Angle between segments resting against a circle

Question

Motivation:

A couple of days ago, when I was solving this question, I had to consider a configuration like this

Now, I didn't intentionally make those two yellow bars stand at what appears to be a $90^{\circ}$-angle but it struck me as an interesting situation, so much so that I thought the following question might be an interesting one to solve.

The Question:

Given a circle of radius $r$, a horizontal line a distance $c>r$ from the circle's centre, and two points $A$ and $B$ on that line located as indicated in the picture below, find the angle $\Theta$ as a function of the parameters given ($a,b,c,r$). The blue and red lines passing through $M$ are tangents to the circle, at $P$ and $Q$ respectively.

Answer 1

I have a solution of the form $\tan\frac{\theta}{2} = f(a,b,c,r)$.

First, note that $\angle QOM = \angle POM$. This comes from the triangles $OPM$ and $OQM$ being similar. They're both rectangular (at $P$ and $Q$), they share the same hypotenuse ($\overline{OM}$), and $\overline{OP} = \overline{OQ} = r$. This also implies that $\overline{PM} = \overline{QM}$. Let $\xi \equiv \overline{PM}$.

Next, note that $\angle POM = \theta/2$. This results from the above and $\theta + (\pi/2 - \angle POM) + (\pi/2 - \angle QOM) = \pi$.

Next, note that $\tan\left(\angle POM\right) = \overline{PM}/\overline{OP} = \xi/r$, so

$$ \xi = r\tan\frac{\theta}{2} $$

We'll now seek an equation for $\xi$ in terms of the various parameters of the problem. Note the following relations: $$ \overline{AM} = \overline{AQ} - \overline{QM} = \overline{AQ} - \xi \qquad (1) $$ $$ \overline{OA}^{\,2} = \overline{OQ}^{\,2} + \overline{AQ}^{\,2} = r^2 + \overline{AQ}^{\,2} $$ $$ \overline{OA}^{\,2} = (a-b)^2 + c^2 $$

Similarly, $$ \overline{BM} = \overline{BP} + \overline{PM} = \overline{BP} + \xi \qquad (2) $$ $$ \overline{OB}^{\,2} = \overline{OP}^{\,2} + \overline{BP}^{\,2} = r^2 + \overline{BP}^{\,2} $$ $$ \overline{OB}^{\,2} = b^2 + c^2 $$

Then $$ \overline{AQ}^{\,2} = (a-b)^2 + c^2 - r^2 \qquad (3) $$ $$ \overline{BP}^{\,2} = b^2 + c^2 - r^2 \qquad (4) $$

Now let

$$ u \equiv \overline{AQ} = \sqrt{(a-b)^2 + c^2 - r^2} \qquad\mbox{and}\qquad v \equiv \overline{BP} = \sqrt{b^2 + c^2 - r^2} $$

From (1)-(4) and the law of cosines, $$ \overline{AB}^{\,2} = \overline{AM}^{\,2} + \overline{BM}^{\,2} - 2\,\overline{AM}\,\overline{BM}\,\cos\theta \qquad (5) $$ we see that we'll need $\cos\theta$ in terms of $\tan\frac{\theta}{2}$. That's easy: $$ \cos\theta = \frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}} = \frac{r^2 - \xi^2}{r^2 + \xi^2} \qquad (6) $$

Combining all of the above, we can get a quadratic equation for $\xi$, after some tedius but not particularly difficult algebra: $$ \left[ (u+v)^2 + 4r^2 - a^2 \right] \xi^2 - 4(u-v)r^2\,\xi + \left[ (u-v)^2 - a^2 \right] r^2 = 0 $$ whose solution is $$ \frac{\xi}{r} = \frac{2(u-v)r \pm \sqrt{ 4(u-v)^2r^2 - \left[ (u+v)^2 + 4r^2 - a^2 \right] \left[ (u-v)^2 - a^2 \right] }}{\left[ (u+v)^2 + 4r^2 - a^2 \right]} $$

The second term inside the radical, sans the negative sign, simplifies to $$ \left[ (u+v)^2 + 4r^2 - a^2 \right] \left[ (u-v)^2 - a^2 \right] = 4r^2(u-v)^2-4a^2c^2 $$ and we find $$ \frac{\xi}{2r} = \frac{(u-v)r \pm ac}{\left[ (u+v)^2 + 4r^2 - a^2 \right]} $$

Thus,

$$ \tan\frac{\theta}{2} = \frac{2\left[ (u-v)r \pm ac \right]}{\left[ (u+v)^2 + 4r^2 - a^2 \right]} $$ where $$ u \equiv \overline{AQ} = \sqrt{(a-b)^2 + c^2 - r^2} \qquad\mbox{and}\qquad v \equiv \overline{BP} = \sqrt{b^2 + c^2 - r^2} $$

It's not surprising that there should be two valid solutions, since $Q$ could also be on the side closer to $B$.

Answer 2

Set Cartesian coordinates on the plane so that the center of the circle is $(0,0)$ and the black line $\overline{AB}$ is parallel to the $x$-axis.

The coordinates of $A$ are $(-(a - b), c)$ and the distance $OA$ is $\sqrt{(a - b)^2 + c^2}$. Therefore \begin{align} \angle OAB & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c^2}}{c} \right) = \arctan \left( \frac{c}{a-b} \right) & \text{and} \\ \angle OAQ & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c^2}}{r} \right). \end{align}

The coordinates of $B$ are $(b, c)$ and the distance $OB$ is $\sqrt{b^2 + c^2}$, so \begin{align} \angle OBA & = \arcsin\left( \frac{\sqrt{b^2 + c^2}}{c} \right) = \arctan \left( \frac cb \right) & \text{and} \\ \angle OBP & = \arcsin\left( \frac{\sqrt{b^2 + c^2}}{r} \right). \end{align}

Since $\angle BAM = \angle OAB + \angle OAQ$ and $\angle ABM = \angle OBA - \angle OBP$, and $\theta = \pi - \angle BAM - \angle ABM$,

\begin{align} \theta & = \pi - (\angle OAB + \angle OAQ) - (\angle OBA - \angle OBP) \\ & = \pi - \arctan \frac{c}{a-b} - \arcsin \frac{\sqrt{(a - b)^2 + c^2}}{r} - \arctan \frac cb + \arcsin \frac{\sqrt{b^2 + c^2}}{r} . \end{align}

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Another approach: consider the figure below, which shows line $\overline{AB}$ and segments $\overline{OP}$ and $\overline{BP}$. It also shows the perpendicular from $O$ to $\overline{AB}$, which intersects $\overline{AB}$ at $C$ and $\overline{BP}$ at $R$.

From the original problem statement, we have $OP = r$, $BC = |b|$, and $OC = c$. Let $OR = |u|$, with $u$ positive if $R$ is between $O$ and $C$ as shown. Then $CR = |c - u|$ and $PR = \sqrt{u^2 - r^2}$, and by similar triangles,

$$ \frac{\sqrt{u^2 - r^2}}{r} = \pm\frac{c - u}{b}.$$

There are actually three cases represented here:

1. $b > 0$, shown in the figure;
2. $-r < b < 0$, in which case $R$ is on the extension of $\overline{OC}$ beyond $C$, $u > c$, and $\triangle BCR$ has (positive) leg lengths $-b$ and $u - c$; and
3. $b < -r$, in which case $R$ is on the extension of $\overline{OC}$ beyond $O$, $u < -r$, and $\triangle BCR$ has (positive) leg lengths $-b$ and $c - u = c + |u|$. This is the case that requires the "$-$" option of the $\pm$ sign.

In all three cases I assume $c > r$.

Squaring both sides of this equation and rearranging terms appropriately, we get:

$$ (b^2 - r^2)u^2 + 2cr^2 u - (b^2 + c^2)r^2 = 0. \tag 1$$

If $b^2 \neq r^2$ this is a quadratic equation in $u$, and it has roots

$$ u = \frac{-cr^2 \pm br \sqrt{b^2 + c^2 - r^2}}{b^2 - r^2}.$$

We want the positive root if $b > -r$ and the negative root if $b < -r$, so

$$ \angle CBP = \arccos \frac ru = \begin{cases} \arccos \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}} & \text{if $b > r$} \\ \arccos \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}} & \text{if $b < r$ and $b \neq -r$.} \end{cases}$$

But if $b = r$, then $ u = \dfrac{c^2+r^2}{2 c} $ and $$ \angle CBP = \arccos \frac ru = \arccos \frac{2cr}{c^2+r^2} ,$$ whereas if $b = -r$ then $\angle CBP = \frac\pi2$.

And oh, look, all of these formulas apply to $\angle BAQ$ in the original figure if we substitute $b - a$ for $b$ (and $b - a < -r$ provided that $\angle BAQ$ is acute, as shown), so if we assume $b > -r$,

\begin{align} \theta & = \pi - \angle BAQ - \angle CBP \\ & \begin{aligned} = \pi & - \arccos \dfrac{(a - b)^2 - r^2} {-cr + (a - b) \sqrt{(a - b)^2 + c^2 - r^2}} \\ & - \begin{cases} \arccos \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}} & \text{if $b > r$} \\ \arccos \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}} & \text{if $-r < b < r$} \\ \arccos \dfrac{2cr}{c^2+r^2} & \text{if $b = r$} \end{cases} \end{aligned} \\ & = \arcsin \dfrac{(a - b)^2 - r^2} {-cr + (a - b) \sqrt{(a - b)^2 + c^2 - r^2}} \\ & \qquad\qquad + \begin{cases} \arcsin \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}} & \text{if $b > r$} \\ \arcsin \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}} & \text{if $-r < b < r$} \\ \arcsin \dfrac{2cr}{c^2+r^2} & \text{if $b = r$} \end{cases} \end{align}

So that's just two trig functions, though there are three cases depending on the value of $b$. As suggested in a comment, we could get this down to one trig function if we could find the three sides of $\triangle ABM$ without using trigonometry; but I think this would involve saying something about the triangles $\triangle OPM$ and $\triangle OQM$, and I do not yet see how to do it.