Question

If equation of the chord joining the points on the circle $r=2a\cos \theta$ whose vectorial angles are $\alpha$ and $\beta$ is $r\cos (\theta -\alpha -\beta )=2\lambda \cos \alpha ,$ then $\lambda =$

• A. $a\cos \beta$
• B. $a\cos \alpha$
• C. $a\cos (\alpha +\beta )$
• D. $a\cos (\alpha -\beta )$

Answer 1

The correct option is A $a\cos \beta$

The equation of any straight line is
$x\cos \gamma +y\sin \gamma =p$
Putting $x=r\cos \theta$ and $y=r\sin \theta$ it becomes,
$p=r\cos \theta \cos \gamma +r\sin \theta \sin \gamma =r\cos (\theta -\gamma )$ .............$\left(1\right)$
when $p$ and $\gamma$ are unknown.
The radius vector corresponding to vectorial angle $\alpha$ is $2a\cos \alpha$ and that for $\beta$ is $2a\cos \beta$ then
From eqn$\left(1\right)$,
$p=2a\cos \alpha \cos (\alpha -\gamma )$ ..............$\left(2\right)$
and $p=2a\cos \beta \cos (\beta -\gamma )$ ..............$\left(3\right)$
From eqns$\left(2\right)$ and $\left(3\right)$, we get
$2\cos \alpha \cos (\alpha -\gamma )=2\cos \beta \cos (\beta -\gamma )$
$\Rightarrow \cos (2\alpha -\gamma )+\cos \gamma =\cos (2\beta -\gamma )+\cos \gamma$
$\Rightarrow \cos (2\alpha -\gamma )=\cos (2\beta -\gamma )$
$\therefore (2\alpha -\gamma )=(2\beta -\gamma )$
or $\gamma =\alpha +\beta$
Substituting the value of $\gamma$ in eqns$\left(1\right)$ and $\left(2\right)$, then we get
$\therefore r\cos (\theta -\alpha -\beta )=2a\cos \alpha \cos \beta$
$\therefore \frac{r\cos (\theta -\alpha -\beta )}{2\cos \alpha }=a\cos \beta =\lambda$
$\therefore \lambda =a\cos \beta$