Find the equation of straight lines through the point $(\dfrac {1}{\sqrt {3}}, 1)$ whose perpendicular distance from the origin is unity. Find the equation of straight lines through the point $\left(\dfrac {1}{\sqrt {3}}, 1\right)$ whose perpendicular distance from the origin is unity.
My Attempt:
Let the equation of line be $ax+by+c=0$.
Its distance from origin is $1$ unit. So,
$$\left|\dfrac {a.0+b.0+c}{\sqrt {a^2+b^2}}\right|=1$$
$$\left|\dfrac {c}{\sqrt {a^2+b^2}}\right|=1$$
What do I solve further?

Using the normal form of straight line, let the equation of the line be
$$x \cos \alpha + y \sin \alpha = 1$$
Now since the line passes through $\left ( \frac{1}{\sqrt{3}}, 1 \right )$, we get
$$\frac{\cos \alpha}{\sqrt{3}} + {\sin \alpha} = 1 \implies \alpha = \frac{\pi}{6} 
 \ \ or \ \ \frac{\pi}{2} $$
Hence we get the equation of line as $\frac{\sqrt{3}x}{2} + \frac{y}{2} = 1$ or $y = 1$.
