CIRCLES

DERIVING STANDARD FORM OF A CIRCLE

OBJECTIVE

• Review over the Pythagorean Theorem
• Review over the distance formula
• Use the distance formula to derive the standard form of a circle
• Derive the standard form of a circle
• Go over the general form of a circle
• Go over how to convert from general form to standard form
• Explain how it is useful
• Do some examples
• Desmos

So, what is the Pythagorean Theorem?

This is the Pythagorean theorem:

Where each side of a triangle is: a, b, and c; and where a and b are the sides of the triangle with c as the hypotenuse, then:

Now, we can actually see this is true when looking at an actual triangle with measurements.

Like:

Now let’s plug in these numbers and see what we get:

a = 3

b = 4

c = 5

Distance Formula

So, let’s say we have a segment (that we’ve maybe added together?), and we would like to find out how long the segment is.

Well, we know that when we are trying to find the distance along a line, it’s usually in a unit of distance (like inches, centimeters, etc.). We also know, we can use a ruler to find the distance between two points on a line, like points G and H.

So, let’s say we take a ruler and try to find the distance between points G and H

Well, we know that the distance that we want to find is actually the absolute value of the difference of the two numbers on the ruler, which corresponds to the two endpoints.

We say absolute value because there’s no such thing in normal day to day conversation as a negative measurement.

So, then we know that:

So, that’s great if it’s an easy thing to measure right?

But what happens when it’s not? Well, we actually have a way to find that out, but first, let’s look at the Pythagorean Theorem (I promise this is not random).

So, the Pythagorean Theorem states: a² + b² = c² when a and b are the sides of a triangle, and c is the hypotenuse.

Well, let’s say we have a line segment:

K B

And let’s say the coordinates for K are (0,1)

and the coordinates for B are (2,8)

Now, we want to know the distance of KB, but how can we do it?

What if we thought of KB as the hypotenuse of a triangle?

Well, this would be a lot easier to see if we had a graph right?

So let’s graph it!

Remember, we said that the point K was (0,1)

And the point B was (2,8)

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Now, let’s see if we can’t make this into a triangle so we can use the Pythagorean theorem to figure out the length of KB

So, first, since we need KB to be the hypotenuse in order to work, we need to create a few sides.

Let’s go ahead and extend the bottom side from K straight to under B.

We’ll label this point E.

Then, we go ahead and connect E to B.

There! Now we have a triangle!

Well, now, since both the legs of the triangle are flat, we can see how long each leg is.

So, we can see that KE travels from 0 to 2, so we know that |2-0| = 2

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And we can also see that EB travels from 1 to 8, so we know that |1 - 8| = |-7| = 7

So, now we can figure out the length of our hypotenuse!

So, we take: 2² + 7² = c²

Or:

4 + 49 = c²

53 = c²

Now we take the square root and we find that:

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WOW, THAT WAS A LOT OF WORK

We found the distance, but that was a lot of work. Maybe there is an easier way?

Well, turns out there is!

Let’s think about what we just did though to make sense of it.

We figured out that KE was 2 units long because we took the x-coordinate in K and subtracted it from the x-coordinate in E.

We also figured out that EB was 7 units long because we took the y-coordinate of E and subtracted it from B.

Well, let’s see if we can’t put that into our Pythagorean Theorem, and find out what it spits out.

Deriving the Distance Formula

So, essentially what we did was we took: x₂ - x₁ and we made that the length of a

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Then, we took: y₂ - y₁ and we made that the length of b.

Now remember, the Pythagorean Theorem is: a² + b² = c²

Replacing a with x₂ - x₁

(x₂ - x₁)² + (y₂ - y₁)² = c²

Now, since we want the actual length of c, let’s take the square root, and what we are left with is:

and b with y₂ - y₁ we get:

So what does this all have to do with circles?

Well, for starters, let’s go over a few shapes:

Line – So, the graph of an equation in x and y is the set of all points (x, y) in a coordinate plane that satisfy the equation (or make a straight line).

Some equations have graphs with precise geometric descriptions. For example, the graph of the equation y = 2x + 3 is the line with a slope of 2, passing through the point (0, 3).

This geometric description uses the familiar concepts of line, slope, and point. The equation y = 2x + 3 is an algebraic description of the line.

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However, what we are trying to look for is how to translate between geometric descriptions and algebraic descriptions of circles.

Now, we know how to use the Pythagorean Theorem to find missing dimensions of right triangles.

All we have left is to see how the Pythagorean Theorem leads us to the distance formula, which leads us to the standard form of the equation of a circle.

Using the Pythagorean Theorem

So, just to refresh (even though we literally just went over it)

The Pythagorean Theorem states that in any right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

Or, in other words:

Now let’s say for fun, we let r represent the distance between the origin and any point (x, y), then

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x can be positive, negative, or zero because it is a coordinate.

y can be positive, negative, or zero because it is a coordinate.

r cannot be negative because it is a distance.

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x² + y² = r².

Creating a circle with a triangle

Okay, so we now know that the hypotenuse, which we are calling r, is going to be:

Now let’s complete this equation and just solve for r.

So, if you want to get r by itself you’re going to want to:

We’re going to go over some vocab, but don’t forget what we did here okay!

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(Seriously, it’s important).

SOME QUICK VOCAB TO GO OVER

• A circle - is the set of all points in a plane that are equidistant from a reference point in that plane, called the center. The set of points forms a 2-dimensional curve that measures 360°.
• The center of a circle - is the point in the plane of the circle from which all points on the circle are equidistant. The center is in the interior of the circle.
• The radius - of a circle is the distance from the center to a point on the circle.

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STILL REMEMBER WHAT WE DID?

Alright so to recap, we found the hypotenuse of the triangle, and named it r right?

Then, we actually found out how to find the length of r with the Pythagorean theorem.

But, what if instead of leaving the hypotenuse as part of the triangle, we detach it, and spin it around?

Well, that would make something like this:

So, what does this mean?

Well, it means that if we have a circle with a center at (0,0), and a radius r, any point (x, y) is on that circle if and only if

x² + y² = r².

Or, like we found before:

OKAY, BUT THAT SEEMS PRETTY SPECIFIC

Well, sorta, but not really.

Actually, if we really look at it, it helps us see how to find the radius of any circle (and therefore the equation).

But to do this, we need to be a little creative.

Deriving the standard form of a circle

So, we know we can use the distance formula to find the radius of a circle right?

But so far we’ve been looking at triangles that have one of their corners at (0,0).

But what if it doesn’t? Maybe something like this:

So, what we want to find is the length of d right?

And we can also see that the measure of AB = d, right?

So, then we know if we add:

(You know, because of the Pythagorean Theorem?)

But wait, if we look at the triangle, we can actually see that:

Becomes:

So, substituting, we can see that:

Which is the same as saying:

And solving for d, we get:

And there is the standard form of a circle

Building on what we just created, we can now say:

For a circle with center (h, k) and radius r, any point (x, y) is on that circle if and only if �

Squaring both sides of this equation will give you the standard form of an equation of a circle with center (h, k) and radius r :

(x – h)² + (y – k)² = r ^2.

Now let’s sum it all up

If a circle has center (0, 0), then its equation is

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The general form of an equation of a circle is

(x – 0)² + (y – 0)² = r ², or x² + y ² = r ².

If the center and radius of a circle are known, then either of the following two methods can be used to write an equation for the circle:

Apply the Pythagorean Theorem to derive the equation.

Or, substitute the center coordinates and radius directly into the standard form.

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Ax² + By ² + Cx + Dy + E = 0

(x – h)² + (y – k)² = r ²

Ax ² + By ² + Cx + Dy + E = 0

The general form of the equation of a circle comes from expanding the standard form of the equation of the circle.

We’ll go over that later though.�

Where:

A = B,

A ≠ 0, and

B ≠ 0.

If any one of the following three sets of facts about a circle is known, then the other two can be determined:

• center (h, k) and radius r
• standard equation:

• general equation:

Completing the Square Review

So, here’s how we do it.

WOW THAT’S UGLY

Yeah that looked pretty intimidating, but it’s okay, we’re going to do this in baby steps.

So for today, all we are going to look at is how to set up the equation to be solved.

We’ll worry about solving some other day.

So, let’s start off by starting some problem, say:

General Form of a Circle

So, we’ve already touched on the general form of a circle, but to recap, the general form of a circle looks like:

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Ax ² + By ² + Cx + Dy + E = 0

So when do we use general form?

Well, honestly, most of the time we don’t, we actually try to convert it to standard form.

However there are many times when you’re working on a problem, and a general form of a circle may pop out.

In which case, you’re gonna wanna know that what you are working with is a circle.

Trust me, it gets annoying if you don’t see it.

So, let’s get started on converting!

Converting from General to Standard

So, let’s say you are given an equation that looks like:

x ² + y ² + 6x -10y + 29 = 0

The first step to solving is the same way we complete the square.

We want to subtract 29 from both sides

-29 -29

And now we are left with:

x ² + y ² + 6x -10y = -29

Now, let’s rearrange the equation so all of our x’s and y’s are together

x ² + 6x + y ² -10y = -29

Next, we complete the square for the x side:

x ² + 6x

Remember, we take our b, divide it by 2a, and square it, then add it.

x ² + 6x + 9 + y ² -10y = -29 + 9

Now let’s complete the square for the y side:

y ² -10y

Remember again, we take our b, divide it by 2a, and square it, then add it.

(x ² + 6x + 9) + (y ² -10y + 25) = -29 + 9 + 25

Now let’s simplify

And finally we are left with:

Ugh

Yeah I know, it’s a beast, but it is how we convert from general to standard.

So to do so, basically you complete the square for both the x and y variables, and simplify that way.

So, let’s do a few more just to make sure you understand how to do it.

(And yes, you will have homework/be quizzed/tested on this).

Example 1

So, let’s say you are given an equation that looks like:

x ² + y ² + 12x -40y + 100 = 0

The first step to solving is the same way we complete the square.

We want to subtract 100 from both sides

-100 -100

And now we are left with:

x ² + y ² + 12x -40y = -100

Now, let’s rearrange the equation so all of our x’s and y’s are together

x ² + 12x + y ² -40y = -100

Next, we complete the square for the x side:

x ² + 12x

Remember, we take our b, divide it by 2a, and square it, then add it.

x ² + 12x + 36 + y ² -40y = -100 + 36

Now let’s complete the square for the y side:

y ² -40y

Remember again, we take our b, divide it by 2a, and square it, then add it.

(x ² + 12x + 36) + (y ² -40y + 400) = -100 + 36 + 400

Now let’s simplify

And finally we are left with:

Example 2

So, let’s say you are given an equation that looks like:

x ² + y ² - 24x +20y - 75 = 0

The first step to solving is the same way we complete the square.

We want to add 75 to both sides

+75 +75

And now we are left with:

x ² + y ² - 24x +20y = 75

Now, let’s rearrange the equation so all of our x’s and y’s are together

x ² - 24x + y ² +20y = 75

Next, we complete the square for the x side:

x ² - 24x

Remember, we take our b, divide it by 2a, and square it, then add it.

x ² - 24x + 144 + y ² +20y = 75 + 144

Now let’s complete the square for the y side:

y ² +20y

Remember again, we take our b, divide it by 2a, and square it, then add it.

(x ² - 24x + 144) + (y ² +20y + 100) = 75 + 144 + 100

Now let’s simplify

And finally we are left with:

Example 3

So, let’s say you are given an equation that looks like:

x ² + y ² + 14x -18y - 10 = 0

The first step to solving is the same way we complete the square.

We want to subtract 100 from both sides

+10 +10

And now we are left with:

x ² + y ² + 14x -18y = 10

Now, let’s rearrange the equation so all of our x’s and y’s are together

x ² + 14x + y ² -18y = 10

Next, we complete the square for the x side:

x ² + 14x

Remember, we take our b, divide it by 2a, and square it, then add it.

x ² + 14x + 49 + y ² -18y = 10 + 49

Now let’s complete the square for the y side:

y ² -18y

Remember again, we take our b, divide it by 2a, and square it, then add it.

(x ² + 14x + 49) + (y ² -18y + 81) = 10 + 49 + 81

Now let’s simplify

And finally we are left with:

So how do we convert from General to Standard?�

Honestly, it’s way easier than you think.

To convert from general to standard, you just need to actually multiply the squares out, and combine like terms.

We’ll do this by using FOIL, but if you would like to use the area method to do this, that’s totally fine.

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Convert the following:

So let’s say we have something like:

The first thing we are going to want to do is expand the parenthesis.

Now we distribute:

Now we distribute the y side:

Now combine them

Now combine like terms

Let’s make sure the equation equals 0:

-16 -16

Lastly make sure to rearrange so it looks like:

Ax ² + By ² + Cx + Dy + E = 0

And there it is!

AND THAT’S IT!

So even though it seems like a lot, it’s basically:

1. Distributing the expressions in parenthesis
2. Adding like terms
3. Make the equation equal to 0
4. Rearrange

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So, let’s try some more:

Example 1

So let’s say we have something like:

The first thing we are going to want to do is expand the parenthesis.

Now we distribute:

Now we distribute the y side:

Now combine them

Now combine like terms

Let’s make sure the equation equals 0:

-200 -200

Lastly make sure to rearrange so it looks like:

Ax ² + By ² + Cx + Dy + E = 0

And there it is!

Example 2

So let’s say we have something like:

The first thing we are going to want to do is expand the parenthesis.

Now we distribute:

Now we distribute the y side:

Now combine them

Now combine like terms

Let’s make sure the equation equals 0:

-64 -64

Lastly make sure to rearrange so it looks like:

Ax ² + By ² + Cx + Dy + E = 0

And there it is!

Example 3

Last one:

The first thing we are going to want to do is expand the parenthesis.

Now we distribute:

Now we distribute the y side:

Now combine them

Now combine like terms

Let’s make sure the equation equals 0:

-100 -100

Lastly make sure to rearrange so it looks like:

Ax ² + By ² + Cx + Dy + E = 0

And there it is!

How to find the center of a circle in Standard Form

So, we’ve gone over this before, but to start off with, let’s look at what standard form is.

Standard Form looks something like this:

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Where: (h, k) is the center of the circle.

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So, how do we find the radius?

Well, that takes a little extra work, so let’s do an example:

Example 1:

Find the radius of a circle whose center is (3, 4) and passes through (12, 7)

So, to find the radius of this circle, first we look at the equation:

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Again, h and k are the center of the circle, so we can substitute them into our equation to find the radius.

So, substituting in:

Now that we have our equation, we look at our other point: (12, 7)

We can see that the x-coordinate is 12,

The y-coordinate is 7

Plugging them in, we get:

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Example 2:

Find the radius of a circle whose center is (5, 8) and passes through (1, 0)

So, to find the radius of this circle, first we look at the equation:

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Again, h and k are the center of the circle, so we can substitute them into our equation to find the radius.

So, substituting in:

Now that we have our equation, we look at our other point: (1, 0)

We can see that the x-coordinate is 1,

The y-coordinate is 0

Plugging them in, we get:

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Example 1:

Find the radius of a circle whose general form is:

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And passes through the point: (12, 7)

The first step to solving is the same way we complete the square.

We want to subtract 25 from both sides

-25 -25

And now we are left with:

Now, let’s rearrange the equation so all of our x’s and y’s are together

Next, we complete the square for the x side:

x ² - 6x

Remember, we take our b, divide it by 2a, and square it, then add it.

Now let’s complete the square for the y side:

y ² -8y

Remember again, we take our b, divide it by 2a, and square it, then add it.

Now let’s simplify

And finally we are left with:

EXAMPLE 1 (CONTINUED)

Now that we have our equation, we look at our other point: (12, 7)

We can see that the x-coordinate is 12,

The y-coordinate is 7

Plugging them in, we get:

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So it’s center is (3, 4), and the radius is

Example 2:

Find the radius of a circle whose general form is:

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And passes through the point: (1, 0)

The first step to solving is the same way we complete the square.

We want to subtract 89 from both sides

-89 -89

And now we are left with:

Now, let’s rearrange the equation so all of our x’s and y’s are together

Next, we complete the square for the x side:

x ² - 10x

Remember, we take our b, divide it by 2a, and square it, then add it.

Now let’s complete the square for the y side:

y ² - 16y

Remember again, we take our b, divide it by 2a, and square it, then add it.

Now let’s simplify

And finally we are left with:

EXAMPLE 2 ( CONTINUED)

Now that we have our equation, we look at our other point: (1, 0)

We can see that the x-coordinate is 1,

The y-coordinate is 0

Plugging them in, we get:

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So it’s center is (5, 8), and the radius is