Question

If $a>2 b>0$, then the 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

• 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)

$x^2+y^2=b^2$ : centre $(0,0)$ radius $=b$
$(x-a)^2+y^2=b^2$ centre $(a, 0)$ radius $=b$

$y=m x-b \sqrt{1+m^2}$ is tangent to both circle
Since, when $x=0 y=-b \sqrt{1+m^2}<0$
when $y=0 x=\frac{b \sqrt{1+m^2}}{m}>0$
Now, perpendicular distance from centre $(a, 0)$ will be equal to radius $b$.
$\left|\frac{m a-0-b \sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=b$
$ \Rightarrow \left|\frac{m a-b \sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=b$
-ve sign $ \frac{m a-b \sqrt{1+m^2}}{\sqrt{m^2+1}}=-b $
$ \Rightarrow m a=0 \Rightarrow m=0 $ or $ a=0$ not possible.
+ve sign $ \frac{m a-b \sqrt{1+m^2}}{\sqrt{m^2+1}}=+b $
$ \Rightarrow m=\frac{2 b}{\sqrt{a^2-4 b^2}}$