Circles - Online Tutor, Practice Problems & Exam Prep

Circles are defined as the set of all points equidistant from a center point, characterized by the radius. The standard equation for a circle centered at the origin is x^2 + y^2 = r^2. For circles not at the origin, the equation takes the form x - h + y - k = r^2. Converting from general to standard form involves completing the square, allowing for effective graphing and understanding of circle properties.

Circles in Standard Form

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. 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 going to 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. 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 (0,0). 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 count. It takes 1, 2, 3, 4 units to get to this point, and so that means our radius is 4.

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, \( h, k \), 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. If I look at the origin it looks like horizontally we've gone 1, 2 units to the right. So our horizontal position is 2, and then our vertical position is going to be 1 unit up, which is 1. So \( (2,1) \) would be the center of our circle. And to find the radius, well, I just need to start at the center and go 1, 2, 3, 4 units. That would get me to this point, so the radius is also 4 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 x2 + y2 = r2 . And the only thing we really care about in this equation is the \( r \) value, which we already calculated to be 4. So the equation is going to be x2 + y2 = 42 .

Now if the circle is not at the origin, the equation is going to look something like this. Notice it's 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 \( (x - h)^2 \), where our \( h \) is 2. We're going to have \( (y - k)^2 \), where our \( k \) is 1, and then this whole thing is going to be equal to \( r^2 \), which we already said \( r \) is 4, so this would be \( 4^2 \). So this would be the equation for our 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 \( h,k \). Well, I can see that our \( h \) value corresponds with this 1, so our horizontal position is 1, and I can see that our \( k \) value corresponds with 2. So that would be our vertical position. So going over here on the graph, our horizontal is at 1 and our vertical is up 2. So this right here would be the center of our circle.

Now step 2 tells us to find the radius of our circle. Well notice for the equation this last term is \( r^2 \). We can see that \( r^2 \) is going to be this 9. So that means if \( r^2 \) is equal to 9, we could 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 9, which is 3. So that means our radius is 3.

Now our third step says to plot 4 points which are a distance \( r \), and we already figured out \( r \) is equal to 3, 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 3 so we're going to go up 1, 2, 3 units, we're going to go down 1, 2, 3 units, we're going to go to the left 3 units, and we're going to go to the right 3 units. And these are going to be the 4 points that we need to plot. So finding these 4 points was our third step. Now our last step is going to be to connect these 4 points with a smooth curve. So if I go ahead and draw a curve that connects all 4 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.

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

• A.
• B.
• C.
• D.

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

• A.
• B.
• C.
• D.

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

Circles in Standard Form Example 1

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 with each of the following characteristics. We're going to 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-h + y-k = r2

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 1, so that means horizontally we're at 0 and vertically we're at 1. So I can put h is 0 and k is 1, and that's going to be our center. Now we're also told that our radius is 3. So what that means is I can go 1, 2, 3 units to the right; 1, 2, 3 units to the left; I can go 3 units up, and I can go 1, 2, 3 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 is write the equation. So to find the equation, I can use the standard equation, which is going to be:

x2 + y-1 9

And that is going to be the equation and graph for this circle. Now for our next circle, we're given the center at 1, -2, and this has a radius of 1. So what I can do is go to our graph, and horizontally we're going to be at 1, and so that's going to be right here, and then vertically we're going to be at -2, which is down here. So that's the center of our circle. I can see that our radius is 1, so I can go 1 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-1 + y+2 1

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 -6, 3 for situation C, and then we have our radius as the square root of 5. So first, I'm going to find our center on this graph, which is going to be horizontally at -6 and vertically at 3. So if I go to the left, I'd be at -6, and then I can go up to 3, which would be right about there, and this is the center. Now this one's going to be a little tricky, because our radius is the square root of 5, and the square root of 5 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 3, right. So what that means is I can go 2 units down which would put us down here at about 1, 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 3, I need to go up 2.2 units which would put us here at 5, 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 gonna go to the left 2 units, so we'll go from -6 to -8, which is right about here, but I need to go a little past that since it's 2.2, and I would go from -6 to -4 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 to draw the equation, all you need to do is use the standard equation, which would be:

x+6 + y-3 5

And 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, I hope you found this video helpful. Thanks for watching.

Circles in General Form

Hey there, everyone, and welcome back. So in the last video, we talked about circles. We got introduced to the general idea of what the equation looks like and how the graph is going to look. In this video, we're going to learn about something called the general form of circles. 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. 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. This might look a little 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. 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 x's and y's 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 x's and y's and group them together. 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 you could do that by taking this 8 we see here and subtracting it on both sides of the equation. When you do this, you'll get all the x² and x here on one side grouped together, and you'll have all the y's grouped together as well. And then you'll have the 8 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. Recall that for completing the square, you take whatever your middle term is, we call b. You divide it by 2, and you square it, and that's what you add. So since we have a 2 right here, we're going to take 2 and divide it in half, and then we're going to square it. 2 over 2 is 1, and one squared is 1, so we're just going to end up with 1 there. Now for this term right here, we would take 6 and divide it by 2 and square it. 6 divided by 2 is 3, and 3 squared is 9, so we would need to add a 9 there. But since we added a 1 and a 9 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 1 and a 9 added to this negative 8 as well. Your last step from here is going to be to factor both this form, the the x's here, and you're also going to need to factor the y's. And then when you do that, you should end up with (x + 1)², and then you should end up with (y + 3)², and then negative eight plus 1 plus 9 is equal to 2. 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. Our first step when converting to standard form should always be to group the x's and y's 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 x² and we have a 2x. So x² + 2x is going to be grouping those x's together.

I see that we have a y² and a negative 4y, 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. Recall to do this, you're going to end up with b/2 squared for the x's and you're going to end up with b/2 squared for the y's.

So to do this, we're going to start with the x's. If I want to complete the square for the x² + 2x, what I can do is take this middle term here, which will do a b/2 squared, and we said that b is 2, so this is going to be 2/2 squared, which is equal to 1 squared, which is just equal to 1. So completing the square here will get x² + 2x + 1. And then what we need to do is complete the square for the y's as well. And I can see here that our b is negative 4, so this is going to be negative 4/2 squared. Negative 4 divided by 2 is negative 2, and negative 2 squared is equal to positive 4. So we're going to have + y² - 4y + 4.

This whole thing is going to equal negative one, but we have to take the 2 constants we added, and we need to add them to the right side as well to make sure this equality stays the same. Our 3rd step is going to be to factor this to become (x + b/2)², or (y + b/2)², 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 2.

We got this. So factoring the x's, I'm going to get (x + 1)². Doing this factoring step for the y's, I'm going to get (y - 2)². And then this whole thing is going to equal negative 1, plus 1, plus 4, which is just equal to 4. So this right here would be our equation in standard form, and we've completed step 3.

Now our last step is going to be to use this equation that we have to graph our circle in standard form. 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 + 1, 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 2, and I see that's y - 2, y minus k. So we're good. So this is going to be positive 2. And if I go over here to the graph, that means the center is going to be at negative 1.

So I can go horizontally to negative 1, and then I can go up 2, 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 r², our radius squared. So if r² is equal to 4, our radius is going to be the square root of 4, which is 2. So if our radius is 2, I can go up 2 units. So I'll go up to right there. I can go down 2 units. I can go to the right 2 units, and I can go to the left 2 units. So this is what our 4 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 4 and the answer to this problem.

Hopefully, you found this video helpful. This is how you can go 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.

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.

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 $r=4$.
• D. Is not a circle.

Here’s what students ask on this topic:

What is the standard equation of a circle?

The standard equation of a circle is given by:

$x^2+y^2=r^2$

where $r$ is the radius of the circle and the center is at the origin (0,0). For a circle not centered at the origin, the equation is:

$(x-h)2^{}+\; (y-k)2^{}=r^2$

where $h$ and $k$ are the coordinates of the center of the circle.

How do you find the radius and center of a circle from its equation?

To find the radius and center of a circle from its equation, you need to identify the standard form of the circle's equation:

$(x-h)2^{}+\; (y-k)2^{}=r^2$

The center of the circle is given by the coordinates $h$ and $k$. The radius $r$ is found by taking the square root of the constant on the right side of the equation. For example, if the equation is:

$(x-2)2^{}+\; (y-3)2^{}=16$

The center is (2, 3) and the radius is $\sqrt{16}=4$.

How do you convert the general form of a circle's equation to the standard form?

To convert the general form of a circle's equation $x^2+\mathrm{Dx}+y^2+\mathrm{Ey}+F=0$ to the standard form, follow these steps:

1. Group the $x$ and $y$ terms together.

2. Move the constant term to the other side of the equation.

3. Complete the square for both the $x$ and $y$ terms.

4. Factor the perfect square trinomials.

For example, given $x^2+4x+y^2-6y-12=0$, the steps are:

1. Group: $(x^2+4x)\; +\; (y^2-6y)\; =12$

2. Complete the square: $(x^2+4x+4)\; +\; (y^2-6y+9)\; =12+4+9$

3. Factor: $(x+2)2^{}+\; (y-3)2^{}=25$

The standard form is $(x+2)2^{}+\; (y-3)2^{}=25$.

How do you graph a circle given its equation in standard form?

To graph a circle given its equation in standard form, follow these steps:

1. Identify the center $\mathrm{(h,\; k)}$ and the radius $r$ from the equation $(x-h)2^{}+\; (y-k)2^{}=r^2$.

2. Plot the center point $\mathrm{(h,\; k)}$ on the graph.

3. From the center, move $r$ units up, down, left, and right to plot four points.

4. Draw a smooth curve connecting these four points to form the circle.

For example, given the equation $(x-3)2^{}+\; (y+2)2^{}=16$, the center is (3, -2) and the radius is $\sqrt{16}=4$. Plot the center at (3, -2), then plot points 4 units away in all directions, and connect them to form the circle.

What is the general form of a circle's equation?

The general form of a circle's equation is:

$x^2+y^2+\mathrm{Dx}+\mathrm{Ey}+F=0$

where $D$, $E$, and $F$ are constants. This form can be converted to the standard form by completing the square for the $x$ and $y$ terms. The standard form is more useful for identifying the circle's center and radius.