Question

lf the distance between the points $(a\cos \theta ,a\sin \theta ),(a\cos \varphi ,asin\varphi )$ is $2a$ then $\theta =$.

• A. $2n\pi \pm \pi +\varphi ,n\in z$
• B. $n\pi \pm \frac{\pi }{2}+\varphi ,n\in z$
• C. $n\pi -\varphi ,n\in z$
• D. $2n\pi +\varphi ,n\in z$

Answer 1

The correct option is A $2n\pi \pm \pi +\varphi ,n\in z$

Given points $(a=cos\theta ,a\sin \theta ),(a\cos \varphi ,a\sin \varphi )$

distance $2a$

distance between $(x_1,y_1),(x_2,y_2)$

$\sqrt{(x_2-x_1{)}^2+(y_2-y_2{)}^2}$

So,

$\sqrt{(a\cos \theta -a\cos \theta {)}^2+(a\sin \varphi -a\sin \theta {)}^2}=2a$

Squaring on both sides

$\Rightarrow a^2({\cos }^2\theta +{\cos }^2\varphi -2\cos \theta \cos \varphi )+a({\sin }^2\varphi +{\sin }^2\theta -2\sin \theta -\sin \varphi )=4a^2$

$\Rightarrow a^2({\sin }^2\theta +{\cos }^2\theta +{\sin }^2\theta +{\cos }^2\theta -2(\sin \theta -\sin \varphi +\cos \theta \cos \varphi \left)\right)=4a^2$

$\Rightarrow a^2(1+1-2(\cos (\theta -\varphi ))=4a^2$

$\Rightarrow 2-2\cos (\theta -\varphi )=4$

$\Rightarrow 2\cos (\theta -\varphi )=-2$

$\Rightarrow \cos (\theta -\varphi )=-1=\cos \left(\pi \right)$

$\Rightarrow \theta -\varphi =2n\pi \pm \pi ,n\in z$

$\Rightarrow \theta =2n\pi \pm \pi +\varphi ,n\in z$