Unit 6, Lesson 4
Distances and Circles
Geometry
4.1
Warm-up
Instructional Routines
NoneMaterials
To Gather
Scientific calculatorsActivity Narrative
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 generalize 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 .
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 to decide if they are both correct, even if they are different. Follow with a whole-class discussion.
Activity
NoneSegment P Q on a coordinate plane, origin O. Horizontal axis scale 0 to 50 by 10’s. Vertical axis scale 0 to 20 by 10’s. Points P(8 comma 12) and Q(53 comma 22) form segment P Q. Point N(53 comma 12) forms segments N Q and P N with dashed segments.
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 and .
Activity Synthesis
Invite a previously identified student, who subtracted coordinates directly, to share that 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 “” and “.”
Lesson Synthesis
Add the following theorem to the class reference chart, and ask students to add it to their reference charts:
A circle with center and radius has equation . (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 that we use to find the distance between the circle’s center, , and a given point, .)
Now, display this graph.
Invite students to describe the graph. (It is a circle with its center at and a radius of 6.) Instruct students to write an equation for the circle: or . 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 -coordinate of the center is 2, the -coordinate of the center is the opposite of 3, and the radius is 6.)
Student Lesson Summary
The diagram shows the point , along with several points that are 5 units away from . The set of all points 5 units away from is a circle with its center at and a radius of 5.
The point appears to be on this circle. To verify, calculate the distance from to . If this distance is 5, then the point is on the circle.
Let stand for the distance, and set up the Pythagorean Theorem: Evaluate the left-hand side to find that . Now is the positive number that squares to make 25, which means really is 5 units away from . This point is on the circle.
Triangle and 6 points on coordinate plane, origin o. Horizontal axes from negative 2 to 9 by 1’s. Vertical axis from negative 5 to 6 by 1’s. Right triangle vertices at 3 comma 1, 7 comma 1, and 7 comma 4. Bottom leg labeled 7 minus 3 equals 4. Right leg labeled 4 minus 1 equals 3. Hypotenuse labeled D. 6 points at 3 comma negative 4, 6 comma negative 3, 3 comma 6, negative 1 comma 4, negative 2 comma 1, and 1 comma negative 2.
The point also looks like it could be on the circle. To find its distance from , we can do a similar calculation: . Evaluating the left side, we get . This means that must be a little more than 5. So does not lie on the circle.
Circle and triangle on coordinate plane, Origin O. Horizontal axes from negative 2 to 9 by 1’s. Vertical axis from negative 5 to 6 by 1’s. Circle center at 3 comma 1, passes through 3 comma negative 4, 7 comma 4, 3 comma 6, negative 1 comma 4, negative 2 comma 1, and negative 1 comma negative 2. Right triangle vertices at 3 comma 1, 3 comma 2, and 8 comma 2. Legs labeled 1 and 5, hypotenuse labeled d.
To check if any point is on the circle, we can use the Pythagorean Theorem to see if is equal to 52 or 25. Any point that satisfies this condition is on the circle, so the equation for the circle is .
By the same reasoning, a circle with center and radius has equation .