Lesson Explainer: Special Segments in a Circle Mathematics

In this explainer, we will learn how to use the theorems of intersecting chords, secants, or tangents and secants to find missing lengths in a circle.

Let’s begin by recalling the names of the various parts of a circle.

We can then focus on some specific parts. If a line segment intersects the circumference of a circle exactly once such that it is perpendicular to the radius at that point and it has an endpoint on the circumference of the circle, it is called a tangent segment. If a line segment has one endpoint outside the circle, one endpoint on the circle, and a point between these points that intersects the circle, it is called a secant segment.

Having recapped, previously, the names of different line segments in a circle and demonstrated how properties of these line segments can help us to solve problems, we will consider two different theorems that will help us to solve further problems involving circles.

Theorem: The Intersecting Chords Theorem

When two chords intersect inside a circle, each chord is divided into two segments. These are called chord segments. In the circle given, these segments are defined by 𝐴𝐸, 𝐸𝐡, 𝐢𝐸, and 𝐸𝐷.

If chord 𝐴𝐡 and chord 𝐢𝐷 intersect at point 𝐸, 𝐴𝐸×𝐸𝐡=𝐢𝐸×𝐸𝐷.

Alternatively, π‘ŽΓ—π‘=𝑐×𝑑,π‘Žπ‘=𝑑𝑏.

This means that if we know any three of these values, we can find the fourth. Let’s demonstrate a simple application of this theorem.

Example 1: Finding the Length of a Chord in a Circle

Given that 𝐸𝐢=4, 𝐸𝐷=15, and 𝐸𝐡=6, find the length of 𝐸𝐴.

Answer

Recall that the intersecting chords theorem tell us that if chord 𝐴𝐡 and chord 𝐢𝐷 of the same circle intersect at point 𝐸, 𝐴𝐸×𝐸𝐡=𝐢𝐸×𝐸𝐷.

We are given 𝐸𝐢=4, 𝐸𝐷=15, and 𝐸𝐡=6, so we can substitute these values into this formula, where 𝐢𝐸=𝐸𝐢 and 𝐴𝐸=𝐸𝐴, to obtain 𝐸𝐴×6=4Γ—156𝐸𝐴=60𝐸𝐴=10.

Hence, the length of 𝐸𝐴 is 10 units.

In our next example, we will demonstrate how to apply this theorem when the ratio of the lengths of two chord segments is given.

Example 2: Finding the Length of Two Segments Drawn in a Circle Using the Ratio between Them

If 𝐸𝐴𝐸𝐡=87, 𝐸𝐢=7cm, and 𝐸𝐷=8cm, find the lengths of 𝐸𝐡 and 𝐸𝐴.

Answer

The first thing we can do is take the information we are given and enter it onto our diagram.

And then, we recall what we know about intersecting chords: 𝐸𝐢×𝐸𝐷=𝐸𝐡×𝐸𝐴.

We can use this to form an equation in terms of 𝐸𝐴 and 𝐸𝐡, where 𝐸𝐢=7cm and 𝐸𝐷=8cm:

7Γ—8=𝐸𝐡×𝐸𝐴56=𝐸𝐡×𝐸𝐴..

At this point, it does not seem as though we have enough information to solve the problem.

However, we do know that 𝐸𝐴𝐸𝐡=87.

So, 𝐸𝐴=8𝐸𝐡7.

We can then substitute this into 56=𝐸𝐡×𝐸𝐴 to give 56=𝐸𝐡×8𝐸𝐡7392=8𝐸𝐡49=𝐸𝐡∴𝐸𝐡=7.

Note: We do not need to include the negative root of 49 since 𝐸𝐡 is a length.

So, we can therefore say that 𝐸𝐴=8𝐸𝐡=7.cmandcm

Next, we will consider two further theorems: the intersecting secants theorem and the tangent–secant theorem.

Theorem: Intersecting Secants Theorem

Given secant segments 𝐴𝐸 and 𝐢𝐸, 𝐡𝐸×𝐴𝐸=𝐷𝐸×𝐢𝐸.

Alternatively, π‘ŽΓ—π‘=𝑐×𝑑.

Theorem: The Tangent–Secant Theorem

This is a special case of the intersecting secants theorem and applies when the lines are tangent segments.

In the diagram, 𝐸𝐡=π‘Ž, 𝐸𝐴=𝑏, and 𝐸𝐢=𝑐. In the case where one line is a secant segment and the other is a tangent segment, π‘ŽΓ—π‘=𝑐.

In our next example, we will use one of these theorems to solve a problem involving two secants that intersect outside the circle.

Example 3: Finding an Unknown Length of a Proportion Resulting from Two Circle Secants Drawn from the Same External Point

If 𝐸𝐢=10cm, 𝐸𝐷=6cm, and 𝐸𝐡=5cm, find the length of 𝐸𝐴.

Answer

When we look at our figure, we see that we have two secant segments that intersect outside the circle at point 𝐸.

We can add the dimensions that we have been given to our diagram.

To enable us to find 𝐸𝐴, we recall the intersecting secants theorem:

π‘ŽΓ—π‘=𝑐×𝑑.

Applying this to our question, we can say that 𝐸𝐴×𝐸𝐡=𝐸𝐷×𝐸𝐢.

Now, if we substitute in the values we know, we get 𝐸𝐴×5=6Γ—105𝐸𝐴=60𝐸𝐴=12.

Hence, the length of 𝐸𝐴 is 12 cm

In the next example, to find a missing length, we will have to use not only information we know about secants and tangents, but also information we know about triangles.

Example 4: Finding the Length of a Tangent to a Circle Using the Application of Similarity in Circles

In the figure shown, the circle has a radius of 12 cm, 𝐴𝐡=12cm, and 𝐴𝐢=35cm. Determine the distance from 𝐡𝐢 to the center of the circle, 𝑀, and the length of 𝐴𝐷, rounding your answers to the nearest tenth.

Answer

The first thing we will do is take the information we are given and add it to our diagram.

The two lengths that we are trying to find are the perpendicular distance from 𝐡𝐢 to the center of the circle, 𝑀, and 𝐴𝐷.

To solve the first part of the question, we will calculate the distance from 𝐡𝐢 to 𝑀.

Let’s recall some facts about triangles.

We know the length of 𝑀𝐢 as this is the radius of the circle, which means the distance from 𝑀 to 𝐡 will also be 12 cm.

What this now gives us is an isosceles triangle for which we can calculate the height; the height of an isosceles triangle is the length of its median, which is the line segment that joins the vertex to the midpoint of the opposite side. This means it divides the base into two equally sized segments.

Next, we can calculate the length of the base of each of the right triangles: 23Γ·2=11.5.cmcm

From here, we can use the Pythagorean theorem to find the length we are looking for:

π‘₯=12βˆ’11.5π‘₯=144βˆ’132.25π‘₯=11.75π‘₯=√11.75π‘₯=3.4278.

Then, if we round this to the nearest tenth, we get 3.4 cm.

Next, we will calculate the length of 𝐴𝐷.

Since 𝐴𝐷 is a tangent and it intersects the secant 𝐴𝐢 at the point 𝐴, we can say that 𝐴𝐷=𝐴𝐡×𝐴𝐢𝐴𝐷=12Γ—35𝐴𝐷=420𝐴𝐷=√420𝐴𝐷=20.4939…𝐴𝐷=20.5().tothenearesttenth

When finding √420, we were only interested in the positive result as we were finding a distance, which cannot be negative.

Therefore, the distance from 𝐡𝐢 to the center of the circle, 𝑀, is 3.4 cm (to the nearest tenth).

The length of 𝐴𝐷 is 20.5 cm (to the nearest tenth).

We will now solve a problem that combines algebraic manipulation with the skills we have explored in this explainer.

Example 5: Finding the Length of the Chords in a Circle Using the Properties of Chords

In the following figure, find the value of π‘₯.

Answer

Inspecting the diagram, we see it consists of a circle with two chords: 𝐴𝐡 and 𝐢𝐷. The two chords intersect at a point 𝐸 inside of the circle. In the question, we are asked to find π‘₯, which has been used in expressions for the lengths of the segments of the two chords.

Therefore, to solve this problem, we need to recall the intersecting chord theorem.

If chord 𝐴𝐡 and chord 𝐢𝐷 intersect at point 𝐸, then 𝐴𝐸×𝐸𝐡=𝐢𝐸×𝐸𝐷.

We can use this to find an equation for π‘₯ by substituting in the expressions we have been given for the dimensions: (π‘₯+8)(π‘₯+3)=π‘₯(π‘₯+12).

This equation can then be solved to find the value of π‘₯. Distributing the parentheses, then rearranging the equation to put all terms on the left-hand side, we obtain π‘₯+8π‘₯+3π‘₯+24=π‘₯+12π‘₯π‘₯+11π‘₯+24βˆ’π‘₯βˆ’12π‘₯=0βˆ’π‘₯+24=0π‘₯=24.

In the final example, we will determine whether four points that define two intersecting line segments can be points on a circle given the lengths of their parts.

Example 6: Understanding the Chords Theorem

Given that 𝐸𝐴=5.2cm, 𝐸𝐢=6cm, 𝐸𝐡=7.5cm, and 𝐸𝐷=6.5cm, do the points 𝐴, 𝐡, 𝐢, and 𝐷 lie on a circle?

Answer

Firstly, we are going to label the diagram with the lengths that we have been given.

In order for these four points to lie on a circle, they would have to satisfy the intersecting chord theorem.

Therefore, to solve this problem, we need to recall the intersecting chord theorem.

If chord 𝐴𝐡 and chord 𝐢𝐷 intersect at point 𝐸, then 𝐴𝐸×𝐸𝐡=𝐢𝐸×𝐸𝐷.

Let’s now see if this is satisfied by the lengths of the line segments in our diagram: 𝐴𝐸×𝐸𝐡=5.2Γ—7.5=39, and 𝐢𝐸×𝐸𝐷=6Γ—6.5=39.

From our calculations, we can see that 𝐴𝐸×𝐸𝐡=𝐢𝐸×𝐸𝐷, as both 𝐴𝐸×𝐸𝐡 and 𝐢𝐸×𝐸𝐷 are equal to 39. Therefore, we can say that yes, the points 𝐴, 𝐡, 𝐢, and 𝐷 lie on a circle.

Let us finish by recapping some key points.

Key Points

  • The Intersecting Chords Theorem
    𝐴𝐸×𝐸𝐡=𝐢𝐸×𝐸𝐷
  • The Intersecting Secants Theorem
    𝐴𝐡×𝐴𝐢=𝐴𝐷×𝐴𝐸
  • The Tangent–Secant Theorem
    𝐸𝐡×𝐸𝐴=𝐸𝐢

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