Graphing Circles

When a circle is centered at the origin (0, 0), the equation for graphing the circle is:

$x^2+y^2=r^2$

In this equation, r represents the radius of the circle. For example, a circle with center (0, 0) and radius 5 has the equation:

$x^2+y^2=5^2$

Or $x^2+y^2=25$

To visualize this, let's take a look at the graph of the circle:

We can graph a circle of any size centered at the origin by changing the right side of the equation.

For example:

• $x^2+y^2=16$
  • A circle with radius $\sqrt{16}$, or 4
• $x^2+y^2=64$
  • A circle with radius $\sqrt{64}$, or 8

We've seen how we can make the circle any size we want by changing the right side of the equation to change the radius. But how do we move it, so it's not centered at the origin?

If we apply what we learned in the "Transformations" section, this is fairly simple.

• To move the circle to the right by $h$ units, we replace $x$ with $(x-h)$.
• To move the circle up by $k$ units, we replace $y$ with $(y-k)$.

As a result, our general equation for a circle becomes:

${(x-h)}^2+{(y-k)}^2=r^2$

This equation can now represent a circle of any size, centered at any point in the coordinate plane.

• (h, k) gives us the coordinates of the center of the circle
• r is the radius of the circle

There are a couple of important things to note about this equation to avoid common mistakes:

• The signs of $h$ and $k$ are opposite of what you might expect.
  • To move the circle to the right (positive x direction), we subtract $h$ from $x$, not add it.
  • To move the circle up (positive y direction), we subtract $k$ from $y$
• The right side of the equation is the radius squared, not the radius itself.
  • To get the radius, we need to take the square root of the right side of the equation.

Let's look at a few examples:

• ${(x-3)}^2+{(y-4)}^2=25$
  • This is a circle with center (3, 4) and radius 5 (since 5² = 25).
• ${(x+2)}^2+{(y-6)}^2=9$
  • This is a circle with center (-2, 6) and radius 3 (since 3² = 9).

The standard form above tells us everything we need to know about the circle: its center and radius. But equations for graphing circles aren't always served to us in such a convenient form. Another common form is called general form.

The general form of the equation of a circle is:

$x^2+y^2+Dx+Ey+F=0$

In the equation above, D, E, and F are constants, so they could represent any number.

A couple examples of general form equations are:

• $x^2+y^2+6x-8y+9=0$
• $x^2+y^2+10x-4y+13=0$

Since general form doesn't directly give us much useful information, we will often want to convert it to standard form in order to find the center and radius of the circle.

To convert from general form to standard form, we will need to use a method called completing the square.

1. Group the x terms and y terms together

   First, we want to get the x terms next to each other, and the y terms next to each other. And we might as well bring the constant term to the other side of the equation while we're at it.

   In doing so, we get:

   $x^2+6x+y^2-8y=-9$

2. Complete the square for the x terms

   Ultimately, we want to make the x terms look like something like (x - h)², so we want to find a binomial that, when squared, gives us x² + 6x.

   So, we want to find a binomial (x + c) such that when we square it, we end up with x² + 6x as part of the result. There will be some additional constant term, but we'll deal with that later.

   The constant that accomplishes this is always half the coefficient of x, which in this case is 3. So, we want to replace the x² + 6x term with (x + 3)².

   But, these two expressions are not equal

   (x + 3)² = x² + 6x + 9, so our new expression is actually 9 greater than the original one.

   In order to keep the equation balanced, we can either subtract 9 from the left side, or add 9 to the right side. We'll do the latter because it will result in fewer steps overall.

   So, we get:

   ${(x+3)}^2+y^2-8y=-9+9$

3. Complete the square for the y terms

   Now we'll want to complete the square for the y terms as well.

   Using half the coefficient of y, we get the expression (y - 4)².

   When we FOIL this out, we get y² - 8y + 16, meaning that we have to add 16 to the right side of the equation to keep it balanced.

   So, we get:

   ${(x+3)}^2+{(y-4)}^2=-9+9+16$

   Now we just carry out the simple addition on the right side of the equation to get:

   ${(x+3)}^2+{(y-4)}^2=16$

And we're done! Now that we have it in standard form, we can easily see that the center is (-3, 4) and the radius is 4.