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 is x - h + y - k = r^2, where (h, k) is the center. Converting from general to standard form involves completing the square for both x and y terms.

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 can 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. 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, 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. And 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 x 2+y 2=r 2. 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 x 2+y 2=4 2. 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-2 2+y-1 2=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 9, which is 3. So that means our radius is 3. Now our third step says to plot 4 points that 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 want to 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: $x^2+\left(y-1\right)^2=9$

• A.
• B.
• C.
• D.

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

• 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)² + (y - k)² = r², 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 (0, 1), so horizontally we're at 0 and vertically we're at 1. I can put h = 0 and k = 1, and that's going to be our center. Now we're also told that our radius is 3. So that means I can go 3 units to the right, 3 units to the left, 3 units up, and 3 units down. And this is going to give us our circle, which is the shape that we're looking for. 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)2 = 9

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, horizontally we're going to be at 1, and then vertically we're going to be at -2. That's the center of our circle. I can see that our radius is 1, so I can go 1 unit up, down, left, and 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:

(x-1)2 + (y+2)2 = 1

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. First, I'm going to find our center on this graph, which is going to be horizontally at -6 and vertically at 3. This one's going to be a little tricky because our radius is the square root of 5, approximately 2.2. So here's how you can draw this circle:

The equation is actually pretty straightforward. So to draw the equation, all you need to do is use the standard equation:

(x+6)2 + (y-3)2 = 5

And that is how you can find the graphs and equations for these circles based on the characteristics that we were given in this problem. 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. 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. 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 8 we see here and subtracting it on both sides of the equation. So, when you do this, you'll get all the x squareds and x here on one side grouped together, and you'll have all the ys 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. And 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 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. Now, your last step from here is going to be to factor both this form, the xs 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 3 squared, 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

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 = $c\left(0,0\right)$, radius $r=2$.
• B. Is a circle, center = $c\left(0,0\right)$, radius $r=3$.
• C. Is a circle, center = $c\left(1,-2\right)$, radius $r=3r=2$.
• 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 = $c\left(0,0\right)$, radius $r=4$.
• B. Is a circle, center = $c\left(0,0\right)c\left(2,-4\right)$, radius $r=4$.
• C. Is a circle, center = $c\left(-2,4\right)c\left(0,0\right)$, 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). If the circle is not centered at the origin, the equation becomes:

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

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

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

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

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

The center of the circle is given by the coordinates $(h,k)$, and the radius is the square root of the constant term on the right side of the equation, $\sqrt{r}$.

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. To convert this to the standard form, you need to complete the square for both $x$ and $y$ terms.

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

To convert the general form $x^2+y^2+\mathrm{Dx}+\mathrm{Ey}+F=\; 0$ to the standard form, follow these steps:

1. Group the $x$ and $y$ terms together.
2. Move the constant term $F$ to the other side of the equation.
3. Complete the square for both $x$ and $y$ terms.
4. Rewrite the equation in the form $(x-h)2^2+\; (y-k)2^2=r^2$.

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

To graph a circle given its equation in standard form $(x-h)2^2+\; (y-k)2^2=r^2$, follow these steps:

1. Identify the center $(h,k)$ and the radius $\sqrt{r}$.
2. Plot the center on the coordinate plane.
3. From the center, move $r$ units up, down, left, and right to mark four points.
4. Draw a smooth curve connecting these points to form the circle.