Equation of a Circle

Once you have worked with functions for a while, inevitably you will begin to wonder "what about circles?"  You have explored all sorts of different equations and their graphs.  Perhaps you've even gone so far as to rotate graphs sideways and learn how to manipulate those equations as well.  But despite these seemingly more involved concepts, you've yet to come across circles.  For such an apparently simple shape, why have these not been included in such rudimentary graphing lessons.

Well, circles are a little different from what you've done so far.  For starters, circles technically aren't functions.  This may surprise you at first, but recall the vertical line test.  Would a circle pass such a test?  Of course not, because passing a straight vertical line through any point on a circle (except for the tangent points on the sides) would also intersect the circle on its opposite side.  Try it and you will see!  So, if a circle is not a function, then how do we handle describing their equations?  That is a little different, but really not much harder than your typical parabola graph.  Follow along and I will explain the equation of a circle in more detail.

For starters, let me just show you the equation for a simple circle.  Consider a circle with a center at the origin (0, 0) with a radius of 1.  We call this a unit circle.  Here is what the equation looks like:

It looks simple, right?  This equation can also be modified using similar concepts to how you would manipulate a function, such as a parabola.  There is a simple change to make to the equation that causes the graph of the function to translate left or right, and a second similar change you can make that results in a vertical translation of the graph.  In this case, depending on where you want to shift your circle, that is the variable you modify.

So, if you want to center your circle at (2, 0), which is a horizontal translation of 2 units to the right, you would change the above formula to include an (x-2)2 term.  Think of it as "if x=2, the whole x term becomes 0.  So 2 is its root."  Same thing applies to vertical translation.  If you change the y term to be (y-5)2, then you can see that the whole y term becomes 0 when y=5, so this means that the circle is centered at a height of y=5.  If we combine these two translations, we shift the circle to have a center of (2, 5), and the final equation looks like this:

Hopefully you can see how easy it can be to locate the center of a circle based on its equation, and how equally simple it can be to determine the equation of a circle just by a visual inspection.  However, to fully describe a circle, there is still something missing.  We haven't looked at what the "1" means.

To fully describe a circle mathematically, the only things that you need to know about it are the coordinates of its center, as well as its radius.  You may think that you should need more information, but think of it as the mathematical equivalent of a geometry set's compass.  To draw a perfect circle with a compass, all you do is put the point down at the center, set the radius, and spin the pencil around the paper.  You only needed those two pieces of information to be able to construct a circle.

Continuing with our example, the radius of our circle is described by the 1 term.  Technically, you can think of it as a 12 term, which provides for our circle to have a radius of 1 (or a unit circle).  To change our radius to 5, we would change the 1 to a 25, because just like the x and y terms, the radius term is also squared.  If we wanted a radius of 7, we would put the term as 49.  In general, this term is r2.

If we put all of these components together, we can come up with the general form for the equation of a circle.  We can include terms that allow for horizontal and vertical translation, as well as whatever radius we want.  Let's call the horizontal shift term "a" and the vertical shift term "b", and the radius "r".  In this case, here is the general form of the equation of a circle:

The equation describes a perfect circle, and doesn't allow for any stretching or compressing along either of the axes.  All that needs to be done is determine the circle's center point and radius, and you can easily fill in the relevant values.

As an example, consider the circle that has radius of 10, and is centered at (-5, 7).  Here is its equation:

Here is the opposite form of the question: what is the center and radius of the circle described by the following equation:

Quick analysis of this by comparison to the general form allows you to find that the center point is (-12, -2) and has a radius of 12.

Hopefully you can see that graphing circles is a little bit different from graphing more familiar functions, though still similar enough that the math concepts we've learned so far can easily be adapted to apply here as well.  In my next post, I would like to explain a little about just "why" the equation of a circle looks the way it does.  For teachers, this is also a good exercise to have students explore when first learning how to graph circles.  

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How to use LaTeX Equation Editor

This is a post aimed at anyone who has a reason to type math symbols.  It is about the LaTeX equation editor, which is an online math utility.  I'm sure that many people know of this already, but it was only recently shared with me, and so I would like to share with anyone else who will find this valuable.  In particular, those who I feel this will benefit are: teachers and education professionals; students doing math homework; graduate students preparing thesis material.  Of course, anyone having to put any kind of math equations into a document will appreciate this.

The LaTex math symbol code generator can be found online at http://www.codecogs.com/latex/eqneditor.php.  It's a little intimidating at first glance, but after a little bit of experimentation, you should be able to generate any math equations that you can write down on paper.

In the upper control panel, you can select the functions that you wish to include.  For example, if you click the a/b button (left side of the panel), this is the generator for a fraction expression.  You can put any level of detail into the equations, and nest them within each other as necessary.  When you select the function, you will see a code for that function appear in the large main rectangle directly underneath the control panel.  When you become a LaTeX professional, you can just type your function codes directly into this box.  Once you click the function buttons to generate code, you will also find sets of brackets {}, in which you can type numbers, letters, or insert additional function codes.  Beneath the box, you can find some formatting options, below which you will see your math expression, rendered as an image .gif file.  You can copy this image, and then paste it into your document.

For example, if I want to generate an image for y=mx+b, rendered in a nice, textbook-like font, here is how you would do it.  This is a very simple example.  All you would do is type this in the box, and then observe the rendered image below.  This one doesn't require specific selection of functions from the control panel.  Here is what it looks like:

Note how much more professional it looks, compared to regular in-line text.

Another example will demonstrate the function abilities.  Let's try to write y=x² + 2x + 1.  First, I would type "y=x" and then click the x^a function button, which denotes exponents.  This would make your code look like "y=x^{}".  Into the brackets, you click and type a 2, and then you can move the cursor to the end and just finish typing the remainder of the equation.  The final expression looks like this: y=x^{2}+2x+1, and here is the image file:

To make a slightly more complicated functions, let's try to write y = x^2/3/4.  You have to build complex equations from the outside-in.  So type "y=" to start, and then we need to select the fraction button, the a / b button.  That displays the code "y=\frac{}{}".  In this case, you type the numerator in the first brackets, the denominator in the second brackets.  However, the numerator has a fraction in it.  So, you click within the first brackets, and then type x and then add the new function, the exponent one again.  Into the additional brackets that appear, you can type 2/3, and then in the final set of brackets, type the 4 for the denominator.  Here is what the code looks like for this one:  y=\frac{x^{2/3}}{4}, and here is the image:

The best part about this code generator is that it updates the .gif file in real-time, so you don't have to type out a huge, complicated expression, and then hope you get it right.  If you make a mistake, you will see it right away, and then you can modify as necessary until you get what you want.  When you have it correct, you can right-click and save the equation image as an individual file, and then immediately erase your code and start over again with a new equation.

I find the LaTex equation editor to be extremely helpful, especially here on my blog when I want to show more complex expressions.  It does require having to go to a separate website, typing your code, and then coming back to your original document and inserting a file.  So, for ease, I often will just type straight text into my documents.  However, for that extra professional appearance, you cannot argue that the generated codes look fantastic!  If you have never used this before, I highly recommend that you give it a try!  If this is new to you, as it was to me recently, and you find value in this, then please click the +1 button below to share this!

Minimum Distance Between Two Parallel Lines

A lot of students seem to have difficulties in determining the minimum distance between two parallel lines.  Conceptually, if you have these two lines next to each other with the same slope, you can draw any number of different lines that can connect the two, though they all would be at different angles.  However, the shortest line that you can draw is the one that is perpendicular to them - the line that has no relative side to side travel to add extra length to the connecting line.  This post will hopefully explain simply how this can be done, while showing that it isn't really anything more difficult than identifying the information that you need, and then using a series of simpler concepts to get to the final solution.  You will hopefully find that these problems really aren't that difficult!  Please do me a favor and click the +1 button at the end to help me share my post!

Sometimes you will be given two lines (or line segments) or an equation to start with.  Here, let's start even further back, just for practice.  Let's begin just with two random ordered pairs that I've selected.

Find the equation of the line AB that crosses through the points A(2,1) and B(4,6).

As I described in my previous post here, you can determine the equation of this line quite easily.

Slope m = (y₂-y₁) / (x₂-x₁) = (6-1) / (4-2) = 5/2

Y-intercept b = y-mx = 1-(5/2)*2 = (-4)

Therefore: y = (5/2)x - 4.  This is the equation of our line.  You shouldn't have had problems following along through this step, though if you'd like some review, check out my post here.  Now that we have our first line, let's develop the rest of our problem.

What is the minimum distance between line AB and a parallel line CD that passes through point C(3,8)?

Recall that parallel lines, by definition, are lines that have the same slope, and so will never cross or come closer together.   Also, though you can fit any length of line in between these two starting lines, the shortest one you can draw between them is perpendicular.  Draw two lines and convince yourself of this!  The minimum distance between the two lines is a line perpendicular to them both. 

Here is a small recap of the information that we already know to help us answer this question, as well as a bit of an outline of how we will go about solving it.

Finding the distance between our line and point

• First line: y = (5/2)x - 4
• Point on second line = (3, 8).  FYI, you now have enough information to determine the equation of the second line, though it's not required right now.
• Shortest distance between the lines is an intersecting line that is perpendicular to both. 
• The distance can be calculated by the distance formula. What does this need?  We need two points. We have one given to us, and now we need the second point, which will lie on our first line.
• We find this second point at the intersection of the starting line and the perpendicular line that passes through the given point. How do we find this intersection?
• The intersection is where the equations of the two lines are equal.  We have the first equation, solved in the first step above.  How do we get the second equation, of the perpendicular line?
• The second equation comes from knowing the given point, and the slope perpendicular to our parallel lines.  Recall: the slope that is perpendicular to slope "m" is "-1/m".
• Solving the system of equations will yield our second point of interest.  And once we have the point, we can simply plug numbers into the distance formula to find our final answer.

It seems like a lot, but if you follow it through, you can see the logic behind all of the steps. Find one thing which leads to another, which leads to another, which leads us to the solution.  This is another good piece of advice for working with more complex problems.  Look at what you need to find, and try to step backwards to identify helpful steps that you can take to progress through the process. Ask yourself "what do I need to find this?" and then "what do I need to find that?" and you will then find that you have outlined your own strategy of how to solve the problem!

So then, let's get to the numbers.  To start, let's find the equation of the perpendicular line.  We have a point (3,8), and we want the slope that is perpendicular to the slope (5/2)... which is (-2/5). Now we can find our equation.

(y-y₁) = m(x-x₁)

y-8 = (-2/5)(x-3)

y=(-2/5)x + (6/5) +8

y=(-2/5)x + (46/5)

Our line and point, with the perpendicular minimum distance shown.

Now we want to find the intersection point of this perpendicular line with our first line (equation y=(5/2)-4). To do this, we solve the system of equations by setting them to be equal to each other, solve for the variable x, and then sub in this x value into either one of the original line equations to find y.  x and y are the ordered pair of our second coordinate.

Solve the following system:

y=(5/2)x - 4
y=(-2/5)x + (46/5)

(-2/5)x + (46/5) = (5/2)x - 4

(-2/5)x - (5/2)x = (-4) - (46/5)

(-4-25)x/10 = (-20-46)/5

(-29)x/10 = (-66/5)

x = (-66/5)(10) / (-29)

x = 660 / (5)(29)

x = 132/29 Here is our x value! (Kinda ugly, but sometimes that's what you get!)

Now we put that value into either one of the first two equations in our system, and solve for y.  Both will give the same answer... they have to! We're talking about the POINT where the two lines cross, and hence are EQUAL in both cases.

y = (5/2)x - 4

y = (5/2)(132/29) - 4

y = 660/58 - 4

y = (660 - 232) / 58

y = 428/58

y = 214/29 Here is our y value! (Just as ugly as x!)

Here's our system, with the intersection point now displayed.  Looks like our calculations are correct!

So, we have our point on the first line as (132/29, 214/29).  Now all we need to do is find the distance between this point, and the given point (3,8).  To do this we can use the distance formula.  I've explained the distance formula in another post, so I'll just go ahead and use it here.

distance, d = sqrt[(x2-x1)² + (y2-y1)²] 

(Again, my apologies for not being able to show a proper square root sign.)

d = sqrt[(3-132/29)² + (8-214/29)²]

d = sqrt[(-45/29)² + (18/29)²]

d = sqrt[(2025+324)/841]

d = sqrt[2349/841]

d = (9/29)*sqrt(29) This is the answer!

Unfortunately, we get a nasty looking answer.  But that doesn't make it any less correct.  Hopefully you were able to follow along with how I approached this problem, and how we arrived at the solution, because these are the kinds of steps and logical thinking that you will need to use.  In all likelihood, most questions that you will encounter in your math studies are going to look a whole lot friendlier than this monster.  As always, feel free to leave any comments or questions below, and I'll do my best to address them. :)  Also, please remember to click the +1 button below if this was helpful.  You can even tweet about it automatically by clicking this link.  Bookmark my site and come back again!

What is Perpendicularity?

"Perpendicular" is the term used in mathematics to describe two lines that intersect at right angles.  I recently introduced this concept in a separate post about the definition of perpendicular lines, but I thought it might be interesting to go into a little bit more detail about this rudimentary and familiar math concept.  For students who are just learning about graphing lines for the first time, this is undoubtedly sufficient, but for those more familiar with the concept, this post might provide additional insight.

I opened above by stating that we are talking about two intersecting lines that form a 90 degree angle with each other, though this definition can and should be expanded.   Though technically correct, there is more to the concept of "perpendicularity," as it's called when talking about this subject.  More correctly, this term applies to not only lines (and line segments), but planes (surfaces, not airplanes!) as well.  At first this may sound advanced, but if you think about it, it is completely obvious. If you stack two books together in a perpendicular arrangement, you essentially are viewing the intersection of two planes.  Lines are drawn on paper, but planes are like the three-dimensional versions that have depth., and you can readily find examples in real life of things that are at 90 degrees to each other.

That leads me to a second point about perpendicular lines and planes.  I have said that they intersect at 90 degrees, which is true again, but more accurately and mathematically you can say that the lines form two "congruent adjacent angles" (Wikipedia link).  This means that if you look at the perpendicular intersection in a T-shape, you see two angles next to each other (adjacent) that are the same (congruent).  And by the rules of geometry, since the angles along a straight line must add up to 180 degrees, this obviously means that the intersection must be composed of 90 degree angles.  Furthermore, if the lines extend through each other, the rules of geometry state that all angles around a point must sum up to 360 degrees, so again we have 90 degrees for each - four right angles.

Perpendicular lines

Perpendicular planes

Here's some additional information that you might find interesting.  The term "perpendicular" itself can be an adjective describing the lines, as in "the perpendicular lines are written in red ink."   Alternatively, it can be the noun, as in "the perpendicular to the ground rises to the sky."   Sorry, those examples aren't very creative!  Also, another word for perpendicular is "orthogonal," which can be used the same way to describe right angle intersections.  "Orthogonality"comes from the Greek for "straight angle" and refers to lines and planes at ninety degree angles, much like "perpendicularity."

I know these math definitions probably aren't going to make solving your graphing problems any easier, but I thought that it was good information to know!  It's a very important but basic concept that is introduced very early in math education, so hopefully I have explained it simply enough even for beginners to grasp.  Please click the +1 button below to share my post, and you can even tweet about my site if you like it!

Definition of Perpendicular Lines

A lot of traffic coming to my site lately has been specifically seeking to learn more about perpendicular lines.  At this point in the new school year, many math classes are just beginning to study graphing, and so perpendicular and parallel lines are often discussed along with other basic graphing concepts, such as slopes and intercepts.  In this post, I would like to give you a definition of perpendicular lines, maybe include a bit of review of some other graphing notes, and go over a few example questions involving them that you are almost guaranteed to encounter in your mathematics courses.  By the end of it, you should be an expert at recognizing and graphing perpendicular lines.

Let's start with the perpendicular lines definition.  It specifically is talking about the relationship between two different lines who intersect at a 90 degree angle.  Two lines that cross at 90 degrees are said to be perpendicular to each other.  A good example of this is the familiar x-y axis.  The y-axis is perfectly up and down with absolutely no slant to the side, whereas the x-axis is perfectly left to right with absolutely no deviation to the vertical.  You can easily see that they form right angles where they meet.  However, it is important to realize that the lines themselves can go in any direction, not strictly up/down and left/right.  The only critical part is that they intersect at 90 degrees.

On a side note, a corollary to this definition is that the four angles created by the intersection of two perpendicular lines are all 90 degrees.  Comparatively, the intersection of two non-perpendicular lines results in the formation of two acute angles (less than 90 degrees) and two obtuse angles (more than 90).  This also demonstrates the geometric law which states that the sum of the angles around a point equals 360 degrees.  This is a useful rule to remember when solving geometric proofs.

So then, now that you understand what this word means and how to visually recognize it on a graph, you may then wish to prove that your two lines do indeed meet the criteria.  How would you even go about that?  How can you determine if two lines are perpendicular?  Well, to do this, you need to know the mathematical equations of the lines… or, more specifically, you need to know the slopes of the two lines.  (Recall that the slope of a line is most simply expressed as "rise over run", which represents the ratio of vertical change to horizontal change.  The slope of the line, often abbreviated by "m", is easily solved by comparing two ordered pairs, and then performing the slope calculation m = y₂-y₁ / x₂-x₁.)  True perpendicular slopes will have the following relationship:

m₁ = -1/m₂

In words, this means that the slope of the first line is equal to the negative inverse slope of the second line.  Looks a bit complicated, but it's not really.  Let's take a look at an example.

Consider the lines described by the equations y = 2x - 1, and y = (-1/2)x + 2.  Are these perpendicular?

This is an example of perpendicular lines.

If you want to get some practice graphing perpendicular lines, you can go ahead and plot these curves.  They are already expressed in standard form, so it is simple to determine the slopes and y-intercepts, and you can also readily generate a table of values to plot points along the lines.  That's a great way to show that you know how to graph the lines, and in the end, you would end up with two lines crossing at 90 degrees.  But that would take an awful lot of time on a test to find a solution that can be found much more quickly and easily.  Just consider the relationship that I explained, and see if it is true in this example.  You can see that the slope of the first line is m = 2, and the slope of the second line is m = -1/2.  This precisely fits the mathematical description of perpendicular lines.  You don't even technically need to graph it out to be able to answer this!  Of course, a wise plan of attack for solving this problem would be to check this relationship first, and THEN graph it out to show that you are correct.  This comes from personal experience.  Always check your work!  ;)

With this information, you should be able to see that all you need to know about lines are their slopes to be able to say whether they are perpendicular or parallel.  Recall, parallel lines have the exact same slope.  The trick when working with this type of question is to realize the the intercept values can be 2 and 3847234, or absolutely anything else at all.  They equation of the line may look completely and extraordinarily different for each, though the only important part of them is their m values.  Keep this in mind, and don't get intimidated by complex and scary-looking equations!

Another type of question might ask you to determine the equation of a line perpendicular to a given line through a specific point.  This takes a bit more work, but it is based on the same concepts.  Let's try a question like this.

Find the perpendicular line to the line y = 2x - 1 that goes through point (4,0).

Here's the approach I would take to solve this.

1. First, recognize that you are given one of the lines' equations, so from that you can easily find its slope. 
2. Second, from the first slope, you can use the perpendicular relationship to determine the slope of the second line.  
3. Third, since you now have the slope and a point that lies on the second line, you can substitute numbers into y = mx + b to solve for b, and then rewrite in in terms of x and y to give the final equation.  

I will leave the actual work for you to try yourself, but the line in this case is the same as above, y = -1/2x + 2.

A third type of question might ask you to determine the perpendicular bisector for a given line segment.  A bisector is a line that perfectly splits another line into two equal pieces, but it can slice through at any angle.  On the other hand, a perpendicular bisector is one the does this at precisely 90 degrees.  If you can first determine what the exact midpoint of your line segment is, you can then apply the approach that I outline above to solve this question as well.

There is one last important point that I would like to make about this topic, and it is about notation.  You are not incorrect to simply state that "line AB is perpendicular to line CD" (or whatever your lines are called), but the shorthand symbol to show this is an upside-down T shape, ⊥.  The keyboard character is called the "up tack", though this term is more applicable to lattice theory, type theory, and logic.  I believe it is more appropriate to simply call it the "perpendicular sign."  So, in this case, you would simply state your answer as AB⊥CD.  That's it.  It's much simpler!

Finally, I thought I would just throw in a bit of trivia that I came across while researching this topic.  Who knows… you might be able to impress your teacher!  The word "perpendicular" originally arose in the late 14th century, and its etymology shows that it came from the Latin word "perpendiculum", which means "plumb line", and "perpendicularis", which means "vertical, as a plumb line".  A plumb line was a simple device which was composed of a small weight on the end of a string, and when holding it up, gravity pulls the weight straight down and the string represents perfectly vertical.  In relation to the ideally perfectly horizontal ground, you can see how they came up with this term.  It's not overly useful information, but you never know where extra trivia might come in handy.

So, with that information, you should now know lots about this subject, and now have no problems graphing perpendicular lines or analyzing and identifying them in either graphs or equations.  There are several different variations to the questions that you may encounter, but if you understand the basics of what it is that defines two perpendicular lines, then you should have no problems in coming up with the appropriate solutions!  Please let me know in the comments below if you would like any further explanation or examples, and don't forget to +1 my post below and follow me on Twitter!  I'm @MathConcepts.  You can even click here to tweet about my post!  Be sure to visit my follow-up post that discusses a bit more of this concept of perpendicularity.