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Objective
Common Core Text:
[G.GPE.1] Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.
Said Differently:
Derive the equation of a circle
Given an equation of a circle, find the center and radius of the circle
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Explanation
If we have a coordinate plane, we can use an equation to describe the position and size of a circle. This is useful if we want to find important points on the circle, do calculus, etc.
Let's see how the equation of a circle was created.
Here we have a circle in a coordinate plane. The center of the circle is at (h, k) the radius is r. We want to find an equation that will give us all points (x, y) which are the points that lie on the circumference of the circle. This picture shows an example point P, but P could be anywhere on the circle.
The trick to finding the equation is that we can make a right triangle by connecting C and P to point A. Now lets use the Pythagorean Theorem.
a2 + b2 = c2
CA2 + PA2 = r2
Now we just need to figure out how long these sides are. Lets start with CA.
CA starts at C and ends at A, and goes in the x direction. So, we can just compare the x-coordinates of C and A.
The x-coordinate of C is h
We aren't told the x-coordinate of A
But we can see A is the same distance along the x-axis as P is.
So we can use the x-coordinate of P, which is x
To find the distance something travels, you simply subtract the starting point from the ending point. (For example, if you started 5 ft from the origin and ended at 8 ft from the origin, you would know that you traveled 3 ft because 8 - 5 = 3). We'll do the same thing for CA, and we get
CA = x - h
If we do the same thing for side PA, we get
PA = y - k
Now let's substitute this into the Pythagorean Theorem equation we started with
CA2 + PA2 = r2
(x - h)2 + (y - k)2 = r2
And that's the famous general equation for a circle
(x - h)2 + (y - k)2 = r2
where r = radius and (h, k) is the center of the circle)
Using the Equation of a Circle
Simplest example
Let's say you're working on a coordinate plane and you want to create a circle with center (3, 5) and a radius of 6. What would your equation be?
Easy, just substitute into the general equation of a circle
(x - h)2 + (y - k)2 = r2
where r = radius and (h, k) is the center of the circle)
We get:
(x - 3)2 + (y - 5)2 = 62
We could leave the answer like this, or expand it out.
Completing the squares
Sometimes you'll be doing some calculus or algebra and you'll get an answer like this:
x2 + y2 - 4x + 2y = 4
You got this answer, but you really don't know what it means because you don't know what shape it is, where it's positioned, or how big it is. Because you know geometry, you think it might be a circle, so you want to make the equation look like (x - h)2 + (y - k)2 = r2
How can you do that? Well, we can use a method called completing the square. Completing the square is a strange method that takes a lot of practice to get the hang of. The bottom line is, we want to factor parts of this equation, but we don't have everything we need yet. Let's start by rearranging the equation, grouping our x terms together and our y terms together
x2 - 4x + y2 + 2y = 4
First, let's just look at the x terms
x2 - 4x
What would we need to add in order to factor this? Well, if we had
x2 - 4x +4
We could factor it into
x2 - 4x +4 = (x - 2)2
Why would we want to do that? Because (x - 2)2 looks a lot like (x - h)2. They're written in the same form. So anyways, we just need that 4. That's ok, because we know from algebra I can add whatever I want, as long as I add it to both sides of the equation. So, I'll add 4 to both sides of the equation
x2 - 4x + y2 + 2y = 4
+4 +4
x2 - 4x +4 + y2 + 2y = 8
Now let's look at the y part
y2 + 2y
If we added a 1, we could have y2 + 2y + 1 = (y + 1)2 So let's add 1 to both sides.
x2 - 4x +4 + y2 + 2y = 8
+1 +1
x2 - 4x +4 + y2 + 2y + 1 = 9
Now let's factor as we planned. But we also want to make the 9 have a square as well. So we ask ourselves, what could we square that would give us 9? Well, x = 9, so let's use that.
(x - 2)2 + (y+1)2 =
But wait, is just 3, so we can just write
(x - 2)2 + (y+1)2 = 32
Notice how similar our equation looks to the equation of a circle.
(x - 2)2 + (y + 1)2 = 32
(x - h)2 + (y - k)2 = r2
In fact, we can see that 2 corresponds to h, 1 corresponds to k, and 3 corresponds to r. We just have 1 little problem left. In the equation of a circle, we have y -, but in our given equation we have y +. Can't have that. We need to make that + 1 into a - (-1), because those are the same thing.
(x - 2)2 + (y - (-1))2 = 32
And now the equations correspond perfectly, so we can say
h = 2
k = -1
r = 3
Remembering that (k, h) is the center of the circle and r is the radius, our final answer would be:
Our shape is a circle with center (2, -1) and radius 3.
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