InterMath
Title
Equation of a Circle
Problem Statement
If a circle is centered at the origin, write an equation describing every point on the circle.
Your equation should use the radius of the circle and the x-coordinate and y-coordinate
of ANY point on the circle.
Problem Setup
I am trying to determine an equation for a circle using both the radius of the circle and
any x-coordinate and y-coordinate of any point on the circle.
Plans to Solve/Investigate the Problem
I will use the Geometer’s Sketchpad to construct a circle with the center located at the
origin (0,0). I will then plot a point somewhere on the circle. Then, I will look for
possible relationships perhaps a right triangle. I predict that the equation of the circle
will in some way be related to the Pythagoras theorem a 2  b 2  c 2 . I will then try to
come up with an equation using both the radius of the circle and the x and y
coordinates.
Investigation/Exploration of the Problem
In order to discuss how I went about solving this problem I will do it in a step by step
process:
1. First, I will construct a circle with the center located at the origin (0,0)
10
8
6
4
2
-10
-5
5
-2
-4
-6
-8
10
2. Next, I will label the origin (0,0) point B and place a point somewhere on my
circle and label that point A. I will then connect points A and B with a line
segment to get the radius of my circle.
8
6
4
2
A
-10
B
-5
5
10
-2
-4
-6
3. BA represents the radius of the circle. I used the snap point feature in GSP so
that it was easy for me to obtain an a close measure for my x and y coordinates
when they were moved.
4. I then noticed that I could create a right triangle within the triangle by dropping a
perpendicular line through point A. This would allow me to use the Pythagoras
theorem ( a 2  b 2  c 2 ) to find the value of r which is represent by BA .
8
6
4
2
A
r
-10
B
-5
-2
-4
-6
y
x
C
5
10
5. The above figure and figure below shows the right triangle. By using the
Pythagoras theorem and the measurements of sides x and y, I can find the
measurement of r. The Pythagoras theorem states a 2  b 2  c 2 . However, I have
restated this so that a is represented by x, b is represented by y and c is
represented by r.
8
6
4
2
A
r
-10
B
-5
y
x
C
5
10
-2
-4
-6
x = 2 cm
y = 2 cm
xA = 2
yA = 2
xB = 0
yB = 0
A: (2, 2)
I will now plug these measurements in to the Pythagoras theorem to get
22  22  8 . Therefore, r 2  8cm 2 . To get r, we would take the square root of both
sides which would give us r=2.83cm
5. You could also find the r by using the distance formula
This would give me the distance between points A ( x1, y1) and B ( x 2, y 2) . This
distance would also represent the radius. Therefore after plugging in the values
for the variables I end up with d  8 . Therefore, 8  2.83
6. This leads me to the conclusion that if I have the x and y coordinates of a circle,
then for any point, I can find the radius by using the formula, x²+y²=r², and
plugging in my values for x and y to find the radius.
Extensions of the Problem
Given the triangle below, find the radius of a circle whose center is located at (0,0).
B
3
4
A
C
First, I will construct a circle with the center located at origin. Then I will inscribe the
triangle in the circle and label all points and lines.
8
6
4
B
2
r
-10
A
-5
y
x
C
5
10
-2
-4
-6
-8
y = 3.0 cm
x = 4.0 cm
xB = 4.0
yB = 3.0
xA = 0.0
yA = 0.0
I can look at the coordinate system and count the number of units to see that x=4cm
and y=3cm. I can now use these measurements and plug them into the formula
x²+y²=r² to get the radius of the circle and also the measure of side r of the triangle.
42  32  25 . Therefore, r 2 =25. So, to find the measure of r, we simply take the square
root of both sides to give me r=5cm.
Author & Contact
Carla McNeely, Middle Grades Education student, concentrating in English/Language
Arts and Math. I am currently a junior at Georgia College and State University.
carlalynnmc@yahoo.com
Link(s) to resources, references, lesson plans, and/or other materials
http://intermath.coe.uga.edu/dictnary