Hey everyone and welcome back. So in the last video, we got introduced to this idea of conic sections basically various ways you could slice a three dimensional cone with a two dimensional plane to get different shapes. Now today we're going to be looking at the first shape which is the circle and you can get a circle from slicing a three dimensional cone directly horizontally. Now, I think a circle is a pretty general shape that we're all familiar with. But what we're going to be looking at in this video is how the equation for a circle will justify both the size and position that that shape is in. Now, that might sound a bit complicated and a little bit scary, but don't sweat it because in this video, we're going to be taking a look at some graphs and examples that I think are gonna make this concept super straightforward. So let's get right into this. Now, the thing that makes a circle unique is that a circle contains all points which are the same distance from the center of the circle. So to understand this, let's say you're looking at a circle and you have the center of it. If you want to get to this point right here on the circle, this would be the same distance as going to this point or going to that point. It's the same distance all the way around. And that's what makes this shape unique. Now, in order to graph a circle, there are two things you need, you need the center and you also need this distance we just described which is called the radius. Now, if you're given the graph already, it's actually pretty straightforward to find both of these. So if you want to find the center of a circle, let's say you're dealing with a circle that's at the origin of your graph. This is a pretty common case. And when this happens, the center of your circle is going to be at 00. This is what it means to be at the origin. Now to find the radius of the circle. Well, we could just start at the center and kept it takes 1234 units to get to this point. And so that means our radius is four. Now, let's say we have a circle that this time is not at the origin. Well, when this happens, your first step should be to find where your new center HK is H being the horizontal position and K being the vertical position. So what we can do is start by looking at the origin of our graph, which is right here. And if I look at the origin, it looks like horizontally, we've gone 12 units to the right. So our horizontal position is two and then our vertical position is going to be one unit up, which is one. So two comma one would be the center of our circle and define the radius. Well, I just need to start at the center and go 1234 units that would give me to this point. So the radius is also four for this circle. So as you can see finding the radius and center is pretty straightforward, but once you have these things, you can write the equation for the circle. If you have a circle at the origin, the equation is going to be X squared plus Y squared equals R squared. And the only thing we really care about in this equation is the R value which we already calculated to be for. So the equation is going to be X squared plus Y squared equals four squared. Now, if the circle is not at the origin, the equation is going to look something like this. Notice is very similar to what we had before, except now we have a minus H and A minus K inside of the equation. This is because we have this new center that the circle is at. So we have to take this new position into account. So writing the equation for this, we're going to have X minus H squared, where our H is two, we're going to have Y minus K squared where our K is one. And then this whole thing is going to be equal to R squared, which we already said RS four. So this would be four squared. So this would be the equation for a circle which is not at the origin. Now, something that I'll mention is that it's actually not too common that you're going to see the graph of the circle already given to you. But rather you're going to see the equation of the circle and have to graph it from there. So let's actually see if we can try an example where we graph a circle from just the equation. Now our first step should be to figure out what our center is. HK Well, I can see that our H value corresponds with this one. So our horizontal position is one and I can see that our K value corresponds with two. So that would be our vertical position. So going over here on the graph, our horizontal is at one and our vertical is up two. So this right here would be the center of our circle. Now step two tells us to find the radius of our circle. We'll notice for the equation, this last term is R squared, we can see that R squared is going to be this nine. So that means if R squared is equal to nine, we can find the radius by simply taking the square root on both sides of this little equation here. This will get the square to cancel, giving us that R is equal to the square root of nine, which is three. So that means our radius is three. Now, our third step is to plot four points which are a distance R and we already figured out R is equal to three. And this is going to be to the left right, up and down from our center point. So if I go over here to the center point, our radius is three, so we're going to go up 123 units, we're going to go down to three units, we're going to go to the left three units and we're going to go to the right three units and these are going to be the four points that we need to plot. So finding these four points was our third step. Now, our last step is going to be to connect these four points with a smooth curve. So if I go ahead and draw a curve that connects all four of these outside points together, this will give me the circle that I desire. So this right here is the graph for our circle. Now, one more thing that I wanna mention is that whenever you're dealing with a circle, the circle is not an example of a function. And this is because the circle would fail the vertical line test we talked about in previous videos, how if you're able to take a line and draw it on a graph and that line crosses more than one point on the curve. This means you are not dealing with a function. And since the circle fails the vertical line test, it's not a function. So that's the basic idea behind the equations and graphs for a circle. Hope you found this video helpful. Thanks for watching and let me know if you have any questions.

2

Problem

Problem

Sketch a graph of the circle based on the following equation: ${x}^{2}+{(y-1)}^{2}=9$

A

B

C

D

3

Problem

Problem

Sketch a graph of the circle based on the following equation: ${(x-2)}^{2}+{(y+3)}^{2}=1$

A

B

C

D

4

Problem

Problem

Find the equation for the following circle:

A

$x^2+y^2=4$

B

$\left(x+1\right)^2+y^2=4$

C

$\left(x-1\right)^2+y^2=4$

D

$\left(x+1\right)^2+y^2=16$

5

example

Circles in Standard Form Example 1

Video duration:

3m

Play a video:

Hey everyone. And welcome back. Let's see if we can give this problem a try. So in this example, we're asked to write an equation for a circle which each of the following characteristics. And we're gonna start with situation a. Now, before I do anything, what I'm going to do is write the standard form for any circle which is X minus H squared plus Y minus K squared is equal to R squared. And this will be how I write the equations. Once I find what the graph looks like, now we're going to start with this first circle here, you can see our center is at +01. So that means horizontally, we're at zero and vertically, we're at one. So I can put H is zero and K is one and that's going to be our center. Now, we're also told that our radius is three. So what that means is I can go 123 units to the right, 123 units to the left. I can go three units up and I can go 123 units down. And this is going to give us our circle, which is the shape that we're looking for. So this is what the graph looks like and what I also need to do right the equation. So to find the equation, I can use the standard equation which is going to be X minus zero squared, which is just X squared plus Y minus one squared. And that's all equal to R squared. The radius is three and three squared is nine. So this whole thing equals nine and that is going to be the equation and graph for this circle. Now, for our next circle, we're given the center at one negative two and this has a radius of one. So what I can do is go to our graph and horizontally, we're gonna be at one. And so that's gonna be right here and then vertically, we're going to be at negative two, which is down here. So that's the center of our circle. I can see that our radius is one. So I can go one unit up down to the left and to the right. And this is going to give us the circle we need right here. Now, in order to find the equation for this circle, I can use this same standard equation. So we're going to have X minus H which is one squared. And then this is going to be plus Y minus K which is negative two squared. Oh Since we have a negative there, that's going to actually cancel the negative sign there giving us A plus two. So we're gonna have Y plus two squared and that's all going to equal R squared and R squared is one. So, or R is one and so one square would just be one. And this is the equation we're looking for for this circle. Now, for this last situation that we have, we have a center at negative 63 for situation C and then we have a radius as the square root of five. So first, I'm gonna find our center on this graph which is going to be horizontally at negative six and vertically at K at three. So if I go to the left, I'd be at six, negative six and then I can go up to three, which would be right about there. And this is the center. Now, this one's gonna be a little bit tricky because our radius is the square root of five and the square root of five is approximately 2.2. So what we can do is look at where the center of our circle is and we know that vertically, the center is at three, right? So what that means is I can go two units down which would put us down here at about one. But I need to go a little bit past that, which is going to be right here. And that would be one of the points. Now, since we're vertically at three, I need to go up 2.2 units, which would put us here at five. But again, since I'm, it's 2.2 I need to go a little bit past that to be right about there. Now, what I can also do is go to the left and to the right now, I'm going to go to the left two units. So we'll go from negative six to negative eight, which is right about here. But I need to go a little past that since it's 2.2 and that would go from negative six to negative four to the right. But again, go a little past to put a point there. And this is going to be where the points line up and that is how you can draw this circle. Now, the equation is actually pretty straightforward. So draw the equation. All you need to do is use the standard equation which would be X minus negative six, which is X plus six squared and this would be plus Y minus three squared. And then this whole thing equals R squared and the square root of five squared is just five. So this is going to be what the equation looks like for our circle. And that is how you can find the graphs and equation for these circles based on the characteristics that we were given in this problem. So hope you found this video helpful. Thanks for watching.

6

concept

Circles in General Form

Video duration:

6m

Play a video:

Hey there everyone and welcome back. So in the last video, we talked about circles and we got introduced to the general idea of what the equation looks like and how the graph is going to look. And in this video, we're going to learn about something called the general form of circles. And when you have the general form instead of the standard form, your equation is going to look something like this that we see on the right here. Now, I'm pretty sure that a lot of us are familiar with this version of the equation down here, which we talked about in the last video, but this is what the general form would look like. Now, this might look a little bit scary and like, we don't really know what to do with this, but don't worry about it because in this video, I'm going to show you how to deal with this equation and we're going to be going over how you can actually take an equation that looks like this and convert it to standard form. So going from this to that, so that's what we're going to learn. So without further ado let's get right into things. Now, in order to convert to standard form, when dealing with these types of equations, what you need to do is know how to complete the square because you're going to have to do this step twice for both the Xs and Ys in your equation. So let's say for example, we have this type of equation right here. This would be an example of a circle in general form. And if we want to get this to standard form, there are a few steps that we need to take. Your first step should be to take all of the XS and Ys and group them together. So we have the XS there and the Ys there. You want to group these together and then you want to take all the constants and move them to the other side of the equal sign. And we could do that by taking this eight, we see here and subtracting it on both sides of the equation. So when you do this, you'll get all the XX squares and X here on one side on group together and you'll have all the Ys group together as well. And then you'll have the eight moved to the other side of the equation, which is what it would look like when you've done the step. Now, our next step from here is going to be completing the square twice and we're called it for completing the square. You take whatever your middle term is we call B, you divide it by two and you square it. And that's what you add. So, since we have a two right here, we're going to take two and divide it in half and then we're going to square it 2/2 is one and one squared is one. So it's going to end up with one there. Now, for this term right here, we would take six and divide it by two and square. It six divided by two is three and three squared is nine. So we would need to add a nine there. But since we added a, a one and a nine in this equation, we will also need to add these numbers over here to keep the equality the same. So in this case, we're going to have a one and a nine added to this negative eight as well. Now your last step from here is going to be to factor both this form the, the X is here and you're also going to need to factor the Ys. And then when you do that, you should end up with this X plus one squared and then you should end up with Y plus three squared and then negative eight plus one plus nine is equal to two. So this is what it would look like in standard form and notice how we went from general to standard by completing these steps. Now, to make sure we understand this process. Let's try an example where we are given an equation of a circle in general form and we have to convert it to standard form and graph the circle. So let's try this. Now, our first step in converting the standard form should always be to group the XS and Ys together and to move all constants to the right side of the equal sign. So I can do that pretty quickly here. I see that we have an X squared and we have a two X. So X squared plus two X is going to be grouping those X's together. I see that we have a Y squared and a negative four Y and then I see we have a constant one so I can take that one and move it to the other side of the equation by subtracting it, giving us this whole thing is equal to negative one. Now, our next step is going to be to add the square terms on both sides that we get when completing the square. And we're called to do this. You're going to end up with B over two squared for the X's and you're going to end up with B over two squared for the Y. So to do this, we're going to start with the XS if I want to complete the square for the X squared plus two X. What I can do is take this middle term here which will do A B over two squared. And we said that B is two. So this is going to be 2/2 squared, which is equal to one squared, which is just equal to one. So completing the square here, we'll get X squared plus two X plus. And then we have this one and then what we need to do is complete the square for the Ys as well. And I can see here that our B is negative four. So this is going to be negative 4/2 squared, negative four, divided by two is negative two and negative two squared is equal to positive four. So we're going to have plus Y squared minus four Y and let me write that square a little better minus four, Y plus four. And then this whole thing is going to equal negative one, but we have to take the two constants we added and we need to add them to the right side as well to make sure this equality stays the same. Now, our third step is going to be to factor this to become X plus B over two squared or Y plus B over two squared depending on which of these we're looking at. So this is actually a simple way to factor things. So we need to keep track of what this middle term was because this is what happened when we took B and divided it by two, we got. So factoring the X squared over here. Or the XS, I should say I'm going to get X plus B over two for X, which was one. So X plus one squared doing this factoring step for the Ys, I'm going to get Y and then we had a negative two. So it's going to be Y minus two squared. And then this whole thing is going to equal negative one plus negative one plus four, which is just equal to four. So this right here would be our equation in standard form. And we've completed step three. Now, our last step is going to be to use this equation that we have to graph our circle in standard form. So I can find the center of our circle. Now, what I first need to do is look at what our equation looked like and notice how there's a minus sign for the H and A minus sign for the K. And so, since there's a minus sign and we see X plus one, I know that H must be negative one since we see a plus sign instead of a minus sign, and then we have K is equal to two. And I see that's Y minus two, Y minus K. So we're good. So this is going to be positive two. And if I go over here to the graph, that means the center is going to be at negative +12. So I can go horizontally to negative one and then I can go up two. And this right here is going to be the center of our circle. And I can see here that this term is going to be associated with R squared, our radius squared. So if R squared is equal to four, our radius is going to be the square root of four, which is two. So if a radius is two, I can go up two units. So I'll go up to right there, I can go down two units. I can go to the right two units. And I think, and I can go to the left two units. So this is what our four points are going to look like. And then from here, I just need to connect these all with a smooth curve. So our circle is going to look something like that. So this is the graph of our circle that was step four and the answer to this problem. So hopefully, you found this video helpful. This is how you can convert from general to standard form and also write the graph. Once you've done that, let me know if you have any questions. And thanks for watching.

7

Problem

Problem

Determine if the equation ${x}^{2}+{y}^{2}-2x+4y-4=0$ is a circle, and if it is, find its center and radius.

A

Is a circle, center = $c(0,0)$, radius $r=2$.

B

Is a circle, center = $c\left(0,0\right)$, radius $r=3$ .

C

Is a circle, center = $c(1,-2)$, radius $r=3$.

D

Is not a circle.

8

Problem

Problem

Determine if the equation ${x}^{3}+{y}^{2}+4x-8y+4=0$ is a circle, and if it is, find its center and radius.

A

Is a circle, center = $c(0,0)$, radius $r=4$.

B

Is a circle, center = $c(2,-4)$ , radius $r=4$.

C

Is a circle, center = $c(-2,4)$, radius $r=4$.