Trigonometry - Find the angle $\alpha$ between center and tangential lines of a circle

Question

I have been solving a few problems involving circles and various intersecting lines lately. Some are actually pretty complicated. Here is a particular one that takes some effort. (It has to do with the design work of an overpass.)

Question: The point P is the cross point between a line that passes through the center of a circle and a line that is tangential to the circle. The point Q is on the center line whose vertical distance to the tangential line is equal to its horizontal distance to the circle. Show that the angle between the center and the tangential lines is: $$\alpha = \tan^{-1}\left[ \frac{1+s-s\sqrt{2s(1+s)}}{s+(1+s)\ \sqrt{2s(1+s)}} \right]$$ where $s=d/r$ is the ratio of the distance $d$ and the radius of the circle with $d<r$. (It is noted that there are various ways to express the angle depending on setups and real effort may be to show the particular expression of interest given above.)

Answer 1

$$\left.\begin{align} \tan\theta &= \frac{d+r}{d} = \frac{1+s}{s} \\[6pt] \tan\phi &= \frac{\sqrt{2d(d+r)}}{r}=\sqrt{2s(1+s)} \end{align}\right\}\quad\to\quad \tan(\theta-\phi) = \frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi} = \cdots$$

Answer 2

Mark the center of circle O, the point the line touches the circle T and the the point the perpendicular from Q crosses the tangent line R.OT is perpendicular on PT and right triangles PQR and PTO are similar and we may write:

$\frac{r}{d}=\frac{PQ+d+r}{PR}=\frac{PQ}{PR}+\frac{d+r}{PR}= Cos(\alpha)+(1+\frac{r}{d}) Sin(\alpha)$ Where $PR=\frac{d}{Sin(\alpha)}$

If $\frac {r}{d}=s$ then we finally have:

$s=\frac{Sin(\alpha)+Cos(\alpha)}{1-Sin(\alpha)}=\frac{1+tan(\alpha}{1/Cos(\alpha)-tan(\alpha)}$

Considering:

$Cos (\alpha)=\frac{1}{\sqrt{1+tan^2(\alpha)}}$

We finally get:

$tan(\alpha)=\frac{s\sqrt{2(1+s)}-(s+1)}{2s+1}$

This formula must be modified; for example for $\alpha=0$ we have:

$s=\frac{Cos(0)+Sin(0)}{1-Sin(0)}=1$

And with formula for s=1 we have:

$tan (\alpha)=\frac{1\sqrt{2(1+1)}-(1+1)}{2\times 1+1}=0$

But for s=2 we have:

$tan(\alpha)=\frac{2\sqrt{2(2+1)}-(2+1)}{2\times 2+1}≈0.38$

And with given formula we have:

$tan (\alpha)=\frac{1+2-2\sqrt{2\times 2(2+1)}}{2+(1+2)\sqrt{2\times 2(1+2)}}≈-0.31$

Which does not seem correct.

Answer 3

I would outline only the steps in analytical geometry that lead you to the required relation:

Abbreviated sin,cos,tan for $\alpha$ as $(s,c,t)$

Equation of tangent $PVT$ to the circle containing variable point $V$

$$ \dfrac{y-r c }{x-r(1-s)} = t \tag1 $$

The Red line shown has equation for equal $d$

$$ (x+y)=0\, \tag2 $$

$$ s=\frac{y}{r}=\frac{d}{r} \tag3 $$

Plug 2) and 3) into 1)

$$ \dfrac{s-c_{\alpha}}{-s-(1-s_{\alpha})} =t\tag4; \dfrac{s-\dfrac{1}{\sqrt{1+t^2}}} {s+(1- \dfrac{t}{\sqrt{1+t^2}})} =-t $$

Cross multiply to bring $t$ terms to left of equation and with algebraic manipulation

$$s - \frac{1}{\sqrt{1+t^2}} = -st-t +\frac{t^2}{\sqrt{1+t^2}} \tag5 $$

$$ t^2 (s^2+2 s)+2 t s (1+s) + (s^2-1) =0 ;\tag6 $$

Quadratic equation solution:

$$ PP= (s+1)/(s+2) ; QQ=(s^2-1)/ (s^2+2s) ;\tag{7a}$$

$$ t = \tan \alpha = \sqrt{PP^2-QQ} - PP \tag{7b}$$

which coincides with formula you gave..

$$\tan \alpha = \frac{1+s-s\sqrt{2s(1+s)}}{s+(1+s)\ \sqrt{2s(1+s)}} \tag{7c}$$

The following sets have been checked in a geometric construction. You can also check them on Geogebra, Desmos etc.

$$ \alpha = ( 30^{\circ} ,36^{\circ} ,45^{\circ} ,60^{\circ});\tag{8a}$$

$$ s \approx (0.366,0.2951, 0.2072, 0.0981 ). \,\,\tag{8b}$$

Answer 4

\begin{align} \tan\alpha&=\frac{|FQ|}{|PQ|} =\frac{|IB_t|}{|PB_t|} \tag{1}\label{1} . \end{align}

Let $|PQ|=x$. Then

\begin{align} |PB_t|&= \sqrt{|PI|^2-|IB_t|^2} =\sqrt{(x+d+r)^2-r^2} . \end{align}

From \eqref{1} \begin{align} \frac dx&= \frac{r}{\sqrt{(x+d+r)^2-r^2}} ,\\ \frac{d^2}{x^2}&= \frac{r^2}{(x+d+r)^2-r^2} =\frac{r^2}{x^2+2x(d+r)+d(d+2r)} , \end{align}

\begin{align} (d^2-r^2)\,x^2+2\,d^2\,(d+r)\,x+d^3\,(d+2r)&=0 , \end{align}

\begin{align} x&= \frac{d^2(d+r+r\sqrt{2(1+r/d)})}{r^2-d^2} ,\\ \tan\alpha&= \frac dx= \frac{r^2-d^2}{d(d+r+r\sqrt{2(1+r/d)})} \\ &= \frac{1-s^2}{s^2+s+\sqrt{2\,s\,(s+1)}} . \end{align} Now, show that

\begin{align} \frac{1-s^2}{s^2+s+\sqrt{2\,s\,(s+1)}} - \frac{1+s-s\,\sqrt{2s(s+1)}}{s+(1+s)\sqrt{2s(s+1)}} \equiv0 . \end{align}