# Lesson Video: Constructing Circles Mathematics

In this video, we will learn how to construct circles given one, two, or three points.

18:37

### Video Transcript

Constructing Circles

In this video, we will learn how to construct circles given one, two, or three distinct points. We will also learn under which conditions can we construct these circles and, in particular, how to find the circumcircle of a triangle. Thatβs a circle which passes through all three vertices of a triangle.

To begin, letβs start by recalling what we mean by a circle. A circle is the set of all points a specific distance from a point known as its center. In particular, the set distance is usually known as the radius of the circle. So, every single point on our circle will be equidistant from the point π. If we call the radius π, then the distance between π and π΄ will be π for any point on our circle. And itβs also worth noting the line segment between π and π΄ is called a radius of the circle. So the length of the radius is π, and we also sometimes refer to this distance just as the radius as well.

Now that weβve recapped the definition of a circle, weβre ready to start discussing the conditions under which we can construct a circle. Letβs start by trying to construct a circle through a given point π΄. We can do this by just choosing any distinct point on the plane to be the center of our circle. Letβs label this point π. Then, since every point on our circle needs to be equidistant from the point π, we can find the radius of our circle. It will be the distance between π and π΄. This then defines our circle mathematically; it has center π and radius π.

However, we can also sketch this circle by using a compass. We put the needle point of our compass at the point π, and we adjust our compass so that the radius is going to be equal to π, so our pencil lies at the point π΄. Moving the compass around then sketches a circle which passes through the point π΄. But remember, we could choose any distinct point in the plane π to be the center of our circle. This means that thereβs infinitely many circles which pass through the point π΄. In fact, an even stronger result holds true. Thereβs infinitely many circles congruent to the circle we just drew which passes through the point π΄. Remember, for two shapes to be congruent, they need to have the same size and shape. So for a circle to be congruent to our given circle, it just needs to be a circle with the same radius.

To see why this is true, letβs start by sketching a circle of radius π centered at the point π΄. Of course, since π is a distance of π away from the point π΄, π lies on this circle. In fact, any point on this circle is a distance of π away from the point π΄. For example, the distance between the point π two and π΄ is π. But now we can sketch a circle of radius π around the point π two. This then gives us a second circle which passes through the point π΄, which also has radius π. Therefore, we sketched two circles of radius π which both pass through the point π΄. These are congruent circles. And we can choose any point on the circle of radius π around the point π΄ to be the center of our circle. This means there is infinitely many congruent circles which pass through the point π΄ which have radius π.

Now that weβve seen how to construct a circle through one given point, letβs move on to trying to construct a circle between two given distinct points. Letβs say the points π΄ and π΅. To construct a circle, weβve seen we need to find two things. We need to determine its center and its radius. Since both points π΄ and π΅ need to be on our circle, they need to be the same distance away from the center of our circle. So the first thing weβre going to need to do is find a point equidistant from π΄ and π΅. And to do this, weβre going to recall the following result. Every point which is equidistant from two points π΄ and π΅ will lie on the perpendicular bisector of the line segment π΄π΅.

Before we see why this might be true, letβs start by constructing the perpendicular bisector of the line segment π΄π΅. Letβs start by sketching the line segment π΄π΅, and weβll call the midpoint of this line segment π. And we know that π΄π and ππ΅ have the same length. And in fact, this is already enough to answer our question. π is equidistant from the point π΄ and π΅, so we could choose this to be the center of our circle. Then all we would do is choose the radius to be the distance between π΄ and π or the distance between π and π΅. However, this is only one circle which passes through π΄ and π΅. Letβs see if we can find more.

We then draw a line perpendicular to the line segment π΄π΅ which passes through π. This is the perpendicular bisector of the line segment π΄π΅. Our property then tells us any point on this line is equidistant from π΄ and π΅, and every point which is equidistant is also on this line. And we can actually prove that this is true. For example, letβs show that π two is equidistant from π΄ and π΅. We have two right triangles, triangle π΄ππ two and triangle π΅ππ two. We can find an expression for the hypotenuse of either triangle by using the Pythagorean theorem.

Remember, this tells us the square of the hypotenuse is equal to the sum of the squares of the two shorter sides in a right triangle. So in our left-right triangle, we have the length of π΄π two squared is equal to the length of π΄π squared plus the length of ππ two squared. But remember, the length of π΄π is equal to the length of ππ΅. So in our expression, we can replace the length of π΄π with ππ΅. This then gives us the length of ππ΅ squared plus the length of ππ two squared.

But then by the Pythagorean theorem, this is actually just the square of the hypotenuse in our other right triangle. So by the Pythagorean theorem, this is the length of π΅π two squared. Then, all we need to do is bring down the left-hand side of our equation. The length of π΄π two squared is equal to the length of π΅π two squared. Then, since both of these are lengths, we can just take the positive square root of both sides of the equation. The distance between π΄ and π two is the same as the distance between π΅ and π two. Therefore, any line on our circle is equidistant from π΄ and π΅, which means it can be the center of a circle which passes through both π΄ and π΅.

For example, to sketch a circle from the point π three which passes through π΄ and π΅, we just put our compass point at the point π three and then choose the radius to be the distance between π three and either π΄ or π΅. Once again, because we can choose any point on the perpendicular bisector, thereβs infinitely many circles which pass through both π΄ and π΅. Letβs go for a question which will allow us to find the smallest radius circle which passes through two points π΄ and π΅.

Consider the two points π΄ and π΅. What is the radius of the smallest circle that can be drawn in order to pass through the two points?

In this question, we need to determine the radius of the smallest circle which passes through two points π΄ and π΅. To answer this question, we start by recalling that every point on a circle will be equidistant from its center. And we also recall another fact. Every point which is equidistant from two points π΄ and π΅ will lie on the perpendicular bisector of π΄ and π΅. So the center of our circle must lie on the perpendicular bisector of π΄ and π΅. So letβs start by constructing this. We need to draw the line segment from π΄ to π΅ and then mark the midpoint of this line segment. Weβll call this π. Then the perpendicular bisector is the line perpendicular to π΄π΅ which passes through π. Any point which lies on this line is equidistant from π΄ and π΅.

And so these can be choices for the center of our circle. For example, π, π two, π three, or π four could be the centers of our circle. Then the radius of this circle would just be the distance between the center and any point on the circle, for example, the distance between the center and point π΄. Graphically, it appears the further we get away from the point π, the larger this radius is. In fact, we can prove this is true. For example, we can see that triangle π΄ππ two is a right triangle. In particular, this means its hypotenuse must be longer than the other sides. π΄ππ three is also a right triangle, and π΄ππ four is another right triangle. So π three is bigger than π, and π four is bigger than π. Therefore, π will be the circle with the smallest radius which passes through π΄ and π΅.

And although we were not asked to in this question, we can even sketch this circle by putting the points of our compass at the point π and then the tip of our compass at either the points π΄ or π΅. We were asked to find the radius of this circle. Remember, we chose π to be the midpoint of the line segment π΄π΅. And this means the value of π is going to be one-half the distance from π΄ to π΅. In other words, the radius of the smallest circle which passes through two distinct points π΄ and π΅ is one-half the distance from π΄ to π΅.

So far, weβve constructed circles through one point and through two distinct points. Now letβs try to construct the circle through three distinct points π΄, π΅, and πΆ. To do this, we need to find the center of our circle which needs to be equidistant from all three points. If the center of the circle is equidistant from all three points, then it must be equidistant from both π΄ and π΅. And we can find the set of all points equidistant from both π΄ and π΅ by finding the perpendicular bisector. So we start by finding the midpoint of the line segment between π΄ and π΅. And then we draw the line through the midpoint π one which is perpendicular to the line segment π΄π΅. Every point equidistant from both π΄ and π΅ lies on the perpendicular bisector between π΄ and π΅. So if the center of our circle exists, it must lie on this line.

But the exact same reasoning is true for points πΆ and π΅. The center of our circle is equidistant from πΆ and π΅, so it must lie on the perpendicular bisector of the line segment π΅πΆ. So weβll also sketch the perpendicular bisector of line segment π΅πΆ. We find the point π two which is the midpoint of this line segment and then draw a line perpendicular to this line which passes through π two. Every point on this line is equidistant from π΅ and πΆ, and we can see thereβs a point of intersection between the two perpendicular bisectors. Since this point is on both perpendicular bisectors, itβs equidistant from π΄, π΅, and πΆ. This means we can construct a circle with center π which passes through all three points π΄, π΅, and πΆ. The radius of this circle will then be the distance between either π and π΄, π and π΅, or π and πΆ.

We can then ask another question. Is there another circle which passes through these three points? And in fact, using the exact same reasoning, we can show that there is not. This is the only circle which passes through these three points. To see why this is true, remember, the center of our circle must lie on the perpendicular bisector between π΄ and π΅ and the perpendicular bisector between π΅ and πΆ. And in the plane, distinct lines can only ever intersect in at most one point. So π is the only point on both of these lines. Therefore, itβs the only possible center of this circle. And so this is the unique circle which passes through π΄, π΅, and πΆ.

In fact, thereβs slightly more we can show. What if we wanted a circle which also passes through another point π·? We can see that π· does not lie on the circle. And since this is the only circle which passes through π΄, π΅, and πΆ, we can conclude that there is not a circle which passes through all four of these points. But weβve only shown that we can construct a circle between these three points. What if we had three different points? Can we do this in general? In our next example, weβll discuss the condition under which we can sketch a circle through three points.

True or false: if a circle passes through three points, then the three points should belong to the same line.

To answer this question, letβs start by sketching a circle. We can choose three points on this circle. Letβs label these π΄, π΅, and πΆ. We can see that these three points do not lie on the same straight line. However, the circle does pass through these three points. This is enough to prove the statement is false. If a circle passes through three points, then the three points do not need to belong to the same line.

This could make us ask an interesting question. What would happen if the three points were in a straight line? For example, can we construct a circle between these three points π, π, and π? The center of the circle must be equidistant from both three of these points. In particular, the center of the circle must be equidistant from both π and π. And we know that the perpendicular bisector of π and π contains all points equidistant from both π and π. So the center of our circle must lie on this line. Similarly, the center of this circle is equidistant from π and π, so it must lie on the perpendicular bisector of π and π.

And then we see a problem. These two lines are parallel. They never intersect, so thereβs no point equidistant from π, π, and π. This leads to a slightly stronger result. There is no circle which passes through three points which lie on the same straight line. But we can answer this question as false. A circle that passes through three points do not need to have the three points lie on the same straight line. The statement is false.

In fact, thereβs an even stronger result still that weβll need to know. If three points are noncolinear, which means they donβt lie on the same straight line, then there exists a unique circle which passes through all three of the points. And we saw how to construct this circle earlier. We just find the intersection of the perpendicular bisectors of two pairs of the points. This gives us the center of our circle, and then we can find the radius of our circle as the distance between the center and any of the three points. Letβs now see an example of how we can use this property on a problem involving triangles.

True or false: a circle can be drawn through the vertices of any triangle.

To answer this question, we might start by drawing a triangle and seeing how we might construct a circle which passes through all three vertices of the triangle. However, this is not necessary since we can just recall a useful fact about circles. We know that any three noncolinear points have a unique circle which passes through all three points, where we remember noncolinear points means they donβt lie on the same straight line. And the vertices of a triangle donβt lie on the same straight line. Thatβs what it means for a shape to be a triangle. For example, if we had three points which lied on a straight line and we tried to connect the vertices together, we would just end up with a straight line.

Therefore, we can just apply this property to show that the answer is true. A circle can be drawn through the vertices of any triangle. This is called the circumcircle of the triangle.

Letβs now see an example where we determine how to construct the circumcircle of a given triangle.

In the following two figures, two types of construction have been made on the same triangle π΄π΅πΆ. Which point will be the center of the circle that passes through the triangleβs vertices?

In this question, weβre asked to determine the center of a circle which passes through all three of the triangleβs vertices. Thatβs the points π΄, π΅, and πΆ. And to do this, we need to remember, the center of a circle will be equidistant from all points on the circumference of the circle. Therefore, we need to determine which points in either of the two diagrams is equidistant from all three points π΄, π΅, and πΆ. To do this, we know in the second diagram weβre given the perpendicular bisectors of all three sides of the triangle. We can see that these are the perpendicular bisectors because they cut the sides of the triangle at right angles. And they cut the sides of the triangle in half represented by either the one, two, or three lines.

This is useful because every single point on the perpendicular bisector between π΅ and πΆ will be equidistant from π΅ and πΆ. For example, the point πΉ is equidistant from both π΅ and πΆ. This result is true for any perpendicular bisector, so we can see that πΉ also lies on the perpendicular bisector between π΄ and π΅. So πΉ is also equidistant from π΄ and π΅. This means itβs equidistant from all three vertices of the triangle. Therefore, if we draw a circle centered at πΉ of radius the distance between π΄ and πΉ, we would get a circle which passes through all three vertices of the triangle.

The same cannot be true for the point π; although itβs found by taking the bisectors of each of the angles of our triangle, π will not necessarily be equidistant from all three vertices of the triangle. Therefore, we can conclude the point πΉ is the center of a circle which passes through all three vertices of the triangle.

In our final example, weβll go for a property involving the number of intersections possible in two distinct circles.

True or false: two distinct circles can intersect at more than two points.

To answer this question, letβs start by thinking about how circles can intersect. Thereβs two different ways that circles canβt intersect. For example, we could have one circle inside another circle, or we could just have two circles next to each other. Itβs also possible for two circles to intersect at a single point, and in fact, thereβs two different ways this can happen. We can have both circles next to each other, or we can have one circle inside of another circle. Both of these only intersect at a single point. Finally, itβs also possible for circles to intersect at two distinct points, for example, the following shape which looks like a Venn diagram.

But what if we wanted two circles which intersect at three distinct points? Well, if the three points of intersection were noncolinear, then we know thereβs a unique circle which passes through all three of these points. This is called the circumcircle of triangle π΄π΅πΆ, and since itβs unique, we canβt have two distinct circles in this case. But weβre not done yet. We also need to consider the case if the three points of intersection lied on the same straight line. But this is also not possible because we recall there is no circle which passes through three points which lie on the same straight line. Therefore, weβve shown the answer is false. Two distinct circles cannot intersect at more than two points.

Letβs now go over the key points of this video. First, we showed that there are an infinite number of circles passing through any single point π΄. We can then construct this circle by choosing any point to be our center π, and then the radius of our circle will be the distance between π΄ and π. Next, we saw that there are an infinite number of circles passing through any two distinct points π΄ and π΅. We can then construct this circle by choosing any point on the perpendicular bisector of the line segment π΄π΅ to be the center of our circle. And the radius of this circle will then be the distance between π΄ and π or the distance between π΅ and π.

Next, we also saw that there is a unique circle passing through any three noncolinear distinct points π΄, π΅, and πΆ. We can then construct this circle by recalling the center of this circle will be the intersection between the perpendicular bisectors of the line segments between any two pairs of points, for example, the perpendicular bisectors of the line segment π΄π΅ and π΅πΆ. And the radius of this circle will be the distance between π΄ and π, π΅ and π, or πΆ and π. And itβs also worth pointing out we showed that the noncolinearity condition is necessary. If the three points lie on the same line, then there is no circle which passes through the three points. Finally, we saw that two distinct circles will intersect at at most two points.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.