Question

lf $a>2b>0$, 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. $\frac{\sqrt{a^2-4b^2}}{\sqrt{a-2b}}$
• C. $\frac{2b}{a-2b}$
• D. $\frac{b}{a-2b}$

Answer 1

Answer: (A)

$\frac{2b}{\sqrt{a^2-4b^2}}$

$y=mx-b\sqrt{(1+m^2)}$ is a tangent to $(x-a{)}^2+y^2=b^2$.
Hence, the distance of $y=mx-b\sqrt{(1+m^2)}$ from the center $(a,0)$ should be equal to the radius $b$.
$\left|\frac{ma-b\sqrt{1+m^2}}{\sqrt{1+m^2}}\right|=b$
$\Rightarrow |\frac{ma}{\sqrt{1+m^2}}-b|=b$
$\Rightarrow \frac{ma}{\sqrt{1+m^2}}=2b$
$\Rightarrow m^2{a}^2=(1+m^2)4b^2$
$\Rightarrow m=\frac{2b}{\sqrt{a^2-4b^2}}$ (taking the positive value)