Question

Given that $a > 2b > 0$ and that the line $y = mx - b \sqrt{ 1 + m^2}$ is a common tangent to the circles $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$ . Then the positive value of $m$ is

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

Answer 1

Answer: (D)

Given equation of circles are
$x^{2}+y^{2}=b^{2} \dots$(i)
and $(x-a)^{2}+y^{2}=b^{2} $
Equation of common tangent
$y=m x-b \sqrt{1+m^{2}} \dots $(i) (given)
Common tangent equation for second circle
$(x-a)^{2}+y^{2}=b^{2} \text { is } y=m(x-a)+b \sqrt{1+m^{2}} \ldots $ (ii)
By Eqs. (i) and (ii), we get
$ m x-b \sqrt{1+m^{2}}=m x-a m+b \sqrt{1+m^{2}} $
$\Rightarrow -b \sqrt{1+m^{2}}=-a m+b \sqrt{1+m^{2}} $
$\Rightarrow a m=2 b \sqrt{1+m^{2}} $
$\Rightarrow a^{2} m^{2}=4 b^{2}\left(1+m^{2}\right) $
$\Rightarrow a^{2} m^{2}=4 b^{2}+4 b^{2} m^{2} $
$\Rightarrow a^{2} m^{2}-4 b^{2} m^{2}=4 b^{2} $
$\Rightarrow m^{2}\left(a^{2}-4 b^{2}\right)=4 b^{2} $
$\Rightarrow m^{2}=\frac{4 b^{2}}{\left(a^{2}-4 b^{2}\right)}$
$ \Rightarrow m=\frac{2 b}{\sqrt{a^{2}-4 b^{2}}}$