# Parameterizing a circle with the intersection point of two perpendicular lines.

I’ve been really taken with Desmos, an online calculator and easy to use graphing tool. My students have been using it for some time, and I’m especially happy with the “slider” tool that it offers. Whenever you put a letter into a function while graphing, it suggests a value to assign it, and lets you tune that value with the slider. This tool is similar to Mathematica’s Manipulate or Animate functions, which I’ve had success using in previous classes to show how a function depends on its parameters.

My year-long teacher’s Mathematica license recently expired, making it a bit tougher to install on a new device. While I do have access to an unsupported copy, Desmos has more than replaced M-ca for any of my presentation needs.

In a recent class, we were playing around with linear systems and intersecting lines. To show that a negative reciprocal slope leads to a perpendicular line, I assigned a slider to the value m, and made two linear equations with slopes of m and -1/m: $y=mx \qquad\text{and}\qquad y=-\cfrac{1}{m}\ x$

The slider has the nice effect of letting you rotate the lines to see that they’re always perpendicular. Play with it yourself, why don’t ya. You can animate or adjust the slope with the m slider on the left.

The kids were delighted by the pinwheel spinning of the lines as the slope was adjusted. To show that we weren’t limited to lines that passed through the origin, I tacked on a y-intercept to both of the equations, and asked the students, what do you think happens when I adjust the slope now?

My point was to show that the lines remain perpendicular. I would have been pleased to hear that the students could also predict that the point of intersection of the two lines would now move around, instead of be fixed at the origin.

One student went further, however: he was able to predict that the point of intersection of the two lines will always be fixed to a circle. You can adjust the slope once again, as well as the points the line are fixed to pass through using the sliders. Only adjusting the slope m keeps the point of intersection on a circle. You can also adjust the points the lines are forced through with the sliders below.

The student had seen a connection to his geometry class from the previous year. An inscribed angle is half the measure of the intercepted arc. An angle inscribing half the circle must then be a right angle. What this student had realized was the converse of this statement: that a right angle, formed by two perpendicular lines each forced to pass through particular points, must lie on a circle, and those two points are the endpoints of a diameter of that circle. I thought this was awfully insightful!

I figured it would be neat to try to show that this must be true on my own. Solving the system $\left\{ \begin{array}{c} y =m (x-a)+c\\\\ y=-\frac{1}{m}\ (x-b)+d \end{array} \right.$

gives the point $\left(\frac{a\thinspace m^2+\left(d-c\right)m+b}{m^2+1},\frac{d\thinspace m^2+\left(b-a\right)m+c}{m^2+1}\right)$

I thought this was really neat. We haven’t shown that this point lies on a circle yet, but assuming it does, it shows a way to parameterize a circle with m as the ratio of quadratics. Maybe this is something a mathematician would immediately recognize, but it’s new to me!

To show this does lie on a circle, I need to find an appropriate transformation that turns the above into the more familiar $\left( R\cos\theta + x_1 , R\sin\theta + y_1\right)$

for a circle of radius and center $(x_1,y_1)$. The obvious choice is to connect the slope of one of the lines to the angle on the circle: $m \rightarrow \tan\theta$

The parameterization becomes $\left(\frac{a\thinspace \tan^2\theta+\left(d-c\right)\tan\theta+b}{\tan^2\theta+1},\frac{d\thinspace \tan^2\theta+\left(b-a\right)\tan\theta+c}{\tan^2\theta+1}\right)$

This is where all your trig identities pay off. Those denominators become squared secants, letting you get rid of the fractions altogether. $\left(\frac{a\thinspace \tan^2\theta+\left(d-c\right)\tan\theta+b}{\sec^2\theta},\frac{d\thinspace \tan^2\theta+\left(b-a\right)\tan\theta+c}{\sec^2\theta}\right)$ $\bigg(\enspace a\thinspace \sin^2\theta+\left(d-c\right)\sin\theta\cos\theta+b\cos^2\theta\quad,\quad d\thinspace \sin^2\theta+\left(b-a\right)\sin\theta\cos^2\theta+c\cos^2\theta\enspace\bigg)$

I’m having a bit of difficulty with formatting here. I’ll have to just write it like so: $\begin{array}{c} x=a\thinspace \sin^2\theta+\left(d-c\right)\sin\theta\cos\theta+b\cos^2\theta \\\\y=d\thinspace \sin^2\theta+\left(b-a\right)\sin\theta\cos\theta+c\cos^2\theta\end{array}$

The middle bits of these should pop out: a sine times a cosine is a part of one of the double angle formulas: $2\sin\theta\cos\theta = \sin2\theta$

While we’re tossing in sine of a double angle, we might as well introduce the cosine of the double angle as well. This shows up from the squares: $\sin^2\theta = \cfrac{1-\cos2\theta}{2} \qquad \text{and} \qquad \cos^2\theta=\cfrac{1+\cos2\theta}{2}$

Our parameterization becomes $\begin{array}{c} x=a\thinspace\left(\cfrac{1-\cos2\theta}{2}\right) +\left(\cfrac{d-c}{2}\right)\sin 2\theta+b\left(\cfrac{1+\cos2\theta}{2}\right) \\\\y=d\thinspace \left(\cfrac{1-\cos2\theta}{2}\right)+\left(\cfrac{b-a}{2}\right)\sin2\theta+c\left(\cfrac{1+\cos2\theta}{2}\right)\end{array}$

What’s neat about this is that the center of the circle now falls out as a constant term at the end, and we’ve maintained some kind of symmetry with the sines and cosines. $\begin{array}{c} x= \left(\cfrac{b-a}{2}\right)\cos2\theta + \left(\cfrac{d-c}{2}\right)\sin2\theta + \left(\cfrac{a+b}{2}\right)\\\\ y= - \left(\cfrac{d-c}{2}\right)\cos2\theta + \left(\cfrac{b-a}{2}\right)\sin2\theta + \left(\cfrac{c+d}{2}\right)\end{array}$

Here’s where my trig knowledge stopped. The sine and cosines can be combined, though: a linear combination of sine and cosine should leave a single sine curve, but with a phase angle tossed in. $w \cos\theta + u\sin\theta = \sqrt{w^2+u^2}\enspace \sin\left(\theta+\arctan\frac{w}{u}\right)$

This is great! We’ve got a way to combine the a, b, c, ds to get something looking like a radius. $\begin{array}{c} x = R \sin\left(2\theta + \arctan\left(\cfrac{a-b}{d-c}\right)\right) +\left(\cfrac{a+b}{2}\right) \\\\ y =R \sin\left(2\theta + \arctan\left(\cfrac{c-d}{a-b}\right)\right) +\left(\cfrac{c+d}{2}\right)\end{array}$

with $R = \sqrt{\left(\cfrac{b-a}{2}\right)^2+\left(\cfrac{d-c}{2}\right)^2}$

At this stage, we need to do is turn that first sine into a cosine (using $\sin\theta = \cos\left(\theta-\frac{\pi}{2}\right)$. $\begin{array}{c} x = R \cos\left(2\theta + \arctan\left(\cfrac{a-b}{d-c}\right)-\cfrac{\pi}{2}\right) +\left(\cfrac{a+b}{2}\right) \\\\ y =R \sin\left(2\theta + \arctan\left(\cfrac{c-d}{a-b}\right)\right) +\left(\cfrac{c+d}{2}\right)\end{array}$

We’re left with one remaining question: are the phase angles the same? $\arctan\left(\cfrac{a-b}{d-c}\right)-\cfrac{\pi}{2} \quad \stackrel{?}{=}\quad \arctan\left(\cfrac{c-d}{a-b}\right)$

A couple more identities that I don’t have memorized clears this up: $\arctan(-x) = -\arctan(x) \qquad\text{and}\qquad \arctan\left(\cfrac{1}{x}\right) - \cfrac{\pi}{2} = -\arctan(x)$ $\longrightarrow \arctan(-x) = \arctan\left(\cfrac{1}{x}\right) - \cfrac{\pi}{2}$

This answers the question above: yes! Our parameterization is $\begin{array}{c} x = R \cos\left(2\theta + \arctan\left(\cfrac{c-d}{a-b}\right)\right) +\left(\cfrac{a+b}{2}\right) \\\\ y =R \sin\left(2\theta + \arctan\left(\cfrac{c-d}{a-b}\right)\right) +\left(\cfrac{c+d}{2}\right)\end{array}$

If we wanted to make it a bit nicer, replace: $\phi = 2\theta + \arctan\left(\cfrac{c-d}{a-b}\right)$

and we get a nice $\begin{array}{c} x = R \cos\phi +\left(\cfrac{a+b}{2}\right) \\\\ y =R \sin\phi +\left(\cfrac{c+d}{2}\right),\end{array}$

a circle centered at x=(a+b)/2 and y=(c+d)/2. Woof! Bark bark! Woof woof bark!