Question

If $a>2b>0$, then the 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)

$x^2+y^2=b^2$ : centre $(0,0)$ radius $=b$
$(x-a{)}^2+y^2=b^2$ centre $(a,0)$ radius $=b$

$y=mx-b\sqrt{1+m^2}$ is tangent to both circle
Since, when $x=0y=-b\sqrt{1+m^2}<0$
when $y=0x=\frac{b\sqrt{1+m^2}}{m}>0$
Now, perpendicular distance from centre $(a,0)$ will be equal to radius $b$.
$\left|\frac{ma-0-b\sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=b$
$\Rightarrow \left|\frac{ma-b\sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=b$
-ve sign $\frac{ma-b\sqrt{1+m^2}}{\sqrt{m^2+1}}=-b$
$\Rightarrow ma=0\Rightarrow m=0$ or $a=0$ not possible.
+ve sign $\frac{ma-b\sqrt{1+m^2}}{\sqrt{m^2+1}}=+b$
$\Rightarrow m=\frac{2b}{\sqrt{a^2-4b^2}}$