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Lecture Comments (13)

0 answers

Post by Peggy Chen on July 20 at 09:31:14 PM

Hi. For the last example. I think there can be two answers.


as the center could be both above and below the x axis

1 answer

Last reply by: Dr Carleen Eaton
Sat Apr 19, 2014 12:47 AM

Post by Robert Monett on March 31, 2014

How do you which quadrants to plot the points. Couldn't the circle be below the x axis?

1 answer

Last reply by: Dr Carleen Eaton
Sun Mar 11, 2012 7:28 PM

Post by Jeff Mitchell on March 9, 2012

In the "Center not at origin" lecture section, I believe you forgot to square the (y-2) part of the equation at the bottom right side of the slide presentation.

1 answer

Last reply by: Dr Carleen Eaton
Mon Nov 7, 2011 8:40 PM

Post by Jonathan Taylor on November 4, 2011


1 answer

Last reply by: Dr Carleen Eaton
Thu Oct 13, 2011 8:54 PM

Post by Manuela Fridman on October 9, 2011

Can you please explain how you put the other 2 points on the graph after you plotted (1,3) in example:circle. Also is there a way for us to reach the teacher better because i notice it takes weeks for anyone to respond.

1 answer

Last reply by: Dr Carleen Eaton
Fri Aug 12, 2011 6:57 PM

Post by Lee Fulton on July 29, 2011

Your demonstration was impeccable! I have chosen certain lectures from you in preparation for my GRE's to enter Temple University! Thanks! This was much better than that boring GRE Manual!

0 answers

Post by aloosh aloosh on March 20, 2011

help please how can we say in the last example that circle is tangent to two points on the x axis i think unless the circle tangent to the lines x=7 x=-5 not the x axis any one know if there is a mistake in the example ???? other wise the points could be at the bottom of the circle and not the diameter

0 answers

Post by Mohammed Jaweed on August 18, 2010

Great teaching style,

Way better than my teacher.

I like the step by step explanations and examples. Very productive lecture.



  • Use symmetry to help you graph a circle.
  • Review completing the square and do so to put the equation of a circle in standard form.


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
  • What are Circles? 0:08
    • Example: Equidistant
    • Radius
  • Equation of a Circle 0:44
    • Example: Standard Form
  • Graphing Circles 1:47
    • Example: Circle
  • Center Not at Origin 3:07
    • Example: Completing the Square
  • Example 1: Equation of Circle 6:44
  • Example 2: Center and Radius 11:51
  • Example 3: Radius 15:08
  • Example 4: Equation of Circle 16:57

Transcription: Circles

Welcome to

Today we are going to talk about circles, beginning with the definition of a circle.0002

A circle is defined as the set of points in the plane equidistant from a given point, called the center.0008

For example, if you had a center of a circle here, and you measured any point's distance from the center, these would all be equal.0018

And the radius is the segment with endpoints at the center and at a point on the circle.0031

The equation for the circle is given as follows: if the center is at (h,k) and the radius is r,0044

then the equation is (x - h)2 + (y - k)2 = r2.0052

And this is the standard form; and just as with parabolas, the standard form gives you a lot of useful information0060

and allows you to graph what you are trying to graph.0067

For example, if I were given (x - 4)2 + (y - 5)2 = 9, then I would have a lot of information.0071

I would know that my center is at (h,k), so it is at (4,5).0084

And the radius...r2 = 9; therefore, r = √9, which is 3.0091

So, based on this information, I could work on graphing out my circle.0099

Use symmetry to graph a circle, as well as what you discover from looking at the equation in standard form.0108

Looking at a different equation, (x - 1)2 + (y - 3)2 = 4: this equation describes the circle0114

with the center at (h,k), which is (1,3); r2 = 4; therefore, r = 2.0124

So, I have a circle with a radius of 2 and the center at (1,3).0134

So, if this is (1,3) up here, and I know that the radius is 2, I would have a point here; I would have a point up here.0139

Symmetry: I know that, if I divide a circle up, I could divide it into four symmetrical quarters, for example.0156

So, if I have this graphed, I could use symmetry to find the other three sections of this circle.0169

All right, if the center is not at the origin, then we need to use completing the square to get the equation in standard form.0187

Remember that standard form of a circle is (x - h)2 + (y - k)2 = r2.0198

With parabolas, we put those in standard form by completing the square.0210

But at that time, we were just having to complete the square of either the x variable terms or the y variable terms.0213

Now, we are going to be working with both; and as always, we need to remember to add the same thing to both sides to keep the equation balanced.0219

If I was looking at something such as x2 + y2 - 4x - 8y - 5 = 0,0226

what I am going to do is keep all the x variable and y variable terms together, and then just move the constant over to the right.0241

The other thing I am going to do is group the x variable terms together: x2 - 4x is grouped together, just like up here.0249

And then, I am going to have y2 - 8y grouped together, and add 5 to both sides.0258

Now, I have to complete the square for both of these.0266

This is going to give me x2 - 4x, and then here I need to have b2/4.0269

Since b is 4, that is going to give me 42/4 = 16/4 = 4; so, I am going to put a 4 in here.0277

For the y expression, I am going to have...let's do this up here...b2/4 = 82/4, which is going to come out to 16.0292

So, I am going to add 16 here; and I need to make sure I do the same thing on the right.0307

So, I need to add 4, and I need to add 16; if I don't, this won't end up being balanced.0312

Now, I want this in this form; so let's change it to (x - 2)2 it is going to be (y - 4)2 =...0321

4 and 16 is 20, plus 5; so that is 25.0335

This gives me the equation in standard form; and the center is at (h,k), (2,4).0341

And the radius...well, r2 is 25; therefore, the radius equals 5.0350

And as always, you need to be careful: let's say I ended up with something in this form, (x + 3)2 + (y - 2) = 9.0355

The temptation for the center might be just to put (3,2); but standard form says that this should be a negative.0365

So, I may even want to rewrite this as (x - (-3))2 + (y - 2) = 9, just to make it clear that the center is actually at (-3,2).0372

And then, the radius is going to be the square root of 9, which is 3.0387

So, be careful that you look at the signs; and if the signs aren't exactly the same as standard form,0390

you need to compensate for that, or even just write it out--because -(-3) would give me +3, so these two are interchangeable.0396

All right, in this example, we are asked to find the equation of the circle which has a diameter with the endpoints (-3,-7) and (9,-1).0405

So, let's see what we are working with.0413

Just sketch this out at (-3,-7), right about there; over here is (9,-1); there we have the diameter of the circle.0418

We have a circle like this, and we want to find the equation.0433

Recall that the formula for the equation of a circle is (x - h)2 + (y - k)2 = r2.0439

Therefore, I need to find h; I need to find k; and I need to find the radius.0449

Recall that (h,k) gives you the center of the circle.0453

Since this is the diameter of the circle, the center of this line segment is going to be the center of the circle.0460

So, (-3,-7)...over here I have (9,-1); and here I have the center--the center is going to be equal to the midpoint of this segment.0467

Recall the midpoint formula equals (x1 + x2)/2, and then (y1 + y2)/2.0477

So, the center--the coordinates for that are going to be equal to (-3 + 9)/2, and then (-7 + -1)/2,0491

which is going to be equal to 6/2...-7 and -1 is -8/2; which is equal to (3,-4).0507

This means that h and k are 3 and -4; so I have h and k; I need to find the radius.0520

Well, half of the diameter...this is all the diameter; this is my midpoint; and I know that this is at (3,-4).0526

So, I just need to find this length--this is the radius.0535

And I now have endpoints; so I can use either one of these--I have a set of endpoints here and here, and here and here.0539

I am going to go ahead and use these two, and put them into the distance formula.0549

(-3,-7) and (3,-4)--I can use these in the distance formula: the distance here equals the radius,0553

which is the square root of...I am going to make this (x1,y1), and then this (x2,y2).0562

So, this is going to give me...x2 is 3, minus -3, squared, plus...y2 is -4, minus -7, squared.0574

So, the radius equals the square root of 3 + 3; a negative and a negative is a positive; all of this squared,0589

plus -4...and a negative and a negative is a positive, so -4 + 7, squared.0599

So, the radius equals the square root of...3 + 3 gives me 6, squared, plus...7 - 4 gives me 3, squared.0608

So, the radius equals √(36 + 9); 36 plus 9 is 45, so the radius equals √45.0619

But what I really want for this is r2, so r2 is going to equal (√45)2, which is going to equal 45.0632

Putting this all together, I can write my equation, because I now have h; I have k; and I have r.0651

So, writing the equation up here gives me (x - 3)2 + (y...I can either write this as - -4,0658

or I can rewrite this as (y + 4)2 = r2, which is 45.0672

So, this, or a little more neatly, like this: (y + 4)2 = 45--this is the equation for the circle.0679

And I found that information based on simply knowing the diameter.0689

Knowing the diameter, I could use the midpoint formula to find the center, which gave me h and k.0693

And then, I could use the distance formula to find the distance from the center to the end of the diameter, which gave me the radius.0698

And then, I squared the radius and applied it to that formula.0707

Example 2: Find the center and radius of the circle with this equation.0711

In order to achieve that, we need to put this equation in standard form.0716

And recall that standard form of a circle is (x - h)2 + (y - k)2 = r2.0719

So, we need to complete the square: group the x variable terms together on the left; also group the y variable terms on the left.0730

Add 8 to both sides to move the constant over.0742

I need x2 - 8x + something to complete the square, and y2 + 10y + something to complete the square, equals 8.0746

So, for the x variable terms, I want b2/4, and this is going to be 82/4, or 64/4, equals 16.0759

So, I am going to add 16 here.0773

For the y variable terms, b2/4 is going to equal 102/4, which is 100/4, which is 25.0774

I need to be careful that I add the same thing to both sides to keep this equation balanced, so I am also going to add 16 and 10 to the right side.0785

Correction: it is 16 and 25--there we go.0805

(x - 4)2 equals this perfect square trinomial, and I am trying to get it in this form; that is what I want it to look like.0809

Plus...(y + 5)2 comes out to this perfect square trinomial.0820

On the right, if I add 8 and 16 and 25, I am going to end up with 49.0830

8 and 16 is going to give me 24, plus 25 is going to give me 49.0840

Now, I have this in standard form: because the center of a circle is (h,k), I know I have h here,0848

and I know I have k here, this is going to give me...h is 4; k is -5.0861

Be careful with the sign here, because notice: this is (y + 5), but standard form is - 5, so this is equal to (y - -5)2--the same thing.0869

It is just simpler to write it like this; but make sure you are careful with that.0880

The radius here: well, I have r2; the radius, r2, I know, is equal to 49.0884

Therefore, r = √49, so the radius equals 7.0892

Therefore, the center of this circle is at (4,-5), and the radius is equal to 7.0898

Example 3: Find the radius of the circle with the center at (-3,-4) and tangent to the y-axis.0909

This one takes more drawing and just thinking, versus calculating.0919

The center is at (-3,-4), right about here.0924

The other thing I know about this circle is that it is tangent to the y-axis.0933

That means that, if I drew the circle, it is going to extend around, and it is going to touch this y-axis.0939

Well, the radius is going to have one endpoint on the circle, and the other endpoint at the center.0948

Therefore, the radius has to extend from (-3,-4) over here.0954

And at this point, we are able to then find the length, because, since x is -3, and it has to go all the way to x = 0,0965

then this distance must be 3; therefore, the radius equals 3.0976

And again, that is because I know the center is here at -3, and I know0984

that the other endpoint of the radius is going to be out here, forming the circle,0988

and that, because it is tangent to the y-axis, x is going to be equal to 0 right here.0996

So, I know that x is equal to 0 here; and I know that x is equal to -3 over here; so it is just 1, 2, 3 over--this distance here is going to be 3.1003

And that is going to be the same as the radius, so the radius is equal to 3.1013

Example 4: Find the equation of the circle that is tangent to the x-axis, to x = 7, and to x = -5.1018

We are given a bunch of information about this circle and told to put it in standard form.1026

The first thing we are told is that it is tangent to the x-axis; so this circle is touching the x-axis; let's just draw a line here to emphasize that.1040

It is also tangent to x = 7; x = 7 is going to be right here--it is tangent to this.1049

It is also tangent to x = -5, out here.1061

I am going to end up with a circle that is touching, that is tangent to, these three things.1068

Let's think about what that tells me.1074

I need to find h and k (I need to find the center).1077

I also need to find the radius, so I can find r2.1081

If this extends from -5 to 7, that gives me the diameter.1086

So, the diameter goes from 7 all the way to -5; so if I just add 7 and 5 (the distance from here to here,1097

plus the distance from here to here), I am going to get that the diameter equals 12.1104

The radius is 1/2 the diameter, so the radius equals 6.1110

I found that the radius equals 6.1117

The radius is going to extend from these endpoints to the center.1122

And I know that it is 6, so I know that the radius is going to go from 7 over here, 6 away from that.1127

7 - 6 is 1; it is going to go up to x = 1.1136

Again, that is because the length of the radius is 6, so the distance between the center and this endpoint has to be 6.1141

7 - 6 is 1; the radius is going to extend from there to there.1152

Therefore, the x-value of the center has to be 1.1156

Now, what is the y-value of the radius? The other thing I know is that this circle is tangent to this x-axis.1164

So, I know that it is going to have an endpoint on the x-axis.1171

And then, if it is going to extend from here to here, it is going to have to go up to 6; therefore, the center is at (1,6).1175

All right, so the radius equals 6, and the center is at (1,6); and that gives me an equation:1193

(x - 1)2 + (y - 6)2 = the radius squared.1200

If r = 6, then r2 = 36.1208

Again, that is based on knowing that this is tangent to x = -5, x = 7, and the x-axis.1212

So, I had the diameter, 12; I divided that by 2 to get the radius; I know that I have an endpoint here and an endpoint at the center.1221

So, that gives me the x-value of the center, which is 1.1231

I know I have an endpoint here, and also an endpoint at the center; so it has to be up at 6.1233

That gives me (1,6) for my value.1238

OK, and this is just drawn schematically, because the center would actually be higher up here.1243

This is just to give you...the center is actually going to be up here, now that I have my value: it is going to be at (1,6).1247

OK, that concludes this lesson of on circles; thanks for visiting!1258