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### Equations of Circles

• The standard equation of a circle with center (h, k) and a radius of r units is

### Equations of Circles

Determine the coordinates of the center and the radius for the following equation.
(x − 8)2 + (y + 4)2 = 16
Center (8, − 4), Radius: 4.
Determine the coordinates of the center and the radius for the following equation.
(x + [5/9])2 + (y − 1.5)2 = 25
Center: ( − [5/9], 1.5), Radius: 5
Write an equation based on the given information.
(x − 4)2 + y2 = 9
Write an equation based on the given information.
Center: ( − 5, 2.5), Radius: 12
(x + 5)2 + (y − 2.5)2 = 144
Write an equation based on the given information.
Center: (3, 5), Point on the Circle (6, 2)
• (x − 3)2 + (y − 5)2 = r2
• (6 − 3)2 + (2 − 5)2 = r2
• r2 = 18
so the equation is: (x − 3)2 + (y − 5)2 = 18.
Write an equation based on the given information.
Endpoints of a Diameter ( − 2, 3), (5, 8)
• Center: ([( − 2 + 5)/2], [(3 + 8)/2])
• Center: (1.5, 5.5)
• (x − 1.5)2 + (y − 5.5)2 = r2
• ( − 2 − 1.5)2 + (3 − 5.5)2 = r2
• r2 = 18.5
(x − 1.5)2 + (y − 5.5)2 = 18.5.
Determine the coordinates of the center and the radius for the following equation.
x2 + (y − 3)2 − 64 = 0
• x2 + (y − 3)2 = 64
Graph the circle. (x − 2)2 + y2 = 16.
Graph the circle. (x + 3)2 + (y − 2)2 = 9.

Write a equation for the circle.
• Center: ( − 2, − 2)
• Point (1, − 1) is on the circle
• (x + 2)2 + (y + 2)2 = r2
• (1 + 2)2 + ( − 1 + 2)2 = r2
• r2 = 10
(x + 2)2 + (y + 2)2 = 10

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Equations of Circles

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Equation of a Circle 0:06
• Standard Equation of a Circle
• Example 1: Equation of a Circle
• Example 2: Equation of a Circle
• Extra Example 1: Determine the Coordinates of the Center and the Radius 4:56
• Extra Example 2: Write an Equation Based on the Given Information 7:53
• Extra Example 3: Graph Each Circle 16:48
• Extra Example 4: Write the Equation of Each Circle 19:17

### Transcription: Equations of Circles

Welcome back to Educator.com.0000

For this lesson, we are going to go over equations of circles.0002

The standard equation of a circle when there is a center (h,k) (and that is the coordinates of the center)0009

and a radius of r units is going to be (x - h) (this is the x-coordinate of the center), squared, plus (y -...0020

and this is the y-coordinate)...squared, equals r2, the radius squared.0033

Now, be careful: the x-coordinate is going to go with the x, and the y-coordinate is going to go with the y.0039

And don't forget: this number here is the radius squared; it is not the radius itself; it just has to be that number, squared.0047

Example: If I give you the center as (1,1) and the radius is 2, then to write an equation, it is going to be (x -...0061

the x-coordinate is 1), squared, plus (y - 1)2, equals...not 2; be careful with that;0077

it is 2 squared, so that is going to be 4; and this would be the equation of that circle.0087

Now, if I give you an equation, (x - 2)2 + y2 = 16 (that is the equation of a circle),0097

and I ask you for the center and the radius, the center, (h,k)...the x-coordinate is going to be the number that is grouped with the x;0111

that is 2; be careful--this is not -2, because the standard is going to be x minus the number.0122

So, if there is a minus there, then it is just the 2 that is going to be the x-coordinate.0130

Keep in mind, also: if this is a negative number, it is going to change to a plus; so instead of x - 2, it would be x + 2, if this was a -2.0137

And then, for the y-coordinate, there is no number that is grouped with the y to become a binomial.0149

So, this right here is going to be 0, because this is like saying (y - 0)2.0159

So then, my y-coordinate is 0, and my radius is the square root of 16, which is 4.0170

Given the equation, you can solve for your center, or you can find your center and the radius.0187

And also, if you are given the equation of a circle, and you are asked to graph, then you would first plot the center,0196

which is (2,0); (2,0) is right there--that would be your center; and then, the radius is 4.0205

Now, we cannot count diagonally on the coordinate plane, so we can only count up, down, left, and right.0214

So, I can count 1, 2, 3, 4, up 4; I can count down 4: 1, 2, 3, 4, because then, from the center, that would make 4 units; the radius is 4.0221

1, 2, 3, 4: the same thing works with that; 1, 2, 3, 4.0236

And then, what you are going to do is connect the dots, not like a square, but you know that it is going to be a circle--0244

it is the equation of a circle--so it is going to be something like that.0250

Again, after finding the center, (h,k), the radius would be this number's square root; that will be 2.0265

If you are given the equation, that is how you would find the center and the radius.0277

And if you are given the center and the radius, then you would have to make sure that the x-coordinate is paired up with the x,0281

the y-coordinate with the y, and then square the radius to give you the constant and then the equation.0287

Our examples: Determine the coordinates of the center and the radius for each equation.0297

The first one: there is the equation; I want to find the center and the radius.0303

The center would be...3 is my x-coordinate, because that is the one that is paired with the x; 3, comma, and 7.0308

If it has a minus, that means that the coordinate is a positive number.0323

And the radius is going to be the square root of that number; so the square root of 9 is 3.0330

For this one, the center is going to be 3/4...and for this one, it says y + 5/8.0340

If it is a plus, if it changed to a plus, that means that this number is a negative number; so that is negative 5/8.0353

And then, the radius is the square root of 4, which is 2.0364

OK, and then, the third one: here, this is in the correct form...plus y2...and then, the constant itself should be on the other side.0379

This is where the constant should go, on the right side by itself.0395

Since it is not there, I need to move it there; so I am going to add the 100; that way,0401

my equation is going to be (x - 10)2 + y2 = 100; and that would be standard form of a circle.0409

Now that it is in standard form, my center will be 10, comma...and don't forget that this right here0426

is the same thing as (y - 0)2; so my y-coordinate is 0, and then the radius is √100, so that is 10.0438

Otherwise, your radius would be 0; so be careful with that, just to make sure; you have to add the constant over to this side.0462

For the next example, we are going to write an equation for each of these, based on what information is given to us.0473

The first one: they give us the center, and they give us the radius.0481

In that case, we can just go ahead and plug each of these in: this is (h,k), and this is r.0486

Remember that the standard form of the equation of a circle is (x - h)2 + (y - k)2 = the radius squared.0495

The equation is going to be (x - -3)...minus a negative...that means I have a plus 3, squared,0513

plus (y + 2)2 (it is a negative number), equals...and then for the radius squared,0524

the constant, don't forget to square these numbers: this is going to be 25/4.0533

And that is the equation of that circle.0542

The next one is the same thing: (x - 0)2 + (y - 1/2)2 = √8, squared.0545

Here, you want to simplify this; if it is x - 0, then that is the same thing as x2, plus (y - 1/2)2,0563

equals...the square root of 8, squared, is just 8; remember that the square root and the square cancel each other out.0574

So, that is the equation of that circle.0581

And then, for the next one, they give us the center, and instead of giving us the radius, they give us a point on the circle.0585

The point on the circle is (4,0); now, in order for me to find the equation of a circle, I need the center and the radius.0593

I need these two things; instead, I have the center, and I have a point on the circle.0610

I have this; now, I have to find the radius...so how do you find the radius?0617

If you have the center, and you have a point on the circle, then you just have to find the distance between those two.0622

So then, if they just give you the center, and they give you a point on the circle, if you find the distance, then that will give us the radius.0630

So, you can use the distance formula; that is going to be (x2 - x1)2 + (y2 - y...0640

oh, I said y, and I wrote x...y1)2.0655

That will be x2...that is 4, minus the 4, the two x's...the difference of them is 0, squared is 0.0666

Plus...0 minus -5; that is 5; squared is 25; so distance equals √25, which is 5.0680

That means that the radius is 5; the center is that; the radius is 5.0692

Now, I can go ahead and write my equation: (x - 4)2 + (y + 5)20699

(and that is a plus because it is a negative) = 52; that is 25.0712

And that is it; everything is simplified; that is the answer.0720

And then, this last one, endpoints of a diameter: instead of giving us the center, and then a point on the circle,0725

they are giving us the two points on the circle that make up a diameter.0733

So then, it is like giving us this and this; they are giving us these two endpoints.0739

So, how do you find the center? Now again, we still need the center, and we still need the radius.0747

We have to look for both of those things; so if they give us the endpoints of the diameter, to find the center,0752

we have to find the midpoint of those two points; the midpoint of a diameter is the center.0761

How do you find the midpoint? You use the midpoint formula, which is x2 + x1,0768

or x1 + x2, divided by 2, and y2 + y1, divided by 2.0778

You are just going to add up the x's and divide it by 2, add up the y's and divide it by 2.0789

It is also like average; you only have two x's, so you add them up and divide by 2; that is the average of the x's, comma, the average of the y's.0793

The midpoint of this is going to be 1 + 5, over 2, comma, the y's: -1 + 3, over 2; that is 6/2, which is 3, and this is 2/2, which is 1.0804

So, this is the midpoint; that means that this is the center; we found the midpoint from this point to this point: that is (3,1).0824

The center is (3,1); now, I have to find the radius.0835

So, just like we did for this problem here, if you have the center, and you have a point on the circle0842

(which is both of those), then you can just find the distance, and then find the radius, like that.0847

You can also find the distance between these two (find the distance of this to this) and divide it by 20855

(diameter divided by 2 is radius), or you can just take the center (since we found the center, (3,1)),0862

and use one of these to find the distance, because from the center to this point is a radius; or the center to this point is a radius.0869

So, you can just pick whichever one looks easier; this one looks easier to me, so I am going to use this point and the center, and find the distance between them.0879

The distance is going to be...I am going to use this one, and I am going to use this one;0893

so (1 - 3)2 + (-1 - 1)2 is going to be the square root of...22 is 4, plus (-2)2 is 4.0900

The distance equals the square root of 8, or it can be 2√2.0925

That is the distance between the center and then one of the points; that makes it the radius: 2√2,0936

or it can also be √8, because I know that I am going to have to square it to give me the equation.0945

My equation, then, is going to be x minus the x-coordinate of the center; that is that; squared--don't forget the squared;0954

plus y minus 1, squared; equals the square root of 8, squared (is 8); that is the equation of that circle.0962

So again, if you are given the endpoints of a diameter, you have to find the midpoint to get the center.0981

And then, you have to use the distance formula to find the distance between the center, and then one of these points, to find the radius.0986

Just, whatever they give you, always make sure that you have the center and the radius.0994

If they don't give you either one of these, then you have to look for it; but you need the center,0999

and you need the radius, in order for you to write the equation of the circle.1003

The third example: we are going to graph each of the circles.1010

So again, I need center, and I need radius; the center, this one, is 0, the x-coordinate; this one is 2; and the radius is 2.1015

Graph the center first: (0,2) is right there, and then the radius is going to be up 2, down 2, right 2, left 2.1032

And then, you are going to just draw a circle.1046

The next one: here, our center is (1,-1); the radius is √20, which makes that 2√5.1054

Now, since we are going to have to count this on the coordinate plane, it would just be best to find √20 in decimals.1069

So, you can just use your calculator; I have a calculator on my screen; so √20...you should get about 4.47, which is around 4.5, 4 and 1/2.1079

So, the center is (1,-1), and then I am going to go up 4 and 1/2: 1, 2, 3, 4, 4 and 1/2; 1, 2, 3, 4, 4 and 1/2; 1, 2, 3, 4; and then, 1, 2, 3, 4.1099

Again, draw your circle, something like that.1131

The next one: we are given the graphs of the circles, and then we have to find the equation.1158

The diameter is from here to here; I have a diameter from here to here.1171

That means that the center (the midpoint of the diameter) is going to be right there.1179

Now, you can count it; so then, this is 5 here, and this is -5 here; that makes the diameter 10.1184

I have 10; that is 5; the radius is going to be 5; so this is r, which is 5; so that means it should be 5 each way.1191

The center is (0,0), and the radius is 5; then, the equation becomes (x - 0)2 + (y - 0)2 = radius squared, 25.1205

So, we have to simplify this; we don't want x - 0, so it is going to be x2 + y2 = 25, and that is your equation.1227

For this one, look for your diameter: there, there, there, and there; so this is -1, and this is +7;1242

so that means that the distance from my diameter is 8; half of that is...let's see, this is (7,1), and this is (-1,1);1258

so then, it is 8; that means that half of that is 4; my radius has to be 4;1280

that means it has to go 4 this way, 4 that way, and each of the ways.1286

And then, that means that my center is at (3,1), and then the radius is 4.1296

Once you have the center and the radius, you can write your equation: (x - 3)2 + (y - 1)2 = 16.1307

Is that simplified? Yes, so that is the equation of the circle.1322

OK, we are going to do one more example; we are going to have a fifth example for this one.1329

This one is a little bit challenging, but we are going to just go ahead and try it.1334

The graphs of x = 1 and y = -2 are tangents to the circle with the center in the second quadrant and a diameter of 8.1342

Write the equation of the circle.1358

They give us two equations: these are linear equations--that means two lines are tangent to the circle.1362

We know that tangents are lines that touch the circle (intersect the circle) at one point.1374

So then, let's draw these lines here: x is 1--that is this side right here; y = -2 is right there, so this is x = 1; this is y = -2.1382

And they are tangents; that means that the circle is going to be touching this line and this line one time.1406

The center is in the second quadrant; that means that somewhere in here is going to be the center.1415

The diameter is 8; so if it is going to be touching this line and this line with the center in the second quadrant,1423

that means that it is going to look something like this.1434

If the diameter is 8, then the radius is going to be 4; so tangent to the center is going to be 4 (4: distance away from this line).1443

So, 1, 2, 3, 4; it is going to be somewhere here; the center is going to be somewhere there.1456

And then, the same thing: 1, 2, 3, 4, so it is going to be somewhere here.1464

I know that that is my center; how do I know that?--because, if I draw a radius of the circle, this point is going to be on the circle.1473

This point is going to be on the circle; this has a radius of 4, and this has a radius of 4.1491

So, 1, 2, 3, 4; that is also on my circle; and that is also on my circle.1498

Since the center is in the second quadrant...if it said that it was in the first quadrant, then you would count 4 this way and count 4 upwards.1508

So, the circle will be tangent to these two lines, but still be in the first quadrant.1518

I can just draw my circle like this; keep in mind: make sure that...1524

if it is tangent, then it has to be touching at one point, and that would be the point of tangency.1536

And then, you can just draw a radius to both tangent lines; it has to be 4 away from this tangent and 4 away from this tangent.1540

And this would be the center, to make that 4 away from each of those.1553

Then, our center (because again, we are just writing the equation) is 1, 2, 3...(-3,2); our radius is 4.1558

So, our equation is going to be (x + 3)...a negative...squared, plus (y - 2)2 = 42 (is 16).1574

That is the equation for this circle here.1590

If you ever get this type of problem, draw your tangents first; see what quadrant it is in.1598

Figure out what your radius is, and then start finding the distance away from the tangents; and you use radius for that.1606

That is it for this lesson; thank you for watching Educator.com.1618