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

Answer 1

Answer: (A)

$\frac{2b}{\sqrt{(a^2-4b^2)}}$

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