Question

The value set of $\theta$ in $[0,2\pi ]$ for which $(1+\cos \theta ,\sin \theta )$ lies inside $x^2+y^2=1$, is

• A. $\left(\frac{2\pi }{3},\frac{5\pi }{3}\right)$
• B. $\left(\frac{2\pi }{3},\frac{4\pi }{3}\right)$
• C. $\left(\frac{\pi }{6},\frac{5\pi }{6}\right)$
• D. $(0,\frac{\pi }{2})$

Answer 1

The correct option is B $\left(\frac{2\pi }{3},\frac{4\pi }{3}\right)$

$S:x^2+y^2+2gx+2fy+c=0$

If $P(x_1,y_1)$ lie inside circle $S$, then $S_1<0$

$S_1:x_1^2+y_1^2+2g{x}_1+2f{y}_1+c<0$

$S:x^2+y^2=1$ and

$P(1+\cos \theta ,\sin \theta )$

Now,

$S_1:(1+\cos \theta {)}^2+{\sin }^2\theta -1<0$

$\Rightarrow 1+2\cos \theta <0$

$\Rightarrow \cos \theta <-\frac{1}{2}$

$\Rightarrow \theta \in (\frac{2\pi }{3},\frac{4\pi }{3})$

Hence, option B.