hence, the correct answer is option: 'A'.