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

Option: 1

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


Option: 2

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


Option: 3

\frac{2 b}{a-2 b}


Option: 4

\frac{b}{a-2 b}


Answers (1)

best_answer

The equation of common tangent is

\mathrm{ m x-y-b \sqrt{1+m^2}=0 \text { (given) } }                  ...(1)

The equation of circles are \mathrm{x^2+y^2=b^2}                ...(2)

and \mathrm{(x-a)^2+y^2=b^2}                                       ...(3)

The centre is (0,0) and radius is b for circle (2)

\mathrm{ \begin{aligned} & \Rightarrow \text { distance from center=Radius. } \\\\ & \Rightarrow \frac{\left|b \sqrt{1+m^2}\right|}{\sqrt{1+m^2}}=b \end{aligned} }

Similarly for circle (3)

\mathrm{ \begin{aligned} & \Rightarrow \frac{\left|a m-0-b \sqrt{1+m^2}\right|}{\sqrt{1+m^2}}=b \\\\ & \Rightarrow\left|a m-b \sqrt{1+m^2}\right|^2=b^2(\sqrt{1+m})^2 \\\\ & \Rightarrow m^2 a^4=4 a^2 b^2\left(1+m^2\right) \Rightarrow m^2 a^2=4 b^2\left(1+m^2\right) \\\\ & \Rightarrow m=\frac{2 b}{\sqrt{a^2-4 b^2}} \end{aligned} }

Posted by

himanshu.meshram

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