Completing the Square

Completing the Square (In Circle Equations)

In general, any equation of the form \(Ax^2 + Ay^2 + Bx + Cy + D = 0\) will produce a circle. Notice that the square terms have matching coefficients (A). 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.

We can use a technique called completing the square to rewrite such an equation so that we can quickly identify the circle's center point (h,k) and the radius.

OUR GOAL: To find the standard form of the given circle equation by factoring.

In order to factor the original equation, we will need to add a "magic number" to BOTH sides of the equation. Remember, of course, that we can always add something to both sides of an equation without unbalancing it. In this case it will help us get the equation into a more useable form. The magic number comes from treating the x-term and y-term separately.

Example: Complete the square given the equation \(4x^2 + 4y^2 - 24x + 32y - 4 = 0\)

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: \(x^2 + y^2 - 6x + 8y - 1 = 0\)

3) Group the x's and y's together:

We rewrite our equation to get: \(x^2 - 6x + y^2 + 8y - 1 = 0\)

4) Complete the square for the x and y-terms SEPARATELY.

\((x^2 - 6x + 3^2) + (y^2 + 8y + 4^2) - 1 = 3^2 + 4^2\)

See what happened? We took each of those X and Y sections and turned it into an easily factorable quadratic. By dividing the linear term coefficients (the 6 and the 8) by two and then squaring that, we found good numbers to add that make each section easily factorable. That's where the \(3^2\) and \(4^2\) terms came from on each side of the equation.

5) Thanks to these magic numbers, we can now factor our equation.

Our equation becomes: \((x-3)^2 + (y+4)^2 = 26\)

Center Point and Radius

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 \(\sqrt {26}\).

NOTE: Step 2 above is the most important to remember.

Now on to part 2:

OUR GOAL: To Understand Completing the Square.

What exactly did we just do in that problem? Where did all the extra numbers come from? Consider the quadratic equation \(x^2 + 12x - 3 = 0\). Can you easily factor that into an expression of the form \((x-h)^2\)? We need to add what I call a "magic number" to BOTH sides of the equation so that we can factor more easily.

1) Move the constant term (the 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 result. Why? Because that gives us the perfect value for a (x-h)^2 form!

Doing this we get: 12/2 = 6, and 6^2 = 36. Here, 36 is our magic number.

3) Next: Add 36 to BOTH sides of the equation.

\(x^2 + 12x + 36 = 39\)

4) Factor the left hand side of the equation.

\(x^2 + 12x + 36 = 39\) becomes \((x+6)(x+6) = 39\)

\((x+6)^2 = 39\)

And that will be one-half of the circle formula! Now see why we divided by 2 and then squared to find the "magic number"? When we multiply out the factored form we square the 6 to make the constant 36, and multiply it by 2 to get the linear term (the \(12x\)).

Special Note: Before trying to compute the magic number, you need to make sure that the coefficient on the \(x^2\) term is 1!

For example, if the original equation is \(5x^2 - 2x - 4 = 0\), we first divide EACH TERM by 5.

Doing this we get: \(x^2 - \frac{2x}{5} - \frac{4}{5} = 0\).

Then the magic number will compute as follows:

\(\frac{1}{2}\frac{2}{5} = \frac{1}{5}\).

So, \((\frac{1}{5})^2 = \frac{1}{25}\).