For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

### Circles

- 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.

### 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
- 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

### Math Analysis Online

### Transcription: Circles

*Welcome to Educator.com.*0000

*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} = r^{2}.*0052

*And this is the standard form; and just as with parabolas, the standard form gives you a lot of useful information*0060

*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...r ^{2} = 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 circle*0114

*with the center at (h,k), which is (1,3); r ^{2} = 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} = r^{2}.*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 x ^{2} + y^{2} - 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: x ^{2} - 4x is grouped together, just like up here.*0249

*And then, I am going to have y ^{2} - 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 x ^{2} - 4x, and then here I need to have b^{2}/4.*0269

*Since b is 4, that is going to give me 4 ^{2}/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...b ^{2}/4 = 8^{2}/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} +...here 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, r ^{2} 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} = r^{2}.*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 (x _{1} + x_{2})/2, and then (y_{1} + y_{2})/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 (x _{1},y_{1}), and then this (x_{2},y_{2}).*0562

*So, this is going to give me...x _{2} is 3, minus -3, squared, plus...y_{2} 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 r ^{2}, so r^{2} 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} = r^{2}, 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} = r^{2}.*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 x ^{2} - 8x + something to complete the square, and y^{2} + 10y + something to complete the square, equals 8.*0746

*So, for the x variable terms, I want b ^{2}/4, and this is going to be 8^{2}/4, or 64/4, equals 16.*0759

*So, I am going to add 16 here.*0773

*For the y variable terms, b ^{2}/4 is going to equal 10^{2}/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 r ^{2}; the radius, r^{2}, 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 know*0984

*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 r ^{2}.*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 r ^{2} = 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 Educator.com on circles; thanks for visiting!*1258

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.

(X-1)^2+(Y-6)^2=36

(X-1)^2+(Y+6)^2=36

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

(x-1)squared+(Y-3)SQUARED 3SQUARE IS 3*3=9 ARE AM I CONFUSED BY THE FORMULA

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!

Lee

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.

Thanks