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Course: Geometry (FL B.E.S.T.) > Unit 8
Lesson 11: Graphs of circles introGetting ready for conic sections
Practicing finding measurements in a circle, using the Pythagorean theorem, and completing the square will help us get ready for reasoning about conic sections (such as circles and parabolas).
Let’s refresh some concepts that will come in handy as you start the conic sections unit of the high school geometry course. You’ll see a summary of each concept, along with a sample item, links for more practice, and some info about why you will need the concept for the unit ahead.
This article only includes concepts from earlier courses. There are also concepts within this high school geometry course that are important to understanding right triangles and trigonometry. If you have not yet mastered the Distance and midpoints lesson, it may be helpful for you to review that before going farther into the unit ahead.
Radius and diameter
What is this, and why do we need it?
A circle is the collection of all points that are a certain distance from its center. We use the word radius both to mean that distance (a number) and to mean any segment (a geometric figure) with one endpoint on the center of the circle and one endpoint on its circumference.
Similarly, the diameter can mean either the widest distance across the circle, which is times the radius, or any segment that passes through the center of the circle with both endpoints on the circumference of the circle.
Practice
For more practice, go to Radius and diameter.
Where will we use this?
We use this vocabulary throughout the unit. Here is the first exercise where reviewing the radius and diameter might be helpful:
Pythagorean theorem
What is this, and why do we need it?
The Pythagorean theorem is , where and are lengths of the legs of a right triangle and is the length of the hypotenuse. The theorem means that if we know the horizontal and vertical distances between any two points, we can find out the distance between the points. We can use the Pythagorean theorem to find the length of the radius of a circle, to derive the equation of a circle, and to derive the equation of a parabola.
Practice
For more practice, go to Use Pythagorean theorem to find right triangle side lengths.
Where will we use this?
Here are a few of the exercises where reviewing the Pythagorean theorem might be helpful:
Completing the square
What is this, and why do we need it?
There is a pattern when we square a binomial:
We complete the square when we have an equation like , and we find the value to add to both sides. Then the left side of the equation becomes a perfect square.
Rewriting equations for circles by completing the square puts it back in the form of the Pythagorean theorem so that we can see the coordinates of the circle's center and the square of the radius.
Practice
For more practice, go to Completing the square (intro),
Completing the square (intermediate), and
Completing the square.
Where will we use this?
Here are a few of the exercises where reviewing completing the square might be helpful:
Want to join the conversation?
- I dpont rememebr learning how to complete the square(24 votes)
- within this unit, are we graphing circles and ellipses on a cartesian plane? Also, would this mean that the ellipses aren't functions (as they don't pass the vertical line test) but hyperbolas and parabolas are? Or are we viewing these graphs as completely separate from functions?(20 votes)
- become unconfused ;)(0 votes)
- Why do I find out the missing sqare just by with the other one? It only has -20, that's it!(2 votes)
- You're not trying to find the missing square. You're trying to find the missing constant term.
It should make sense, if you watch the "completing the square" video in Algebra 1, but I'll try to help anyway.
To get a perfect square, we want the equation to look like this:
x^2 + 2bx + b^2
If we plug the current equation into it, we will get this:
x^2 - 20x + b^2
So, we're missing the last constant term which is b^2.
But we can find it!
From the form, we know that -20x = 2bx, so we divide both sides by 2 and get:
-20x / 2 = 2bx / 2
-10x = bx
Then we divide both sides by x, and we get
-10x / x = bx / x
-10 = b
Now we can go back to the equation, where we were missing b^2, and plug that in
x^2 - 20x + -10^2
x^2 - 20x + 100
So the missing constant term is 100!(5 votes)
- why is there a comments section
this is goffy(2 votes)- could you just stop speaking this is for serious people only fricking idk man(1 vote)
- Are we doing all 20 assignments(0 votes)
- let me change my avatar I don't like this guy(1 vote)
- To change your avatar, go to Learner Home > Click on your avatar.(1 vote)
- Why do I need to know this(1 vote)
- I don't understand how to solve for the completing the square part(1 vote)