This Preamble is to explain why we are investigating a particular Problem (see next tab) and how it ties into the course goals as well as the current course materials.

Course Goal(s) Targeted:

1. To help students develop and refine basic algebra skills by way of an integrated review of these skills as they are needed in the course.
2. To help strengthen students' general academic skills.
3. To promote algebraically, numerically, and graphically addressing a problem.

Course Material Connections

1. Coordinate Plane Geometry
2. Algebra
3. Numerical Computation
4. Symbolic Manipulation
5. Table of values

Create a table of values and then graph the circle:

\[ (x-2)^2 + (y-1)^2 = 4 \]

For my scratch work, I will start by calculating the input/output using either \(x=0\) or \(y=0\) to get some relatively easy calculations. \[ (x-2)^2 + (y-1)^2 = 4 \] \[ \begin{array}{rcl} \text{Let: }x=0&&\\ (0-2)^2 +(y-1)^2 &=& 4\\ 4 + (y-1)^2 &=&4\\ (y-1)^2 &=&0\\ y-1&=&0\\ y&=&1 \end{array} \]

This first set of algebraic calculations yields one point: \((0, 1)\)

\[ \begin{array}{rcl} \text{Let: }y=0&&\\ (x-2)^2 +(0-1)^2 &=& 4\\ (x-2)^2 +1 &=&4\\ (x-2)^2 &=&3\\ x-2 &=&\pm \sqrt{3}\\ x &=&2\pm \sqrt{3} \end{array} \]

This one is a bit more tricky, here I still get my answers, two points \((2-\sqrt{3}, 0)\) and \((2+\sqrt{3},0)\), but they are not easy for me to figure out how to graph since those exact values do not correspond to nice grid lines.

Of course, our goal is still to fill out a table. This is where experimenting numerically is going to prove more helpful. I tried the "easy" or "obvious" choices of the zeros, but we can also just systematically plug in numbers to figure things out. Let's build out a table of values using the calculations we have already performed as guides. I am going to put some \(?\)s in whenever there is an unknown corresponding value, we will work these out in a later step.

As for why start and end at the values I included in my table, there are simply relatively small values and are a starting point. I may refine, remove, or add more values as I perform my numerical computations.

\[ \begin{array}{|c|c|} \hline x&y\\\hline -2&?\\\hline -1&?\\\hline 0&1\\\hline 1&?\\\hline 2&?\\\hline ?&-2\\\hline ?&-1\\\hline 2-\sqrt{3}&0\\\hline 2+\sqrt{3}&0\\\hline ?&1\\\hline ?&2\\\hline \end{array} \]

Notice that we can let either \(x\) or \(y\) take on a value and still calculate the other. With this much calculation, you may be tempted to find some shortcuts; this is encouraged! The practice of mathematics is identifying patterns and looking for consistent rules that allow us to better understand a problem and get to the heart of things.

Let's do some of those calculations now: \[ (x-2)^2 + (y-1)^2 = 4 \] \[ \begin{array}{rcl} \text{Let: }x=-2&&\\ (-2-2)^2 +(y-1)^2 &=& 4\\ (-4)^2 + (y-1)^2 &=&4\\ 16+(y-1)^2 &=&4\\ (y-1)^2 &=&-12\\ \end{array} \]

Alright, seem to have hit a bit of a snag; what is going on here? I cannot square a value and end up with a negative (at least amongst the real number values), so I must conclude that when \(x=-2\), there are no corresponding \(y\) values! On the plus side, that does mean I can remove that row from my table! Let's keep going:

\[ \begin{array}{rcl} \text{Let: }x=-1&&\\ (-1-2)^2 +(y-1)^2 &=& 4\\ (-3)^2 + (y-1)^2 &=&4\\ 9+(y-1)^2 &=&4\\ (y-1)^2 &=&-5\\ \end{array} \]

Whelp. It happened again. If I haven't done so already, I am definitely checking my work to look for errors, but everything looks neat, tidy, and correct, so no help for it but to press on:

\[ \begin{array}{rcl} \text{Let: }x=1&&\\ (1-2)^2 +(y-1)^2 &=& 4\\ (-1)^2 + (y-1)^2 &=&4\\ 1+(y-1)^2 &=&4\\ (y-1)^2 &=&3\\ y-1 &=&\pm\sqrt{3}\\ y&=&1\pm\sqrt{3} \end{array} \]

Nice! Now I have two more points I can add to my table, \((1, 1\pm\sqrt{3})\) and \((1,1-\sqrt{3})\). Still, not great for graphing/sketching due to the lack of landing on a nice gridline, but progress! Let's do one more and see what we get:

\[ \begin{array}{rcl} \text{Let: }x=2&&\\ (2-2)^2 +(y-1)^2 &=& 4\\ (0)^2 + (y-1)^2 &=&4\\ 0+(y-1)^2 &=&4\\ (y-1)^2 &=&4\\ y-1 &=&\pm\sqrt{4}\\ y&=&1\pm 2\\ y=3&\text{or}&y=-1 \end{array} \]

Finally! We landed on some nice whole number values. I recommend you review this set of steps; what about this \(x\) value in particular helped us to land on such a nice value? Look at all the steps and see how each one contributed to this final answer. We now have two more points to add to our table: \((2,-1)\) and \((2,3)\)

Okay, now I have done this calculation so many times, I am quite tired of copying/pasting and re-writing my work. Looking back, I basically repeat the same few steps over and over again. This is where algebra can really save us some of this tedium. Instead of picking an \(x\) value and then solving for \(y\), what if I just started by solving for \(y\), and leave the \(x\) variable alone?

\[ \begin{array}{rcl} \text{Let: }x=x&&\\ (x-2)^2 +(y-1)^2 &=& 4\\ (y-1)^2 &=&4- (x-2)^2\\ y-1 &=&\pm\sqrt{4- (x-2)^2}\\ y &=&1\pm\sqrt{4-(x-2)^2} \end{array} \]

By doing the "harder" work of solving this symbolically for \(y\) (or \(x\)!) we can now more easily plug in \(x\) values to see what the corresponding \(y\) values are! Note that we still need that \(\pm\) in front of the square root, so this equation is not a function! But that's okay! We are just graphing this equation, so no need to worry about whether the equation represents a function.

If we do the same for \(x\) we can pretty quickly use those values to fill out our table:

\[ \begin{array}{rcl} \text{Let: }y=y&&\\ (x-2)^2 +(y-1)^2 &=& 4\\ (x-2)^2 &=&4- (y-1)^2\\ x-2 &=&\pm\sqrt{4- (y-1)^2}\\ x &=&2\pm\sqrt{4-(y-1)^2} \end{array} \]

Using these, we can see pretty quickly how we can get our table filled out: \[ \begin{array}{|c|c|} \hline x&y\\\hline -2&DNE\\\hline -1&DNE\\\hline 0&1\\\hline 1&1-\sqrt{3}\\\hline 1&1+\sqrt{3}\\\hline 2&3\\\hline 2&-1\\\hline DNE&-2\\\hline DNE&-1\\\hline 2-\sqrt{3}&0\\\hline 2+\sqrt{3}&0\\\hline 0&1\\\hline 4&1\\\hline 2+\sqrt{3}&2\\\hline 2-\sqrt{3}&2\\\hline \end{array} \]

For my solution, I will take out all those pesky radicals and DNEs and just focus on the nice grid values. Also, just by doing the work in the order I did, I actually ended up with a duplicated entry: \((0,1)\), so I will consolidate that into just one.

After doing some scratch work, I found at least 4 nice points to help figure out the graph of the circle: \[ \begin{array}{|c|c|} \hline x&y\\\hline 2&3\\\hline 2&-1\\\hline 0&1\\\hline 4&1\\\hline \end{array} \]

If I plot these on a nice grid and connect them in a circle, I should get something like this: