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

The correct option is A

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

Explanation for the correct option:

Find the value of $m$:-

$y=mx-b\sqrt{1+m^2}$ touches both the circles. Since $y=mx-b\sqrt{1+m^2}$ is a common tangent to $x^2+y^2=b^2$and${(x-a)}^2+y^2=b^2$

So, Distance from center $=$ radius of both the circles

$\frac{|ma-0-b\sqrt{1+m^2}|}{\sqrt{m^2+1}}=b$ and $\frac{|-b\sqrt{1+m^2}|}{\sqrt{m^2+1}}=b$

$$\begin{gathered} \Rightarrow \frac{\left|ma-b\sqrt{1+m^2}\right|}{\sqrt{m^2+1}}=\frac{\left|-b\sqrt{1+m^2}\right|}{\sqrt{m^2+1}} \\ \Rightarrow \left|ma-b\sqrt{1+m^2}\right|=\left|-b\sqrt{1+m^2}\right| \\ \Rightarrow {\left(ma-b\sqrt{1+m^2}\right)}^2={\left(-b\sqrt{1+m^2}\right)}^2 \\ \Rightarrow m^2{a}^2-2abm\sqrt{1+m^2}+b^2\left(1+m^2\right)=b^2\left(1+m^2\right) \\ \Rightarrow ma-2b\sqrt{1+m^2}=0 \\ \Rightarrow m^2{a}^2=4b^2(1+m^2) \\ \Rightarrow m^2\left(a^2-4b^2\right)=4b^2 \\ \Rightarrow m^2=\frac{4b^2}{\left(a^2-4b^2\right)} \\ \Rightarrow m=\frac{2b}{\sqrt{a^2-4b^2}} \end{gathered}$$

Hence, the correct option is A.