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)

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