Question

The polar equation of the circle on the line joining the points $(2,\frac{\pi }{3})$ and $(1,\frac{\pi }{6})$ as diameter is

• A. $r^2+r\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]+2\cos \frac{\pi }{6}=0$
• B. $r^2-r\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]+2\cos \frac{\pi }{6}=0$
• C. $r^2-r\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]-2\cos \frac{\pi }{6}=0$
• D. $r^2+r\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]-2\cos \frac{\pi }{6}=0$

Answer 1

The correct option is B $r^2-r\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]+2\cos \frac{\pi }{6}=0$

The polar equation of the circle on the line joining of $(r_1,{\theta }_1)$ and $(r_2,{\theta }_2)$ as diameter is

$r^2-\left[r_1\cos \right(\theta -{\theta }_1)+r_2\cos (\theta -{\theta }_2\left)\right]r+r_1{r}_2\cos ({\theta }_1-{\theta }_2)=0$
$\therefore$ for the points $(2,\frac{\pi }{3})$ and $(1,\frac{\pi }{6})$
So the equation becomes,
$r^2-\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]r+2\cos (\frac{\pi }{3}-\frac{\pi }{6})=0$
$r^2-r\left[2\cos \right(\theta -\frac{\pi }{3})+\cos (\theta -\frac{\pi }{6}\left)\right]+2\cos \left(\frac{\pi }{6}\right)=0$