Mastering Mathematics - The Equation of the Circle
Posted: Tuesday, April 26, 2011
by Joe Pagano
Math by Joe
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.
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)^2and 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 xand 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.
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