Lesson Video: Equation of a Circle | Nagwa Lesson Video: Equation of a Circle | Nagwa

# Lesson Video: Equation of a Circle Mathematics

In this video, we will learn how to find the equation of a circle using its center and a given point or the radius and vice versa.

17:36

### Video Transcript

Equation of a Circle

In this video, weβll discuss what a circle is, how to derive the equation of a circle, and weβll learn how to find the equation of a circle given its center and a point on the circle and how to find the center and radius of a circle given its equation. Before we move on to finding the equation of a circle, letβs briefly recall the mathematical definition of a circle. A circle is the set or locus of all points at given distance from a point. In other words, we have a center of our circle, and every single point on our circle is the same distance away from the center. We call this the radius of the circle.

Now, to help us find an equation for our circle, weβre going to need to plot it on a Cartesian graph. To do this, we need to pick a point for our center; letβs choose the origin. This is the point with coordinates zero, zero, and we represent this with an π. We want to find the equation of a general circle, so weβll just refer to the radius as the letter π. So letβs pick a point on our circle. Weβll call this the point π₯, π¦. We need to find a relationship between π₯ and π¦.

The first thing we can notice from this sketch is that π¦ is not a function of π₯. The reason for this is consider an input of a value of π₯. For π¦ to be a function of π₯, we need one specific output for our function. However, we can see that there will be two outputs in this case. Therefore, this canβt be represented π¦ as a function of π₯, so we canβt use our normal method. Weβre going to try using a geometric method. To start, weβll connect a vertical line from our π₯-axis to the point π₯, π¦. And because this is a vertical line, this means weβll get a right-angled triangle. In fact, we can find the height of this triangle. Since this goes from the π₯-axis to the point π₯, π¦, its height will be π¦.

Itβs also worth pointing out if our point π₯, π¦ had been below the π₯-axis, we would instead need to take the absolute value of π¦. So in both cases, this length will just be the absolute value of π¦. We can, of course, do exactly the same thing to find the width of our triangle. It goes from the π¦-axis to the point π₯, π¦. So its width will be the absolute value of π₯. We now have a right-angled triangle where we know all three of the lengths. And weβre going to apply the Pythagorean theorem to this triangle. Remember, this tells us the width squared plus the height squared will be equal to the hypotenuse squared.

In our case, the width of our triangle was equal to the absolute value of π₯. The height of our triangle was equal to the absolute value of π¦. And the length of our hypotenuse was equal to π. So by the Pythagorean theorem, we must have the absolute value of π₯ squared plus the absolute value of π¦ squared is equal π squared. But remember, it doesnβt matter if a number is positive or negative when we square it; weβll still get the same answer. So in actual fact, we can simplify the absolute value of π₯ squared to be π₯ squared and the absolute value of π¦ squared to just be π¦ squared. So we can just simplify this equation to get π₯ squared plus π¦ squared is equal to π squared. And this is the equation of our circle centered at the origin of radius π.

Itβs also worth pointing out that, technically, since our π₯- and π¦-intercepts do not form a right-angled triangle in this way, we need to check these points satisfy our equation separately. However, using the radius of the circle is equal to π, we can just point the coordinates and see that all of these satisfy our equation. So every point on our circle satisfies the equation π₯ squared plus π¦ squared is equal to π squared. Therefore, weβve derived that the equation of a circle centered at the origin of a radius π is π₯ squared plus π¦ squared is equal to π squared.

But this then begs the question, what would have happened if we didnβt choose the origin as our center? What if weβd picked a different point? Well, we can do very much the same thing. Letβs consider this example where we pick the center to be the point β, π. Once again, weβll pick a point on our circle and call it π₯, π¦. Remember, weβre having the radius of our circle equal to π. And just like we did before, weβll form the same right-angled triangle. This time we need to be a little bit more careful when we find the length and width of our right-angled triangle. For example, to find the height of our right-angled triangle, we go from the point with π¦-coordinate π to the point with π¦-coordinate π¦. In other words, the height of this right-angled triangle will be π¦ minus π.

But remember, sometimes our value of π¦ will be smaller than our value of π. This would give us a negative answer. So weβll take the absolute value of π¦ minus π. And we can do something very similar to find the width of our right-angled triangle. This time we go from β to π₯. This means the width of our right-angled triangle will be the absolute value of π₯ minus β. And just like we did before, we now apply the Pythagorean theorem to this right-angled triangle. This gives us the absolute value of π₯ minus β squared plus the absolute value of π¦ minus π squared is equal to π squared.

And remember, if weβre squaring an absolute value sign, we donβt need the absolute value symbol. So weβll write this equivalently as π₯ minus β all squared plus π¦ minus π all squared plus π squared. And this gives us an equation for our circle centered at the point β, π with a radius of π. But remember, technically, there are four points on our circle which donβt give us a right-angled triangle in this form. But just as we did before, we can use the fact that our radius is π to find the coordinates of each of these points. We can then see that these also satisfy our equation. So weβve shown every point on our circle satisfies this equation.

Therefore, weβve shown a circle centered at the point β, π with a radius π will have the equation π₯ minus β all squared plus π¦ minus π all squared is equal to π squared. So given the center of our circle and the radius of our circle, we can find the equation of the circle. But the opposite is also true. If weβre given the equation of our circle, we can just find the center and the radius. Before we move on to some examples, thereβs one more thing we need to talk about: the general form for the equation of a circle. To do this, we need to distribute the squares over our parentheses in our equation for a circle. Doing this, we get π₯ squared minus two βπ₯ plus β squared plus π¦ squared minus two ππ¦ plus π squared is equal to π squared.

But remember, β, π, and π are all just constants, so we can call negative two β π, negative two π π, and β squared plus π squared minus π squared π. So by using these and then rearranging, we get π₯ squared plus π¦ squared plus ππ₯ plus ππ¦ plus π is equal to zero. This is called the general form for the equation of a circle. Letβs now move on to an example of how to find the center and radius of a circle given its equation.

Find the center and radius of the circle π₯ plus four squared plus π¦ minus two squared is equal to 225.

Weβre given the equation for a circle. We need to use this to find the center of our circle and the radius of our circle. To start, letβs recall the equation of a circle. We know a circle with a center at the point β, π and a radius of π will have the equation π₯ minus β all squared plus one minus π all squared is equal to π squared. And we can see the equation weβre given is almost in this form. We do have to be careful, however. For example, weβre not subtracting a constant from π₯; weβre adding the constant four. But remember, adding four is the same as subtracting negative four. So we can in fact write this as π₯ minus negative four all squared plus π¦ minus two all squared is equal to 225.

Now, itβs really easy to see the center of our circle. Our value of β is negative four, and our value of π is two. All we have to do now is find the radius of our circle. In this case, the radius squared will be equal to 225. So we want π squared is equal to 225. Thereβs a few different ways of doing this. For example, we could take the square roots of both sides of this equation. Normally, we would get a positive and a negative square root. But remember, in this case, this represents the radius. This is a length, so it must be positive. So we get that π is equal to the positive square root of 225. We can calculate this; itβs equal to 15. So we can write 225 as 15 squared. This means the radius of our circle must be equal to 15.

Remember, the center of our circle will be the point β, π. Weβve shown that β is equal to negative four and π is equal to two. And of course, we already showed the radius was 15. Therefore, given the equation of the circle π₯ plus four all squared plus π¦ minus two all squared is equal to 225, we were able to show the center of this circle was the point negative four, two and the radius of this circle was 15.

Letβs now move on to an example where weβre given the graph of a circle and we need to find the equation of this circle.

In the figure below, find the equation of the circle.

Weβre given the graph of a circle. We need to find the equation of this circle. Letβs start by recalling what we know about the equation of a circle. We know that a circle centered at the point β, π and that has a radius of π will have the equation π₯ minus β all squared plus π¦ minus π all squared is equal to π squared. So to find the equation of a circle, we just need to find its center and its radius. Remember, every point on our circle will be equidistant from the center of our circle. In our case, the center of our circle is labeled. We just need to find the coordinates of this circle.

When moving up in a vertical direction to our π₯-axis, we can see the π₯-coordinate of this center is negative five. And by doing the same in the horizontal direction, we can see the π¦-coordinate of this center is negative four. So the center of our circle is negative five, negative four. But how are we going to find the radius of our circle? Remember, this will be the length of any line from the center of our circle to our circle. Thereβs a lot of different choices for which radius we can pick. One such example is to pick the following line. We can see this is a horizontal line which goes from the center of our circle to the point with π₯-coordinate zero. In other words, our radius will be the length of the horizontal line from π₯ is equal to negative five to π₯ is equal to zero.

Of course, the length of this line is just equal to five. So our value of π is five. We now need to substitute our values of β, π, and π into our equation for a circle. Substituting in β is equal to negative five, π is equal to negative four, and π is equal to five, we get π₯ minus negative five all squared plus π¦ minus negative four all squared is equal to five squared. And we could leave our answer like this. However, we can also simplify π₯ minus negative five to π₯ plus five and π¦ minus negative four to π¦ plus four. And of course we, can also evaluate five squared to give us 25. Therefore, we were able to show the equation of the circle given to us in the figure is π₯ plus five all squared plus π¦ plus four all squared is equal to 25.

Letβs now look at an example of finding the equation of a circle in the general form.

Write, in the form ππ₯ squared plus ππ¦ squared plus ππ₯ plus ππ¦ plus π is equal to zero, the equation of the circle of radius 10 and center four, negative seven.

The question wants us to find the equation of a circle of radius 10 and center four, negative seven. In particular, we can see weβre asked to give this in the general form of a circle. To do this, weβll start with the equation of a circle. We know a circle of radius π centered at the point β, π will have the equation π₯ minus β all squared plus π¦ minus π all squared is equal to π squared. And in this case, weβre already told the center of our circle, and weβre already told the radius of our circle. Our circle is centered at the point four, negative seven. So the value of β is four, and the value of π is negative seven. And our circle has a radius of 10, so our value of π is equal to 10.

So we just need to substitute these values into our equation for a circle. Substituting in the center of our circle and the radius of our circle, we get the equation π₯ minus four all squared plus π¦ minus negative seven all squared is equal to 10 squared. Of course, this is not in the general form of the equation of a circle. To do this, we need to distribute our squares over our parentheses. Thereβs a few different ways of doing this. For example, we could use binomial expansion or the FOIL method. Either way, distributing the square over our first set of parentheses, we get π₯ squared minus eight π₯ plus 16.

To distribute the square over our second set of parentheses, it might be easier to rewrite this as π¦ plus seven all squared. Then, if we distribute this, we get π¦ squared plus 14π¦ plus 49. The last thing weβll do is evaluate 10 squared to give us 100. Finally, all we need to do is rewrite this in the form given to us in the question. To start, we rearrange these four terms to give us π₯ squared plus π¦ squared minus eight π₯ plus 14π¦. Then, we want to subtract 100 from both sides of the equation. Doing this, we can see we get a constant term of 16 plus 49 minus 100. And we can calculate this; itβs equal to negative 35.

So we get the equation π₯ squared plus π¦ squared minus eight π₯ plus 14π¦ minus 35 is equal to zero. And we can now see this is in the general form given to us in the question. Therefore, we were able to show the general form of the equation of a circle of radius 10 centered at the point four, negative seven is given by the equation π₯ squared plus π¦ squared minus eight π₯ plus 14π¦ minus 35 is equal to zero.

Letβs now look at an example about finding the equation of a circle, given its center and a point which lies on the circle.

Determine the equation of a circle that passes through the point π΄: zero, eight if its center is π: negative two, negative six.

The question wants us to find the equation of a circle. Weβre told that this circle passes through the point π΄ which has coordinates zero, eight. And the center of this circle is the point π which has coordinates negative two, negative six. Letβs start by sketching what we know about this circle. To start, weβll plot the center of our circle on our graph. Thatβs the point π. It has coordinates negative two, negative six. Next, weβll also plot the point π΄ on our graph. Remember, the circle passes through this point and it has coordinates zero, eight. We donβt actually need to sketch our circle to answer this question; however, weβll sketch a segment of this circle to make it easier to visualize.

Letβs now recall what we know about the equations for circles. We know that a circle centered at the point β, π with a radius of π will have the equation π₯ minus β all squared plus π¦ minus π all squared is equal to π squared. In other words, to find the equation of a circle, we just need to know the coordinates of its center and its radius. Of course, in this case, weβre already told the coordinates of the center of the circle. Weβre told its center is negative two, negative six. So we can set our value of β equal to negative two and our value of π equal to negative six. This means to find the equation of this circle, all we need to do is find its radius.

Remember, π will be the length from the center of our circle to any point on our circle. We only know the coordinates of one point on our circle. Thatβs the point π΄. So in our case, to find the value of π, we need to find the length of the line between negative two, negative six and zero, eight. And thereβs a few different ways of approaching this problem. Weβre going to solve this by drawing the following right-angled triangle. Weβre going to connect the point π to the π¦-axis by using a horizontal line. The width of this triangle will then be the absolute value of the π₯-coordinate of π, which is two. The height of this section will be the absolute value of the π¦-coordinate of π, which is six. And the height of this section of the triangle will be the π¦-coordinate of π΄, which is eight.

Then, we can just combine these two lengths together to find that the height of our triangle will be eight plus six, which is 14. Now, we have a right-angled triangle where we know its width and we know its height. So we can apply the Pythagorean theorem to find the length of its hypotenuse. This tells us π squared is equal to two squared plus 14 squared. If we calculate two squared plus 14 squared, we get 200. And at this point, we can calculate the radius. It would be the positive square root of 200 because, remember, the radius is a length, so it must be positive. However, in our formula, we only need the value of π squared, which we know is 200.

So we can just substitute π squared is equal to 200, β is equal to negative two, and π is equal to negative six into our equation for a circle. This gives us π₯ minus negative two all squared plus π¦ minus negative six all squared is equal to 200. Finally, we can simplify π₯ minus negative two to give us π₯ plus two and π¦ minus negative six to give us π¦ plus six. And this gives us our final answer. Therefore, we were able to show given that a circle has its center at the point π, which is negative two, negative six, and the circle also passes through the point π΄, which has coordinates zero, eight, we were able to show the equation of this circle must be π₯ plus two all squared plus π¦ plus six all squared is equal to 200.

Letβs now go over the key points of this video. First, by using the definition of a circle and the Pythagorean theorem, we were able to derive that a circle of radius π with its center at the point β, π will have the equation π₯ minus β all squared plus π¦ minus π all squared is equal to π squared. We also saw that we can find the center and radius of a circle just from its equation. If a circle has the equation π₯ minus β all squared plus π¦ minus π all squared plus π squared, then its center will be the point β, π and its radius will be π, of course, where π is a positive number because it represents a length.

We also showed that we can find the equation of a circle given only the coordinates of its center and a point on the circle. And this is because we can find the radius in this case because itβs just the distance between the center and the point that weβre given. And we could find this distance by using the Pythagorean theorem. Finally, by distributing the squares over the parentheses in our equation for a circle, we were able to show that the equation of a circle in its general form is given by the equation π₯ squared plus π¦ squared plus ππ₯ plus ππ¦ plus π is equal to zero for some constants π, π, and π.