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?
Best Answer
Hint. You are on the right track. Now the straight lines passes through the given point. Hence by plugging $(x,y)=(\frac {1}{\sqrt {3}}, 1)$ into the equation of the line we get $$\frac{a}{\sqrt{3}}+b+c=0.$$ Note that you may assume that $c=1$ (this line does not pass through the origin!). Then we solve the system $$\begin{cases} a^2+b^2=1\\\frac{a}{\sqrt{3}}+b+1=0 \end{cases}$$ We find TWO solutions. One solution is $a=0$ and $b=-1$, that is the line $y=1$.
What is the second solution?
P.S. From a geometric point of view, here we are looking for the lines which pass through $P=(\frac {1}{\sqrt {3}}, 1)$ and that are tangents to the circle centred at the origin of radius $1$. Since $P$ is outside the circle there are two such lines.