# Tangent Line to the Unit Circle (What an Uncreative Title)

During our bowl team practice last night, we ran into a 'tough' problem.

The problem appears in a contest designed to allow competitors 30 seconds to answer. I am wondering whether anyone reading this blog can find a 'fast' way to do this problem. I will summarize the work we did together in the classroom and see if someone has a more efficient method. We began by re-writing the line in slope-intercept form.

We know this line with slope -1 must be tangent to the unit circle. To help the kids visualize the problem (and because my drawings on the dry erase board aren't always to scale), we constructed the setting in Geogebra.

The slider c is tied to the red line. The red line in the photo is obviously NOT tangent, but we can use the slider to move c to a value that gets pretty close. We can empirically find a value of c that gets the line close to being tangent to the unit circle, but there should be another way to obtain a more precise result through other means. We can also tell there should be two possible values of c per the picture below.

We then began trying some algebra to see if we could find some relationships. Here is what I wrote on the board in typed form:

This didn't seem to be leading us anywhere. We abandoned that work temporarily to return to the graph and began making constructions. See the work below.

We know the height of both of the two colored triangles has to be (√2)/2, so we can infer the value of c that works for the tangent line entering the first quadrant is 2√2. Then by inspection, we can see the c value for the tangent line entering the third quadrant is the opposite, -2√2.

Does anyone have a different approach they would like to share? Or maybe a suggestion on how to speed this process up so the answer can be obtained in under 30 seconds?

## 2 thoughts on “Tangent Line to the Unit Circle (What an Uncreative Title)”

1. Here are some of the approaches I tried out: https://www.dropbox.com/s/eohtf8ych9gsb9a/Tangent%20to%20Unit%20Circle%20Problem%20-%20Aaberg.pdf

Probably the simplest approach I used included the following reasoning:
The tangent line to a circle is perpendicular to to the radius.
The slopes of perpendicular lines are opposite reciprocals, so if the tangent line has a slope of -1, the radius of the unit circle will have a slope of 1.
The points on the unit circle where the radius has a slope of 1 (where change in y and change in x are equal) are ( sqrt(2)/2, sqrt(2)/2 ) and ( -sqrt(2)/2, -sqrt(2)/2 ).
Once you have the slope and a point on a line you can write the equation of the line in point-slope form like y - sqrt(2)/2 = -1 ( x - sqrt(2)/2 ).
Once you have an equation of the line you can manipulate it into the form 2x + 2y = c like this:
y - sqrt(2)/2 = -x + sqrt(2)/2
x + y = sqrt(2)/2 + sqrt(2)/2
x + y = sqrt(2)
2x + 2y = 2*sqrt(2)
Thus c = 2*sqrt(2) and you can follow a similar argument with the other point to find c could be -2*sqrt(2).

I don't know if that is really a "fast" way of solving it that I would think of in 30 seconds, but it is the first way I thought of solving it and it seemed to be the most efficient.

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