Question

The equation of the tangents to the circle $x^2+y^2=9$ which make an angle of $\frac{\pi }{3}$ with the positive $x$-axis

• A. $x-\sqrt{3}y-6=0$
• B. $x-\sqrt{3}y+6=0$
• C. $\sqrt{3}x-y+6=0$
• D. $\sqrt{3}x-y-6=0$

Answer 1

Answer: (C), (D)

$\sqrt{3}x-y-6=0$

Let $y=mx+c$ be the tangents to the circle $x^2+y^2=9$ which make an angle of $\frac{\pi }{3}$ with the $x$-axis.
So, slope, $m=\tan \frac{\pi }{3}=\sqrt{3}$
So, equation of tangent become
$y=\sqrt{3}x+c$
$\Rightarrow \sqrt{3}x-y+c=0$
Distance of the above line from the centre $(0,0)$ of the circle is equal to the radius $\left(3\right)$ of the circle.
$\therefore \left|\frac{c}{\sqrt{3+1}}\right|=3$
$\Rightarrow c=\pm 6$

Hence, equation of the tangents are $\sqrt{3}x-y\pm 6=0$