Question

A point on the line $(x,y)=(-3+r\cos \frac{\pi }{6},2+r\sin \frac{\pi }{6})$ which is at a distance of $2$ units from the point $(-3,2)$ is

• A. $(-3+\sqrt{3},3)$
• B. $(-3-\sqrt{3},1)$
• C. $(-3-\sqrt{3},3)$
• D. $(-3+\sqrt{3},1)$

Answer 1

Answer: (A), (B)

The correct options are
A $(-3+\sqrt{3},3)$
B $(-3-\sqrt{3},1)$


$(x,y)=(-3+r\cos \frac{\pi }{6},2+r\sin \frac{\pi }{6})$
$\Rightarrow x=-3+r\cos \frac{\pi }{6}$
$y=2+r\sin \frac{\pi }{6}$

There exist two points which are at a distance of $2$ units from $(-3,2)$
$\therefore x=-3\pm 2\cos \frac{\pi }{6}$
$y=2\pm 2\sin \frac{\pi }{6}$

$\Rightarrow (x,y)=(-3+\sqrt{3},3)\text{or}(-3-\sqrt{3},1)$