Question

If the line $L:3x-4y=0$ is rotated about the centre of the circle $(x-4{)}^2+(y-3{)}^2=25$ through an acute angle of $\theta$ in anticlockwise sense such that after rotation it becomes one of the members of the family of lines $x+\lambda y-3-\lambda =0,\lambda \in R$, then $\theta$ equals

• A. ${\tan }^{-1}2$
• B. ${\cot }^{-1}2$
• C. ${\tan }^{-1}1$
• D. ${\sec }^{-1}2$

Answer 1

The correct option is B ${\cot }^{-1}2$


$x+\lambda y-3-\lambda =0$
$\Rightarrow (x-3)+\lambda (y-1)=0$ always passes through $(3,1)$
Slope of the rotated line is $\frac{3-1}{4-3}=2$
Slope of the original line $L$ is $\frac{3}{4}$
$\therefore \tan \theta =\frac{2-\frac{3}{4}}{1+2\times \frac{3}{4}}=\frac{1}{2}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{2}\right)={\cot }^{-1}2$