Question

If $y=mx-b\sqrt{1+m^2}$ is a common tangent to $x^2+y^2=b^2$ and $(x-a{)}^2+y^2=b^2$, where $a>2b>0$, then the positive value of $m$ is

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

Answer 1

The correct option is A $\frac{2b}{\sqrt{a^2-4b^2}}$

Given common tangent is $y=mx-b\sqrt{1+m^2}$,
$\Rightarrow mx-y-b\sqrt{1+m^2}=0$
Given circles are
$x^2+y^2=b^2$ and
$(x-a{)}^2+y^2=b^2$
The centre and radius are
$C_1=(0,0),r_1=b{C}_2=(a,0),r_2=b$
Now, the distance from centre to the tangent is equal to radius, so
$\left|\frac{b\sqrt{1+m^2}}{\sqrt{1+m^2}}\right|=b\Rightarrow b=b$
And
$\left|\frac{ma-b\sqrt{1+m^2}}{\sqrt{1+m^2}}\right|=b\Rightarrow |ma-b\sqrt{1+m^2}|=b\sqrt{1+m^2}\Rightarrow ma-b\sqrt{1+m^2}=\pm b\sqrt{1+m^2}\Rightarrow ma=0,2b\sqrt{1+m^2}\Rightarrow ma=2b\sqrt{1+m^2}(\because m>0)$
Squaring both sides, we get
$\Rightarrow a^2{m}^2=4b^2+4b^2{m}^2\therefore m=\frac{2b}{\sqrt{a^2-4b^2}}$