Lesson Explainer: Circles and Triangles | Nagwa Lesson Explainer: Circles and Triangles | Nagwa

# Lesson Explainer: Circles and Triangles Mathematics

In this explainer, we will learn how to identify inscribed angles in semicircles and circumcircles of triangles and find the equation of a circle given three points on the circumference.

Given three noncolinear points , , and in the plane, there is a unique triangle with those points as vertices.

Now, consider the perpendicular bisector of the line segment . Every point on lies at an equal distance from point and point . Similarly, every point on the bisector is equidistant from and .

This means that the point where and intersect is the same distance from all three vertices, , , and .

If we now trace a circle with center passing through the three vertices , , and we get the unique circle with these properties. This circle is called the circumcircle of the triangle.

The point where the three perpendicular bisectors of the sides of the triangle intersect is the center of the circumcircle and is called the triangle’s circumcenter. This is one type of triangle center; there are many others such as the orthocenter, where the three altitudes of the triangle meet, and the centroid, where the three lines that connect each vertex to the midpoint of the opposite side meet.

Be aware that the circumcenter of a triangle does not always lie inside the triangle. When is acute, which is to say all of its angles are smaller than a right angle, then the circumcenter lies inside . When contains an angle larger than a right angle, then the circumcenter lies outside the triangle.

Now, suppose is a right triangle with hypotenuse . By looking at the angles, we can see that the the perpendicular bisector of makes a smaller similar triangle inside , scaled down by a factor of .

Similarly, the perpendicular bisector of cuts out a similar triangle of scale factor inside . This means that both of these lines intersect the hypotenuse at the point halfway along its length. Of course, the perpendicular bisector of the hypotenuse, by definition, intersects the hypotenuse at its midpoint too. Altogether, this shows that if is a right triangle, then its circumcenter is the midpoint of its hypotenuse.

In this situation, the hypotenuse of is a diameter of its circumcircle and its right angle lies on the circumference.

Recall the circle theorem, which states that the angle in a semicircle is always , or, in other words, if two points and form the diameter of a circle, then subtended at any third point on the circumference must be a right angle.

What we have just established is the converse to this theorem: if is a right triangle with hypotenuse , then is a diameter of its circumcircle.

### Properties: Triangles and Circumcircles

Let , , and be three noncolinear points in the plane.

• There is a unique triangle with vertices , , and .
• The perpendicular bisectors of the sides , , and meet at a single point , called the circumcenter of the triangle.
• The circumcenter of the triangle is the center of the circumcircle, which is the unique circle passing through the vertices , , and .
• If is a right triangle, then its circumcenter lies at the midpoint of its hypotenuse and its hypotenuse is a diameter of its circumcircle.

Let us put some of these properties to use in identifying a diameter of a circle.

### Example 1: Identifying the Correct Point That Makes 𝐴𝐵 A Diameter of A Circle

, , , , and are five points on a circle.

Which of the chords , , and is a diameter?

To identify a diameter, we will use the fact that two points and on the circumference of a circle are a diameter if and only if they form a right triangle with any other point on the circumference. In this example, we proceed by trial and error: we pick one of the given chords and check to see if it forms a right triangle with one of the other points. If not, we pick another chord and try again.

Let us test the chord first. We will check to see if is a right triangle using the Pythagorean theorem, which states that is a right triangle if and only if

Let us calculate these lengths. We have and but so is not a right triangle and is not a diameter.

Let us check . We want to know if is a right triangle, so we check if

We have already calculated , so we only need to work out the other two lengths. We have and and, indeed, confirming that is a right triangle and that is a diameter of the circle. Since we have identified the diameter and the question tells us that only one of the given chords is a diameter, there is no need to check .

We will now see how to work out the equation of a circle, given some points on its circumference, by establishing that the chord connecting two of those points is a diameter of the circle.

### Example 2: Showing That Three Points Form a Right Triangle Inscribed in a Semicircle and Finding the Equation of the Circle

The points , , and lie on a circle.

1. What type of angle is ?
2. Write down an equation for the circle in the form , where , , and are constants to be determined.

Part 1

We can find what type of angle is using the Pythagorean theorem, which states that if and only if

Calculating lengths, we find and

So, indeed, which shows that is a right angle.

Part 2

Since is a right angle, we know that must be a diameter of the circle. This allows us to identify the center of the circle as the midpoint of and its radius as half of .

We calculate the coordinates of the midpoint of using the formula

Now, we can write down the equation for the circle: where is the length of the radius, which is, of course, half the length of the diameter . We have already calculated , so and and the equation of the circle is

In the previous example, we saw how to find the center of circle, given some points on its circumference, when two of those points lie on a diameter. We will now use a different technique to find a circle’s center, given points on its circumference, in the situation where no two of those points lie on a diameter.

As discussed in the introduction, if we have three noncolinear points , , and , then the point of intersection of any two of the perpendicular bisectors of the sides , , and is equidistant from all three vertices. The point is therefore the circumcenter of , , and , that is, the center of the circumcircle, the unique circle passing through , , and .

If we are given that the points , , and lie on a circle, then the circumcenter must be the center of that circle. We can sum this up in the following property.

### Property: Perpendicular Bisectors of Chords

The perpendicular bisectors of any two distinct chords of a circle intersect at the circle’s center.

### Example 3: Finding the Perpendicular Bisector of a Chord of the Circle and Using It to Find the Center of the Circle

The points , , and lie on the circumference of a circle. The equation of the perpendicular bisector of is . Find the equation of the perpendicular bisector of and the coordinates of the center of the circle.

To find the equation of the perpendicular bisector of a chord, we need two pieces of information: the slope of the chord and the coordinates of its midpoint. The slope of the chord is given by

The slope of the perpendicular bisector is then the negative reciprocal of the slope of the chord, so, in this case, it is . We can now write down the following equation for the perpendicular bisector and we can calculate the value of the constant using the coordinates of the midpoint of the chord, which are given by

Substituting into , we have giving us the equation for the perpendicular bisector of .

The center of the circle lies at the intersection of the two perpendicular bisectors and . Solving for , we get and substituting back into , we have

Thus, the coordinates of the center of the circle are .

In fact, once we have used the perpendicular bisectors of chords of a circle to find its center, we simply need to find the square of its radius to write an equation for the circle. This is what we are going to do in the next example.

### Example 4: Using the Bisectors of Two Chords of a Circle to Find the Equation of the Circle

The points , , and lie on a circle. Given that the equations of the perpendicular bisectors of and are and , write down an equation for the circle in the form , where , , and are constants to be determined.

Since we are given the equations and of the perpendicular bisectors of two distinct chords of the circle, we can immediately find the center of the circle at their intersection. Substituting into , we have and since , the coordinates of the center are . The only other thing we need to write down an equation for the circle is the square of the length of its radius, . This is simply the square of the distance from the center to any point on the circumference, say, . This square distance is given by

We can therefore write down the following equation for the circle:

We are now in a position to calculate an equation of a circle given any collection of at least three distinct points on its circumference using the perpendicular bisectors of any pair of chords connecting them.

### Example 5: Finding the Equation of the Circle given Four Points on the Circumference

The points , , , and lie on a circle. By finding the perpendicular bisectors of and , write down an equation for the circle in the form , where , , and are constants to be determined.

To write down equations of the perpendicular bisectors of and , we need to work out the slopes and midpoints of these chords. The slope of the chord is given by and the slope of is given by

We have two equations for bisectors, where the slopes are the negative reciprocals of the corresponding slopes of the chords and the constants and are to be determined using, for each bisector, a point lying on it: its midpoint.

The midpoint of is given by and the midpoint of is given by

Substituting the coordinates of into , we find

A similar calculation yields . So, the two perpendicular bisectors are given by

Since we are looking for the intersection of these two lines (that is, the center of the circle), we equate the -coordinates and solve for : and then, we substitute back into for

Hence, the coordinates of the center of the circle are and we have the following equation for the circle: where is the square of the radius of the circle. We can calculate as the square of the distance from the center to any point on the circumference, say, , by

So, the equation of the circle is given by

Let us finish by recapping a few important concepts from this explainer.

### Key Points

• There is a unique circle passing through any three noncolinear points , , and in the plane. It is called the circumcircle of .
• The center of the circumcircle is called the circumcenter. It is equidistant from the points , , and and lies at the intersection point of the perpendicular bisectors of the sides , , and .
• Three points in the plane form the vertices of a right triangle if and only if the hypotenuse is the diameter of the circumcircle through the vertices. We can use this fact to calculate the equation of the circumcircle.
• The perpendicular bisectors of any two distinct chords of a circle intersect in the circle’s center. We can use this fact to calculate the equation of a circle given any collection of at least three distinct points on its circumference.