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?
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.
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 $y = 1$.
let $$y=mx+n$$ the equation of a line then we have $$1=\frac{m}{\sqrt{3}}+n$$ amd we get $$y=mx+1-\frac{m}{\sqrt{3}}$$ the Hessian normalfor is given by $$\frac{y-mx-1+\frac{m}{\sqrt{3}}}{ \pm\sqrt{1+m^2}}=0$$
Consider the family of tangents to the unit circle:
$x = cos (t)$; $y = sin (t)$.
The problem is to find the tangent(s) to the above circle passing through (1/√3,1).
Tangent to circle at $ (cos(t), sin(t))$:
$y - sin (t) = - cot (t) ( x - cos (t))$.
It passes through $(1/√3, 1)$:
$1 - sin (t)$ =
$- cot (t) (1/√3 - cos (t))$.
A bit of trickery:
$sin(t)( 1- sin (t))$ =
$- cos (t) (1/√3 - cos (t))$;
$sin (t) = - (1/√3)cos (t) + 1$;
$(√3/2) sin(t) +(1/2) cos (t) $=
$√3/2$;
$cos(π/6) sin (t) + sin (π/6)cos(t)$ =
$ sin (π/3)$;
$sin (t +π/6) = sin (π/3)$;
$t + π/6 = π/3$ ; or
$t + π/6 = π - π/3$;
$t_1 = π/6$, and $t_2 = (1/2)π$.
We find 2 tangents to the circle that pass through the given point:
1) $y - 1/2 = $
$- (3/√3) ( x - (1/2)√3)$;
2) $y - 1 = 0$;
