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# 10.4: Equation of a Circle

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write the equation of a circle.

## Equations and Graphs of Circles

A circle is defined as the set of all points that are the same distance from a single point called the center. This definition can be used to find an equation of a circle in the coordinate plane.

• A circle is the set of all points equidistant from the ________________________.

Look at the circle shown below. As you can see, this circle has its center at the point (2, 2) and it has a radius of 3.

All of the points \begin{align*}(x, y)\end{align*} on the circle are a distance of 3 units away from the center of the circle.

We can express this information as an equation with the help of the Pythagorean Theorem. The right triangle shown above has legs of lengths \begin{align*}(x - 2)\end{align*} and \begin{align*}(y - 2)\end{align*}, and hypotenuse of length 3. We can write:

\begin{align*}(x-2)^2 + (y-2)^2 &= 3^2 \qquad \text{or}\\ (x-2)^2 + (y-2)^2 &= 9\end{align*}

We can generalize this equation for a circle with center at point \begin{align*}(x_0 , y_0)\end{align*} and radius \begin{align*}r\end{align*}:

\begin{align*}(x- x_0)^2 + (y- y_0)^2 = r^2\end{align*}

Example 1

Find the center and radius of the following circles:

A. \begin{align*}(x-4)^2 + (y-1)^2 = 25\end{align*}

B. \begin{align*}(x+1)^2 + (y-2)^2 = 4\end{align*}

A. We rewrite the equation as: \begin{align*}(x-4)^2 + (y-1)^2 = 5^2\end{align*}. Compare this to the standard equation. The center of the circle is at the point (4, 1) and the radius is 5.

B. We rewrite the equation as: \begin{align*}(x-(-1))^2 + (y-2)^2 = 2^2\end{align*}. The center of the circle is at the point (–1, 2) and the radius is 2.

Example 2

Graph the following circles:

A. \begin{align*}x^2 + y^2 = 9\end{align*}

B. \begin{align*}(x+2)^2 + y^2 = 1\end{align*}

In order to graph a circle, we first graph the center point and then draw points that are the length of the radius away from the center in the directions up, down, right, and left. Then connect the outer points in a smooth circle!

A. We rewrite the equation as: \begin{align*}(x-0)^2 + (y-0)^2 = 3^2\end{align*}. The center of the circle is at the point (0, 0) and the radius is 3.

Plot the center point and a point 3 units up at (0, 3), 3 units down at (0, –3), 3 units right at (3, 0) and 3 units left at (–3, 0) :

B. We rewrite the equation as: \begin{align*}(x-(-2))^2 + (y-0)^2 = 1^2\end{align*}. The center of the circle is at the point (–2, 0) and the radius is 1.

Plot the center point and a point 1 unit up at (–2, 1), 1 unit down at (–2, –1), 1 unit right at (–1, 0) and 1 unit left at (–3, 0) :

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

2. In the general equation of a circle \begin{align*}(x- x_0)^2 + (y - y_0)^2 = r^2\end{align*}, the variables \begin{align*}x_0\end{align*} and \begin{align*}y_0\end{align*} stand for a special point. What point is this?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

3. How can you find the radius of a circle from its equation? What do you need to do to the right side of the equation?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

Example 3

Write the equation of the circle in the graph below:

From the graph, we can see that the center of the circle is at the point (–2, 2) and the radius is 3 units long, so we use these numbers in the standard circle equation:

\begin{align*}(x+2)^2 + (y-2)^2 &= 3^2\\ (x+2)^2 + (y-2)^2 &= 9\end{align*}

Example 4

Determine if the point \begin{align*}(1, 3)\end{align*} is on the circle given by the equation:

\begin{align*}(x-1)^2 + (y+1)^2 = 16\end{align*}

In order to find the answer, we simply plug the point (1, 3) into the equation of the circle given.

Substitute the number _________ for \begin{align*}x\end{align*} and the number _________ for \begin{align*}y\end{align*}:

\begin{align*}(1-1)^2 + (3+1)^2 &= 16\\ (0)^2 + (4)^2 & = 16\\ 16 &= 16\end{align*}

Since we end up with a true statement, the point (1, 3) satisfies the equation. Therefore, the point is on the circle.

## Concentric Circles

Concentric circles are circles of different radii that share the same center point.

• Circles with the same ___________________ but different _________________ are called concentric circles.

Example 5

Write the equations of the concentric circles shown in the graph:

All 4 circles have the same center point at (3, 2) so we know the equations will all be:

\begin{align*}(x-3)^2 + (y-2)^2\end{align*}

Since the circles have different radius lengths, the right side of the equations will all be different numbers.

The smallest circle has a radius of 2:

\begin{align*}(x-3)^2 + (y-2)^2 & = 2^2 \quad \text{or}\\ (x-3)^2 + (y-2)^2 &= 4\end{align*}

The next larger circle has a radius of 3: \begin{align*}(x-3)^2 + (y-2)^2 = 9\end{align*}

The next larger circle has a radius of 4: \begin{align*}(x-3)^2 + (y-2)^2 = 16\end{align*}

The largest circle has a radius of 5: \begin{align*}(x-3)^2 + (y-2)^2 = 25\end{align*}

Look at the word concentric:

In Spanish, the word “con” means “with.”

The second part of the word, “-centric” looks very similar to the word “center.”

When we put these two parts together, “concentric” means “with” the same “center.”

1. If you are given a point and an equation of a circle, how can you tell if the given point is on the circle? Describe what you would do.

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

2. What are concentric circles?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

3. If you are given two equations of two different circles, how can you tell if the circles are concentric? Describe what the two equations would have to have in common.\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

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Date Created:
Feb 23, 2012