Tangent, pencil of lines and angle on the unit circle

Problem

On the unit circle $x^{2}+y^{2}=1$ the point $P$ is given by the angle $\alpha=\dfrac{\pi}{6}$; let $\mathcal{O}=(1,0)$ and let $T$ be the intersection of line $OP$ with the vertical tangent $x=1$. a) Find the equation of line $OP$. b) Prove that $\sin^{2}\alpha+\cos^{2}\alpha=1$ for every $\alpha$. c) Write the coordinates of $T$ and prove that its ordinate equals $\tan\alpha$ (for $0\le\alpha\le\tfrac{\pi}{2}$). d) Find the pencil of lines through $T$ and the two tangents from $T$ to the circle. e) Let $P'$ be the second point of tangency (other than $\mathcal{O}$); what is $\alpha'=P'\hat{O}\mathcal{O}$?

Solution

a) $P=(\cos 30^\circ,\sin 30^\circ)=\left(\tfrac{\sqrt3}{2},\tfrac12\right)$; line $OP$ has slope $\tan 30^\circ$: $\;y=\dfrac{\sqrt3}{3}\,x$. b) By definition $P$ lies on the circle of radius $1$: $\overline{OP}^{2}=\cos^{2}\alpha+\sin^{2}\alpha=1$. c) $T$ has abscissa $1$ and lies on $y=x\tan\alpha$, so $T=\left(1,\tan\alpha\right)=\left(1,\dfrac{\sqrt3}{3}\right)$; its ordinate is $\tan\alpha\approx 0{,}5774$. d) Pencil through $T$: $y-\dfrac{\sqrt3}{3}=m(x-1)$. Imposing distance $1$ from $O$ gives $m=-\dfrac{\sqrt3}{3}$; the other tangent is the vertical line $x=1$ (tangent at $\mathcal{O}$). Tangents: $x=1$ and $y=\dfrac{\sqrt3}{3}(2-x)$. e) The foot of the perpendicular from $O$ to the second tangent is $P'=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$, corresponding to the angle $60^\circ$. $\boxed{\,y=\tfrac{\sqrt3}{3}x\,;\quad T=\left(1,\tfrac{\sqrt3}{3}\right)\,;\quad \alpha'=60^\circ=\tfrac{\pi}{3}\,}$