Main content

## SAT (Fall 2023)

### Course: SAT (Fall 2023) > Unit 6

Lesson 6: Additional Topics in Math: lessons by skill- Volume word problems | Lesson
- Right triangle word problems | Lesson
- Congruence and similarity | Lesson
- Right triangle trigonometry | Lesson
- Angles, arc lengths, and trig functions | Lesson
- Circle theorems | Lesson
- Circle equations | Lesson
- Complex numbers | Lesson

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Circle equations | Lesson

## What are circle equations, and how frequently do they appear on the test?

We can describe circles in the $xy$ -plane using equations in terms of $x$ and $y$ . Circle equations questions require us to understand the connection between these equations and the features of circles.

For example, the equation $(x+2{)}^{2}+(y-3{)}^{2}={4}^{2}$ is graphed in the $xy$ -plane below. It is a circle with a center at $(-2,3)$ and a radius of $4$ .

In this lesson, we'll learn to:

- Relate the standard form equation of a circle to the circle's center and radius
- Rewrite a circle equation in standard form by completing the square

On your official SAT, you'll likely see $xy$ -plane.

**1 question**that tests your knowledge of circles in theThis lesson builds upon the Manipulating quadratic and exponential expressions skill.

**You can learn anything. Let's do this!**

## What is the standard form equation of a circle?

### Features of a circle from its standard equation

### Representing a circle in the $xy$ -plane

Just like the equations for lines and parabolas, the standard form equation of a circle tells us about the circle's features.

In the $xy$ -plane, a circle with center $(h,k)$ and radius $r$ has the equation:

For example, the circle above has a center located at $(1,2)$ and a radius of $3$ . For ${h=1}$ , ${k=2}$ , and ${r=3}$ , its equation is:

### Try it!

## How do I rewrite equations of circles in standard form?

### How do I complete the square?

### Rewriting circle equations in standard form by completing the square

The standard form equation of a circle contains the squares of two binomials. Sometimes, we'll be asked to determine the center or radius of a circle represented by an equation in which the squares of the binomials are expanded.

Let's use the expanded equation ${x}^{2}+2x+{y}^{2}-10y+22=0$ to guide us through how to rewrite an expanded equation in standard form.

First, we have to make sure the coefficients of ${x}^{2}$ and ${y}^{2}$ are both $1$ . For most questions that require completing the square on the SAT, the coefficients will be $1$ .

Next, we need to find the constants that complete the square for $x$ and $y$ . For $x$ , this means we need to find a constant that, when added to ${x}^{2}+2x$ , lets us rewrite the expression as the square of a binomial.

You'll naturally develop a sense for constants that complete the square as you work on polynomial multiplication and factoring. A shortcut is to remember that the constant term of the binomial is equal to $\frac{1}{2}$ the coefficient of the $x$ - or $y$ -term, and the constant that needs to be added to complete the square is equal to the square of $\frac{1}{2}$ the coefficient.

The coefficient of the $x$ -term is $2$ . Therefore, the constant ${\left({\displaystyle \frac{2}{2}}\right)}^{2}=1$ completes the square for $x$ :

The coefficient of the $y$ -term is $-10$ . Therefore, the constant ${\left({\displaystyle \frac{-10}{2}}\right)}^{2}=25$ completes the square for $y$ :

We can rewrite the equation as shown below. Remember that when we add constants to one side of the equation, we must also add the same constants to the other side of the equation to keep the two sides equal.

Now that we have our completed squares, we just need to subtract $22$ from both sides of the equation. The resulting constant on the right side of the equation is equal to the square of the radius.

The equation represents a circle with a center at $(-1,5)$ and a radius of $2$ .

To rewrite an expanded circle equation in standard form:

- If necessary, divide both sides of the equation by the same number so that the coefficients of both the
-term and the${x}^{2}$ -term are${y}^{2}$ .$1$ - Find the constant the completes the square for
.$x$ - Repeat step 2 for
.$y$ - Add the constants from steps 2 and 3 to both sides of the equation.
- Rewrite the expanded expressions as the squares of binomials.
- Combine the remaining constants on the right side of the equation. It is equal to the square of the radius.

### Try it!

## Your turn!

## Things to remember

In the $xy$ -plane, a circle with center $(h,k)$ and radius $r$ has the equation:

## Want to join the conversation?

- How do we know if a given point is inside the circle, outside the circle, or tangent?(4 votes)
- A circle is the collection of all points that are a certain distance (the radius) away from a point. Using this definition, something inside the circle would be less than that distance away from the center point, and something outside the circle would be greater. Something tangent to the circle would be touching it, or its distance would be exactly the same.

To put this idea into a problem, we can just calculate the distance using the distance formula between the center of the circle and the point we're checking. If we get an answer that's less than the radius, the point is inside the circle, and so on.

Hope this helps!(11 votes)

- in the 1st try it question, I do not understand why the circle in the answer C does not fall only in the 1st quadrant but has it's sides in the other three quadrants(1 vote)
- For the circle whose equation is in answer C, the radius is 8 units, and the center is at the point (6,5). We can easily add and subtract the radius to the center point in the x and y directions to find four points that are on the circle. These are (-2,5), (14,5), (6,-3), and (6,13). Because (-2,5) and (6,-3) are well into the second and fourth quadrants, respectively, we know that some of this circle is outside of the first quadrant.(4 votes)

- For the last "Your Turn!" question, how is 4 the radius and not the sqrt(3)?(1 vote)
- Remember that you can only get the radius of a circle from its equation if it's in the proper form: (x - h)^2 + (y - k)^2 = r^2

Here, "r" would be your radius. Unfortunately, the question doesn't give us an equation in that form, so we have to complete the square to get our equation into the standard form:

x^2 + 6x + y^2 - 4y = 3

x^2 + 6x + 9 + y^2 - 4y + 4 = 3 + 9 + 4

(x + 3)^2 + (y - 2)^2 = 16

Now that we have our equation in the right form, the radius is the square root of the right hand side, or sqrt(16) = 4.(3 votes)

- where do we learn the distance formula for this topic?(1 vote)
- If all points on a circle are in Quadrant I in the xyxyx, y-plane, which of the following could be the equation of the circle?

In this particular question,there are two close answers which seems right and I'm confused how I'd get the right one. The two answers are...

(X-6)²+ (y-5)²=64

And

(x-6)²+(y-5)²=16.

The second is the answer. Why isn't it the first. Please break it down for me. Thanks.(1 vote)- Well... 16 and 64 are not simplified to radius form. So in order to know the radius of the equations, those two numbers must be square rooted. So the sqr rt of 16 is 4 and the sqr rt of 64 is a 8. Now we have our radiuses. To compare them to see which answer is correct, 8 is twice the size of 4, making that circle on the graph pretty big and not in Quadrant I. So the answer would be the equation (x-6)^2 + (y-5)^2= 16, because a radius of 4 would keep the circle in Quadrant I.

I hope that all made sense to you. Lol!(1 vote)