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If \mathrm{ a>2 b>0, }then positive value of \mathrm{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

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



 


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Given \mathrm{m>0}..........(i)

\mathrm{a>2 b>0}.............(ii)

\mathrm{y=m x-b \sqrt{\left(1+m^2\right)}} or \mathrm{m x-y-b \sqrt{\left(1+m^2\right)}=0}........(iii)

(iii), is tangent to the to the circle \mathrm{x^2+y^2=b^2}...........(iv)

But (iii) is common tangent to another circle

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

Length of perpendicular from centre \mathrm{ (a, 0)= \pm\: radius = \pm b }

\mathrm{ \Rightarrow \quad \frac{m a-b \sqrt{\left(1+m^2\right)}}{\sqrt{\left(1+m^2\right)}}= \pm b \quad \Rightarrow m a-b \sqrt{\left(1+m^2\right)}= \pm b \sqrt{\left(1+m^2\right)} }...(vi)
if we take ' + ' sign in (vi), then \mathrm{m a=2 b \sqrt{\left(1+m^2\right)}}

\mathrm{\Rightarrow \quad m^2 a^2=4 b^2\left(1+m^2\right) \Rightarrow m^2=\frac{4 b^2}{a^2-4 b^2}}

\mathrm{\Rightarrow \quad m=\frac{2 b}{\sqrt{\left(a^2-4 b^2\right)}}}  by (i)

if we take '-' sign in (vi), we get \mathrm{m a=0\: or \: m=0} which is not possible by (i).

Hence option 1 is correct.

Posted by

rishi.raj

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