# Lesson 4

Distances and Circles

## 4.1: Going the Distance (5 minutes)

### Warm-up

In this activity, students practice subtracting coordinates as part of a method for calculating the distance between two points. This will be helpful in an upcoming activity in which students build the equation of a circle by generalizing the process for finding distance on a coordinate grid.

Monitor for students who subtract pairs of coordinates directly without first finding the coordinates of point \(N\).

### Launch

Arrange students in groups of 2. Provide access to scientific calculators. After quiet work time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Follow with a whole-class discussion.

### Student Facing

Andre says, “I know that I can find the distance between two points in the plane by drawing in a right triangle and using the Pythagorean Theorem. But I’m not sure how to find the lengths of the legs of the triangle when I can’t just count the squares on the graph.”

Explain to Andre how he can find the lengths of the legs in the triangle in the image. Then, calculate the distance between points \(P\) and \(Q\).

### Student Response

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### Anticipated Misconceptions

If students are stuck, suggest that they find the coordinates of point \(N\).

### Activity Synthesis

Invite a previously identified student who subtracted coordinates directly to share their process for finding the lengths of the legs. Emphasize the use of subtraction to find the lengths of the legs. Consider displaying the image from the task and labeling the legs \(22-12=10\) and \(53-8=45\).

## 4.2: Circling the Problem (15 minutes)

### Activity

Students use the Pythagorean Theorem to test whether points are on a circle with a given center and radius. They will build on this process in the next activity to develop a general equation for a circle.

### Launch

*Representation: Internalize Comprehension.*Use color coding and annotations to highlight connections between representations in a problem. For example, highlight connections between representations by using color to highlight \(x\)-coordinates and the \(x\)-axis in one color, and \(y\)-coordinates and the \(y\)-axis in a different color.

*Supports accessibility for: Visual-spatial processing*

### Student Facing

The image shows a circle with center \((6, 10)\) and radius 17 units.

- The point \((14, 25)\) looks like it might be on the circle. Verify if it really is on the circle. Explain or show your reasoning.
- The point \((22, 3)\) looks like it might be on the circle. Verify if it really is on the circle. Explain or show your reasoning.
- In general, how can you check if a particular point \((x, y)\) is on the circle?

### Student Response

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### Student Facing

#### Are you ready for more?

The image shows segment \(AB\) and several points.

- Calculate the distance from each point to the endpoints of segment \(AB\).
point \(A\) point \(B\) point \(G\) point \(C\) point \(E\) point \(F\) -
What do the distances tell you about points \(G,E,C,\) and \(F\) ?

### Student Response

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### Anticipated Misconceptions

If students have trouble getting started, remind them of the definition of a circle: It is the set of points the same distance away from a given center.

### Activity Synthesis

Display this image for all to see.

Draw a right triangle on the image that could be used to find the distance between the center of the circle and the point shown. Ask students how they would find the lengths of the legs. As students describe the process, label the lengths of the legs \(x-6\) and \(y-10\).

*Speaking: MLR8 Discussion Supports.*As students share how they would find the distance between the center of the circle \((6, 10)\) and the point \((x, y)\), press for details by asking students how they know that the lengths of the legs are \(x-6\) and \(y-10\). Show concepts multi-modally by drawing and labeling the lengths \(x\), \(6\), and \(x-6\). Also draw and label the lengths \(y\), \(10\), and \(y-10\). This will help students justify their reasoning for the lengths of the legs of the right triangle.

*Design Principle(s): Support sense-making; Optimize output (for justification)*

## 4.3: Building an Equation for a Circle (15 minutes)

### Activity

Students build the general equation of a circle. As students generalize the work of the previous activities, they are making sense of a problem (MP1).

Monitor for students who write \((\text{-}3 - x)^2\) instead of \((x - (\text{-}3))^2\) in the first equation. Explain that either one is correct, but we traditionally write the variable as the first term in the binomial.

### Launch

Tell students that we traditionally use the letter \(r\) to stand for the radius of a circle, and we use the letters \((h,k)\) to stand for the center of a circle.

*Representation: Internalize Comprehension.*Activate or supply background knowledge about subtracting negative numbers.

*Supports accessibility for: Memory; Conceptual processing*

### Student Facing

The image shows a circle with center \((\text-3, 6)\) and radius 13 units.

- Write an equation that would allow you to test whether a particular point \((x,y)\) is on the circle.
- Use your equation to test whether \((9,1)\) is on the circle.
- Suppose you have a circle with center \((h,k)\) and radius \(r\). Write an equation that would allow you to test whether a particular point \((x, y)\) is on the circle.

### Student Response

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### Anticipated Misconceptions

If students struggle to write the first equation, suggest that they draw a right triangle that could be used to find the distance between \((x,y)\) and the center of the circle.

### Activity Synthesis

Display this image for all to see.

Ask students to connect their general equation to the circle. As students give their descriptions, plot a point \((x, y)\) and draw a right triangle that can be used to find the distance between \((x,y)\) and \((h,k)\). Label the legs \((x-h)\) and \((y-k)\), and label the hypotenuse \(r\).

Tell students that just like an equation in the form \(y=mx+b\) defines a line, an equation in the form \((x-h)^2+(y-k)^2=r^2\) defines a circle. The equation can be thought of as a point tester—any point \((x,y)\) that makes the equation true is a point on the circle, and conversely, all points on the circle make the equation true.

*Speaking: MLR8 Discussion Supports.*As students describe the connection between their general equation and the circle, press for details by asking students how they know that the lengths of the legs are \(x-h\) and \(y-k\). Also ask students how they know that the length of the hypotenuse is \(r\). Show concepts multi-modally by drawing and labeling the lengths \(x\), \(h\), and \(x-h\). Also draw and label the lengths \(y\), \(k\), and \(y-k\). This will help students make connections between the general equation and graph of a circle.

*Design Principle(s): Support sense-making; Optimize output (for generalization)*

## Lesson Synthesis

### Lesson Synthesis

Ask students to add this theorem to their reference charts as you add it to the class reference chart:

A circle with center \((h,k)\) and radius \(r\) has equation \((x-h)^2 + (y-k)^2 = r^2\). (*Theorem*)

Ask students why there are two subtractions in this equation. (The subtractions are used to find the lengths of the legs of the right triangle we use to find the distance between the circle’s center, \((h,k)\), and a given point \((x,y)\).)

Now, display this graph for all to see.

Invite students to describe the graph. (It is a circle with center \((2,\text-3)\) and radius 6.) Instruct students to write an equation for the circle: \((x-2)^2+(y+3)^2=6^2\) or \((x-2)^2+(y+3)^2=36\). Point out that we can write the right hand side of this equation either way. Ask a student to explain what the equation tells you about the graph. (The \(x\)-coordinate of the center is 2, the \(y\)-coordinate of the center is the opposite of 3, and the radius is 6.)

## 4.4: Cool-down - Writing Circle Equations (5 minutes)

### Cool-Down

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## Student Lesson Summary

### Student Facing

The diagram shows the point \((3,1)\), along with several points that are 5 units away from \((3,1)\). The set of *all* points 5 units away from \((3,1)\) is a circle with center \((3,1)\) and radius 5.

The point \((7,4)\) appears to be on this circle. To verify, calculate the distance from \((7,4)\) to \((3,1)\). If this distance is 5, then the point is on the circle. Let \(d\) stand for the distance, and set up the Pythagorean Theorem: \((7-3)^2+(4-1)^2=d^2.\) Evaluate the left hand side to find that \(25=d^2\). Now \(d\) is the positive number that squares to make 25, which means \((7,4)\) really is 5 units away from \((3,1)\). This point is on the circle.

The point \((8,2)\) also looks like it could be on the circle. To find its distance from \((3,1)\), we can do a similar calculation: \((8-3)^2+(2-1)^2=d^2\). Evaluating the left side, we get \(26=d^2\). This means that \(d\) must be a little more than 5. So \((8,2)\) does not lie on the circle.

To check if any point \((x,y)\) is on the circle, we can use the Pythagorean Theorem to see if \((x-3)^2 + (y-1)^2\) is equal to 5^{2} or 25. Any point that satisfies this condition is on the circle, so the equation for the circle is \((x-3)^2 + (y-1)^2 = 25\).

By the same reasoning, a circle with center \((h,k)\) and radius \(r\) has equation \((x-h)^2 + (y-k)^2 = r^2\).