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