Secant to a Circle

What is the equation of a line that is secant to a circle with radius \(r\) and center \((0,0)\)?

This question started as a challenge with a student. She wanted to draw a pentagram on a graphing calculator, and while she could do the five lines freehand, she needed the equation of a circle. So I wondered: How could we create the five lines using linear equations?

By the end of the day, I’d worked that out. Here are the seven functions I came up with, including two for a unit circle:

\[y = \sin\frac{\pi}{10}\\y = \pm\left(\sec\frac{3\pi}{10} + \tan\frac{3\pi}{10}\right)x + 1\\y = \pm\tan\frac{\pi}{5} + \sin\frac{\pi}{10}\left(1 – \frac{2}{1 – \tan^2\frac{\pi}{10}}\right)\\y = \pm\sqrt{1 – x^2}\]

But that lead me to the bigger question: Given two points on a circle, how can we create the secant line between them? Let \(\alpha\) and \(\beta\) be the angles of rotation of the two points along the circle, starting on the positive x-axis and going counterclockwise.

The coordinates of a point on a unit circle at a rotation of \(\theta\) are \(\left(\cos\theta, \sin\theta\right)\). So we’re looking for the equation of a line that passes through \(\left(\cos\alpha, \sin\alpha\right)\) and \(\left(\cos\beta, \sin\beta\right)\).

The slope of this line is \[m = \frac{\sin\alpha – \sin\beta}{\cos\alpha – \cos\beta}\] We can use these trig identities: \[\sin\theta – \sin\phi = 2\cos\frac{\theta + \phi}{2}\sin\frac{\theta-\phi}{2}\\\cos\theta – \cos\phi = -2\sin\frac{\theta + \phi}{2}\sin\frac{\theta-\phi}{2}\] to simplify the slope to \[m = -\cot\frac{\alpha+\beta}{2}\]

The equation of the line is hence \(y = -\cot\frac{\alpha+\beta}{2}x + b\). We next solve for \(b\) by using one of the points on the line: \[\sin\alpha =  -\cot\frac{\alpha+\beta}{2}\cos\alpha + b \Rightarrow b = \sin\alpha  + \cos\alpha\cot\frac{\alpha+\beta}{2}\]

Let’s see about simplifying this by first finding a common denominator. First, let \(\gamma = \frac{\alpha+\beta}{2}\). So \[\sin\alpha  + \cos\alpha\cot\frac{\alpha+\beta}{2} = \sin\alpha  + \cos\alpha\cot\gamma \\= \sin\alpha + \frac{\cos\alpha\cos\gamma}{\sin\gamma}\\= \frac{\sin\alpha\sin\gamma + \cos\alpha\cos\gamma}{\sin\gamma}\] Here are some more identities: \[\sin\theta\sin\phi = \frac{\cos(\theta – \phi) – \cos(\theta + \phi)}{2}\\\cos\theta\cos\phi = \frac{\cos(\theta – \phi) + \cos(\theta + \phi)}{2}\]

This gives us: \[\frac{\sin\alpha\sin\gamma + \cos\alpha\cos\gamma}{\sin\gamma} = \frac{\frac{\cos(\alpha – \gamma) – \cos(\alpha + \gamma)}{2} +\frac{\cos(\alpha – \gamma) + \cos(\alpha + \gamma)}{2}}{\sin\gamma} \\ = \frac{\cos(\alpha-\gamma)}{\sin\gamma}\]

Since \(\gamma = \frac{\alpha + \beta}{2}\), \(\alpha – \gamma = \alpha – \frac{\alpha + \beta}{2} = \frac{\alpha – \beta}{2}\). This gives us: \[b = \frac{\cos\frac{\alpha – \beta}{2}}{\sin\frac{\alpha + \beta}{2}}\]

Here is our current line equation: \[y = -\cot\frac{\alpha+\beta}{2}x + \frac{\cos\frac{\alpha – \beta}{2}}{\sin\frac{\alpha + \beta}{2}}\]

Factoring out the common sine denominator and rearranging the expressions to remove the initial negation yields: \[y = \frac{\cos\frac{\alpha – \beta}{2} – x\cos\frac{\alpha+\beta}{2}}{\sin\frac{\alpha + \beta}{2}}\]

A circle with radius \(r\) instead of the unit circle will require a shift in the \(y\)-intercept but not in the slope, so the equation satisfying the first question in this post is: \[y = \frac{r\cos\frac{\alpha – \beta}{2} – x\cos\frac{\alpha+\beta}{2}}{\sin\frac{\alpha + \beta}{2}}\]

What about the tangent to a circle? In this case, \(\alpha = \beta\), so  \[y = \frac{r\cos\frac{\alpha – \alpha}{2} – x\cos\frac{\alpha+\alpha}{2}}{\sin\frac{\alpha + \alpha}{2}} = \frac{r\cos 0 – x\cos\alpha}{\sin\alpha} \\ = r\csc\alpha – x\cot\alpha\]

This is more interesting if we rewrite it in standard form: \[x\cot\alpha + y = r\csc\alpha \Rightarrow \\ x + y\tan\alpha = r\sec\alpha\]

… which shows one relationship between the tangent of a circle and the tangent of a triangle.

We can also rewrite our secant line into standard form: \[x\cos\frac{\alpha+\beta}{2} + y\sin\frac{\alpha + \beta}{2} = r\cos\frac{\alpha – \beta}{2}\]

In summary, here are our tangent and secant lines: \[x + y\tan\alpha = r\sec\alpha\\x\cos\frac{\alpha+\beta}{2} + y\sin\frac{\alpha + \beta}{2} = r\cos\frac{\alpha – \beta}{2} \]

We can use these to make the original functions more consistent. Note that we need points at the following rotations: \(\frac{\pi}{10}, \frac{5\pi}{10}, \frac{9\pi}{10}, \frac{13\pi}{10}, \frac{17\pi}{10}\). We would combine all pairs except consecutive ones. I will leave this as an exercise for the reader.

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