Mastering Algebra - The Equation of the Circle

As one of the conic sections, the circle is probably the most important of these curves. When studying analytic geometry (the relationship between the algebraic formula for a curve and the actual graph) students are required to learn how to recognize the circle as well as to graph it. Here we discuss the simplest way to recognize this curve, put it into suitable algebraic form, and graph it on a coordinate grid.

The formal definition of the circle is the locus, or site, of points that are all equidistant from another fixed point. The set of points forming the circle outline its circumference; the fixed point is the center. The distance from the fixed point, or center, to any point on the periphery of the circle is the radius.

When in standard form, the equation of the circle takes on the following form: (x - h)^2 + (y - k)^2 = r^2. The center is located at the point (h,k) and the radius is r. Once we get the algebraic equation into such form, graphing could not be easier, as we simply plot the point (h,k) on our grid, and then go r units from this point up, down, to the left, and to the right. We then do our best to connect these points by a smooth circle.

To put the equation into standard form often requires a technique known as completing the square. As in life, the things we usually need require some work to get and this is no different in mathematics. Most equations are not so neat and tidy so as to be in standard form at first blush; therefore, we need to manipulate the equation a bit to get it into good form. This is not difficult however, and we shall show by example how this is done. Once in standard form, the center and radius are obvious and the graph becomes readily accessible.

Take the equation x^2 + y^2 + 2x + 4y - 4 = 0. This is obviously not in the form (x - h)^2 + (y - k)^2 = r^2. However, with a little manipulation, we can put this into such form. This procedure works no matter what the equation, as long as the equation is that of a circle. The only things that change are the numbers. Thus once you follow this procedure, you can put any equation which will produce a circle into standard form.

First isolate both the x and y terms and write as such: x^2 + 2x + y^2 + 4y - 4 = 0. Now bring the -4 over to the right side, and write as such: x^2 + 2x + y^2 + 4y = 4. We now complete the square on x and y by taking half of the coefficient of each and squaring both terms. Half of 2 is 1 and half of 4 is 2. Squaring each of these terms give 1 and 4, respectively, and adding them to both sides of the equation results in x^2 + 2x + 1 + y^2 + 4y + 4 = 4 + 5 = 9. Now we have two perfect square trinomials in x and y. These are always factorable into a form which puts both the x and y terms into standard form for the equation of the circle. The x^2 + 2x + 1 becomes (x + 1)^2 and the y^2 + 4y + 4 becomes (y + 2)^2. Notice that the h and k are -1 and -2, the opposite of what is inside parentheses. Notice also that the 1 and 2 are the terms which were derived by halving the coefficients of the x and y terms.

Thus we have x^2 + y^2 + 2x + 4y - 4 = 0 becomes (x + 1)^2 + (y + 2)^2 = 9. Observe that 9 is 3^2. Consequently, we have (x + 1)^2 + (y + 2)^2 = 3^2. Looking at this equation, we see that the center is (-1, -2) and the radius is 3. From this equation, we plot the center and move 3 units up, down, left, and right. We then draw a smooth curve. This procedure is exactly the same for every circle equation. The only things that change are the numbers.

You now have the tools to slay any circle equation or graph. Just follow the simple procedure above and you will be able to conquer any algebra problem that involves putting circle equations into standard form and graphing. After all, you probably have many other things to put your attention to, such as getting that new iPhone. Now you don't have to worry about circles any more. Enjoy.

Mastering Algebra - Recognizing Special Products and Their Factors - Part II

Now that we understand some key algebraic terminology, we are prepared to recognize some special products and to be able to factor them accordingly.  Herein we master how to recognize and factor both differences of perfect squares and perfect square trinomials.

Perfect squares are numbers which have square roots which are integers.  Thus 25, 36, and 49 are all perfect squares because their respective square roots are 5, 6, and 7.  When you have an algebraic binomial which is the difference of two squares, we can always factor this expression in a convenient fashion.  Let us examine a specific example.  Take x^2 - y^2.  Here x^2 and y^2 are the squares.  Because we are taking the difference, this expression is aptly named a "difference of two perfect squares."  We can always factor an expression like this as (x - y)(x + y).  If we insert perfect square coefficients in front of the variables, no matter.  We simply factor the expression by taking the square root of the number and the variable, and placing them in that "-", "+" pattern.  For example 49x^2 - 25y^2 is factored as (7x - 5y)(7x + 5y). 

The difference of two perfect squares has a nice application in short-cut arithmetic.  Namely, we can use this method to perform some lightning multiplications.  For example, take any two numbers that differ from a common "ten" by the same amount.  Specifically, take 36 and 44.  Both these numbers differ from 40, the "common ten," by 4 units.  If we want to multiply 36*44, we can get the answer immediately, which is 1584!  How?  Write 36 as 40 - 4 and 44 as 40 + 4.  Then 36*44 = (40 - 4)(40 + 4) which is 40^2 - 4^2, which is 1600 - 16, which is 1584.  Try some others out on your own to see the beauty of this method.

Perfect square trinomials comprise another set of special products.  These expressions are generated by the product of two identical binomials.  To wit, (x + 1)^2 = (x + 1)(x + 1).  When multiplied out, this yields x^2 + 2x + 1.  All perfect square trinomials are formed this way.  These expressions are such that the middle term coefficient is double the product of the square root of the constant term and the coefficient of first term.  In x^2 + 2x + 1, 2 = 2*1*1, 1 being the square root of the constant term 1, and 1 being the square root of the coefficient of x^2, which is also 1.  Whenever we have a trinomial that meets this condition, we can always factor it using the square of the binomial which meets these conditions. 

Take an example to make this clear.  Look at x^2 + 10x + 25.  The square root of 25 is 5.  The square root of 1, the coefficient of x^2 is 1; 2*5*1 is 10, which is the middle term coefficient.  Therefore x^2 + 10x + 25 can be factored as (x + 5)^2.   Take one more.  Look at x^2 - 12x + 36.  The square root of 36 is 6.  The square root of 1, the coefficient of x^2 is 1; 2*6*1 is 12, which is the middle term coefficient with a "-" sign in front.  No problem.  We adjust our product by writing a "-" instead of a "+" and get  x^2 - 12x + 36 = (x - 6)^2.  That's all there is to it.  You can now handle any perfect square trinomial on the planet!

Perfect square trinomials play a very important role in a process known as "completing the square."  This process is so named because geometrically, this involves making a square from a rectangle by "completing the square on the rectangle."  This method allows us to solve quadratic equations quite easily and also gives us the proof of the famous quadratic formula.  So if you're ever wondering where that strange formula on the cover of your algebra book came from, remember that a perfect square trinomial had something to do with it.  Wow, isn't it great when you learn where things come from?  See you next time...