Completing the Square (In Circle Equations)
In general, any equation of the form Ax2 + Ay2 + Bx + Cy + D = 0 will produce a circle.
Notice that the square terms have matching coefficients. See it?
If the squared terms have different coefficients, the graph won't be a circle. In fact, it will be an ellipse. An ellipse has an oval shape (like an egg).
We can use a technique called "completing the square" to rewrite such an equation so that we can identify the circle's center point (h,k) and the radius.
OUR GOAL: To reclaim the standard form of the given circle equation by factoring.
In order to factor the original equation, we will need to add what I call a "magic number" to BOTH sides of the equation. This magic number comes from treating the x-term and y-term
separately.
Sample: Complete the square given the equation 4x2 + 4y2 -24 + 32 - 4 = 0.
1) Do the squared terms have matching coefficients?
Yes. So, we have a circle here.
2) Can we make the squared terms have a coefficient of 1?
Yes, we can do this. How? By dividing EACH term in the equation by 4.
We now have an equation that looks like this: x2 + y2 - 6x + 8y - 1 = 0.
3) Group the x's and y's together.
We rewrite our equation to get: x2 - 6x + y2 + 8y + 1 = 0.
4) Complete the square for the x and y-terms SEPARATELY.
x2 - 6x + 32 + y2 + 8y + 42 = -1 + 9 + 16.
By the way, the number 3 was produced by division: 6/2 = 3 and 4 came from division:
8/2 = 4. So, 32 = 9 and 42 = 16. Okay?
5) Thanks to these magic numbers, we can now factor our equation.
Our equation becomes: (x - 3)2 + (y + 4)2 = 26
Keep in mind that the factored form of a circle equation reveals the center point (h,k) and the radius. In the above example, (3, -4) is the center point and the radius is sqrt26.
NOTE: Step 2 above is the most important to remember.
Now on to part 2:
OUR GOAL: To Understand "Completing the Square."
For example, consider the quadratic equation x2 + 12x - 3 = 0. We need to add what I call a "magic number" to BOTH sides of the equation so that we can "unmultiply" more easily.
1) Move the number term or 3 to the right hand side of the equation.
x^2 +12x = 3
2) To create the "magic number", we need to focus our attention on the coefficient of the "x term" or "middle term."
In this example, our middle number is 12. See it in the given equation? We now divide 12 by 2 and then square the results.
Doing this we get: 12/2 = 6^2 = 36. Here, 36 is our magic number.
3) Next: Add 36 to BOTH sides of the equation.
x^2 + 12x + 36 = 3 + 36.
x^2 + 12x + 36 = 39
4) Factor the left hand side of the equation.
x^2 + 12x + 36 = 39 becomes (x+6) (x+6) = 39
(x + 6) (x + 6) = 39 becomes (x-6)^2 = 39.
5) To isolate x, we can "unmultiply" BOTH sides of the equation by using the square root.
So, we get two equations.
x + 6 = + - sqrt39.
6) Finally, we can solve each of the two equations by subtracting 6 from BOTH sides.
Doing this we get: x = + -sqrt39 - 6
Often this equation will be written in the form x = -6+ -sqrt39.
NOTE: In general, we create the "magic number" in terms of completing the square by taking one-half of the coefficient of x or the middle term and squaring that quantity.
Special Note: Before trying to compute the magic number, you need to make sure that the coefficient on the x2 term is 1.
For example, if the original equation is 5x2 - 2x - 4 = 0, we first divide EACH TERM by 5.
Doing this we get: x2 -2x/5 - 4/5 = 0.
Then the magic number will compute as follows:
(1/2)(2/5) = 1/5.
So, (1/5)2 = 1/25.
By Mr. Feliz
(c) 2005
