Lesson Explainer: Relationships between Chords and the Center of a Circle | Nagwa Lesson Explainer: Relationships between Chords and the Center of a Circle | Nagwa

Lesson Explainer: Relationships between Chords and the Center of a Circle Mathematics • Third Year of Preparatory School

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In this explainer, we will learn how to identify the relationship between chords that are equal or different in length and the center of a circle and use the properties of the chords in congruent circles to solve problems.

We begin by recalling that perpendicular bisectors of chords go through the center of the circle. Let us draw a diagram portraying this fact.

In the diagram above, the blue line segment perpendicularly bisects chord 𝐴𝐵. We note that this line goes through the center 𝑂 and, hence, defines the perpendicular distance between the center and the chord.

Definition: Distance of a Chord from the Center

The distance of a chord from the center of the circle is measured by the length of the line segment from the center that intersects perpendicularly with the chord.

From the diagram above, let us label the midpoint of chord 𝐴𝐵, which is where the blue line perpendicularly intersect with the chord. Also, we will add radius 𝑂𝐴.

Since 𝑂𝐶𝐴 is a right triangle, we can use the Pythagorean theorem to find length 𝐴𝐶 from radius 𝐴𝑂 and distance 𝑂𝐶. Since 𝐶 is the midpoint of chord 𝐴𝐵, we know that 𝐴𝐵=2𝐴𝐶. Hence, if we are given the radius of the circle and the distance of a chord from the center of the circle, we can use this method to find the length of the chord. Rather than explicitly writing out this computation, we will focus on the qualitative relationship between the lengths of chords and their distance from the center of the circle in this explainer.

Consider two different chords in the same circle as in the diagram below.

Since 𝑂𝐴 and 𝑂𝐷 are radii of the same circle, they have the same length. We want to know the relationship between the lengths of chords 𝐴𝐵 and 𝐷𝐸 if we know that 𝐷𝐸 is farther from the center than 𝐴𝐵. In other words, we assume 𝑂𝐶<𝑂𝐹. Instead of comparing the full lengths of the two chords, we can compare the half-chords 𝐴𝐶 and 𝐷𝐹. Using the Pythagorean theorem, we can write 𝐴𝐶+𝑂𝐶=𝑂𝐴,𝐷𝐹+𝑂𝐹=𝑂𝐷.

We know that 𝑂𝐴=𝑂𝐷, so the left-hand sides of both equations must be equal to each other: 𝐴𝐶+𝑂𝐶=𝐷𝐹+𝑂𝐹.

This equation can be rearranged to say 𝐴𝐶𝐷𝐹=𝑂𝐹𝑂𝐶.

Our assumption that 𝑂𝐹>𝑂𝐶 leads to 𝑂𝐹𝑂𝐶>0, so the left-hand side of this equation must be positive. This means 𝐴𝐶𝐷𝐹>0,𝐴𝐶>𝐷𝐹.whichleadsto

Since 𝐴𝐶 and 𝐷𝐹 are positive lengths, we can take the square root of both sides of the inequality to obtain 𝐴𝐶>𝐷𝐹. This leads to the following statement.

Theorem: Relationship between the Lengths of Chords and Their Distance from the Center

Consider two chords in the same circle whose distances from the center are different. The chord that is closer to the center of the circle has a greater length than the other.

This theorem allows us to compare the lengths of chords in the same circle based on their distance from the center of the circle. In our first example, we will apply this theorem to obtain an inequality involving lengths.

Example 1: Comparing Chord Lengthes based on their Distances from the Center

Supposed that 𝐵𝐶=8cm and 𝐵𝐴=7cm. Which of the following is true?

  1. 𝐷𝑀=𝑋𝑌
  2. 𝐷𝑀>𝑋𝑌
  3. 𝐷𝑀<𝑋𝑌

Answer

We recall that for two chords in the same circle, the chord that is closer to the center of the circle has a greater length than the other. We also know that the distance of a chord from the center of the circle is measured by the length of the line segment from the center intersecting perpendicularly with the chord.

In this example, we have two chords 𝑋𝑌 and 𝐷𝑀. Since 𝐵𝐴 intersects perpendicularly with chord 𝑋𝑌, length 𝐵𝐴 is the distance of this chord from the center. Similarly, length 𝐵𝐶 is the distance of chord 𝐷𝑀 from the center. Based on the given information, we note that 𝐵𝐶>𝐵𝐴, which means that chord 𝑋𝑌 is closer to the center. Hence, the length of chord 𝑋𝑌 is greater than that of the other chord.

The true option is C, which states that 𝐷𝑀<𝑋𝑌.

In the next example, we will find the range of an unknown variable defining lengths using the relationship between chords and the center of the circle.

Example 2: Finding the Range of Values of an Unknown That Satisfy Given Conditions

If 𝑀𝐹>𝑀𝐸, find the range of values of 𝑥 that satisfy the data represented.

Answer

We recall that for two chords in the same circle, the chord that is closer to the center of the circle has a greater length than the other. We also know that the distance of a chord from the center of the circle is measured by the length of the line segment from the center intersecting perpendicularly with the chord.

In this example, we have two chords, 𝐴𝐵 and 𝐶𝐷. Since 𝑀𝐸 intersects perpendicularly with chord 𝐴𝐵, length 𝑀𝐸 is the distance of this chord from the center. Similarly, length 𝑀𝐹 is the distance of chord 𝐶𝐷 from the center. Since we are given 𝑀𝐹>𝑀𝐸, we know that chord 𝐴𝐵 is closer to the center. This leads to the fact that chord 𝐴𝐵 has a greater length than chord 𝐶𝐷.

In the given diagram, we note that 𝐴𝐵=(𝑥+4)cm and 𝐶𝐷=24cm. Hence, the inequality 𝐴𝐵>𝐶𝐷 can be written as 𝑥+4>24,𝑥>20.whichleadsto

However, this only provides the lower bound for 𝑥. To identify the upper bound for 𝑥, we should ask what the maximum length of chord 𝐴𝐵 is. Since the length of a chord is larger when it is closer to the center, the longest chord should occur when the distance from the center is zero. If the distance of a chord from the center is zero, the chord should contain the center. In this case, the chord is a diameter of the circle. Since the radius of the circle is 33 cm, its diameter is 2×33=66cm. This tells us that the length of 𝐴𝐵 cannot exceed 66 cm. Additionally, since 𝐴𝐵 in the given diagram does not contain the center 𝑀, we know that the length of chord 𝐴𝐵 must be strictly less than 66 cm. Hence, 𝑥+4<66,𝑥<62.whichleadsto

This gives us the upper bound for 𝑥. Combining both lower and upper bounds, we have 20<𝑥<62.

In interval notation, this is written as ]20,62[.

In previous examples, we considered the relationship between the lengths of two chords in the same circle and their distances from the center of the circle when the distances are not the equal. Recall that two circles are congruent to each other if the measures of their radii are equal. Since the proof of this relationship only uses the fact that the radii of the circle have equal lengths, this relationship can extend to two chords from two congruent circles.

What can we say about the lengths of chords in the same circle, or in congruent circles, if their distances from the respective centers are equal? It is not difficult to modify the previous discussion to fit this particular case. Consider the following diagram.

We assume that chords 𝐴𝐵 and 𝐷𝐸 are equidistant from the center, which means 𝑂𝐶=𝑂𝐹. We also know that the radii are of the same length, thus 𝑂𝐴=𝑂𝐷. This tells us that the hypotenuse and one other side of the two right triangles 𝑂𝐶𝐴 and 𝑂𝐹𝐷 are equal. Since the lengths of the remaining sides can be obtained using the Pythagorean theorem, the lengths of the third sides, 𝐴𝐶 and 𝐷𝐹, must also be equal. Since these lengths are half of those of the chords, the two chords must have equal lengths. This result can be summarized as follows.

Theorem: Equidistant Chords in Congruent Circles

Consider two chords in the same circle, or in congruent circles. If they are equidistant from the center of the circle, or from the respective centers of the circles, then their lengths are equal.

In the next example, we will use this relationship to find a missing length of a chord in a given diagram.

Example 3: Finding a Missing Length Using Equidistant Chords from the Center of a Circle

Given that 𝑀𝐶=𝑀𝐹=3cm, 𝐴𝐶=4cm, 𝑀𝐶𝐴𝐵, and 𝑀𝐹𝐷𝐸, find the length of 𝐷𝐸.

Answer

We recall that two chords in the same circle that are equidistant from the center of the circle have equal lengths. We also know that the distance of a chord from the center of the circle is measured by the length of the line segment from the center intersecting perpendicularly with the chord.

In this example, we have two chords, 𝐴𝐵 and 𝐷𝐸. Since 𝑀𝐶 intersects perpendicularly with chord 𝐴𝐵, length 𝑀𝐶 is the distance of this chord from the center. Similarly, length 𝑀𝐹 is the distance of chord 𝐷𝐸 from the center. From the given information, we note that 𝑀𝐶=𝑀𝐹, so the two chords are equidistant from the center of the circle. Hence, the two chords must have equal lengths, 𝐷𝐸=𝐴𝐵.

In the diagram above, we are given that 𝐴𝐶=4. We recall that the perpendicular bisector of a chord passes through the center of the circle. Since 𝑀𝐶 is perpendicular to chord 𝐴𝐵 and passes through center 𝑀 of the circle, it must be the perpendicular bisector of chord 𝐴𝐵. In particular, this means that 𝐶 is the midpoint of 𝐴𝐵, which gives us 𝐴𝐶=𝐵𝐶. Since 𝐴𝐶=4cm, we also know that 𝐵𝐶=4cm. Hence, 𝐴𝐵=𝐴𝐶+𝐵𝐶=4+4=8.cm

This tells us that the length of 𝐴𝐵 is 8 cm. Since we know 𝐷𝐸=𝐴𝐵, we conclude that the length of 𝐷𝐸 is 8 cm.

So far, we have discussed implications for the lengths of chords depending on their distance from the center of the circle. We now turn our attention to the converse relationship. More specifically, if we know that two chords in two congruent circles have equal lengths, what can we say about the distance of the chords from the respective centers of the circles? Let us consider the following diagram.

We can label the midpoints of both chords, which are where the blue lines intersect with the chords perpendicularly. Also, we add radii 𝑂𝐴 and 𝑃𝐷 to the diagram. Since the circles are congruent, we know that the lengths of the radii are equal, which leads to 𝑂𝐴=𝑃𝐷 as seen in the diagram below.

We know that 𝐸 and 𝐹 are midpoints of the chords so 𝐴𝐸=12𝐴𝐵𝐷𝐹=12𝐶𝐷.and

Since we are assuming that the chords have equal lengths, we know that 𝐴𝐸=𝐷𝐹 as marked in the diagram above. This tells us that the hypotenuse and one other side of the two right triangles 𝑂𝐸𝐴 and 𝑃𝐹𝐷 are equal. Since the lengths of the remaining sides can be obtained using the Pythagorean theorem, the lengths of the third sides must also be equal. This tells us 𝑂𝐸=𝑃𝐹.

In other words, the distances of the chords from the respective centers are equal. We can summarize this result as follows.

Theorem: Chords of Equal Lengths in Congruent Circles

Two chords of equal lengths in the same circle, or in congruent circles, are equidistant from the center of the circle, or the respective centers of the circles.

Let us consider an example where we need to use this statement together with other properties of the chords of a circle to find a missing length.

Example 4: Finding a Missing Length Using Equal Chords

Given that 𝐴𝐵=𝐶𝐷, 𝑀𝐶=10cm, and 𝐷𝐹=8cm, find the length of 𝑀𝐸.

Answer

We recall that two chords of equal lengths in the same circle are equidistant from the center of the circle. We also know that the distance of a chord from the center of the circle is measured by the length of the line segment from the center intersecting perpendicularly with the chord.

In this example, we have two chords, 𝐴𝐵 and 𝐶𝐷. Since 𝑀𝐸 intersects perpendicularly with chord 𝐴𝐵, length 𝑀𝐸 is the distance of this chord from the center. Similarly, the length 𝑀𝐹 is the distance of chord 𝐶𝐷 from the center. Since we are given 𝐴𝐵=𝐶𝐷, we know that the chords have equal lengths. This leads to the fact that the chords are equidistant from the center: 𝑀𝐸=𝑀𝐹.

Since we are looking for length 𝑀𝐸, it suffices to find length 𝑀𝐹 instead. We note that 𝑀𝐹 is a side of the right triangle 𝑀𝐶𝐹, whose hypotenuse is given by 𝑀𝐶=10cm. If we can find the length of side 𝐶𝐹, then we can apply the Pythagorean theorem to find the length of the third side, 𝑀𝐹.

To find length 𝐶𝐹, we recall that the perpendicular bisector of a chord goes through the center of the circle. Since 𝑀𝐹 perpendicularly intersects chord 𝐶𝐷 and goes through center 𝑀, it is the perpendicular bisector of the chord. Hence, 𝐶𝐹=𝐷𝐹. Since 𝐷𝐹=8cm, we obtain 𝐶𝐹=8cm.

Applying the Pythagorean theorem to 𝑀𝐶𝐹, 𝑀𝐹+𝐶𝐹=𝑀𝐶.

Substituting 𝑀𝐶=10cm and 𝐶𝐹=8cm into this equation, 𝑀𝐹+8=10,𝑀𝐹=10064=36.whichleadsto

Since 𝑀𝐹 is a positive length, we can take the square root to obtain 𝑀𝐹=36=6.cm

Remember that since 𝑀𝐸=𝑀𝐹, we conclude that the length of 𝑀𝐸 is 6 cm.

In our final example, we will use the relationship between lengths of chords and their distances from the center of the circle to identify a missing angle.

Example 5: Finding the Measure of an Angle in a Triangle inside a Circle Where Two of Its Vertices Intersect with Chords and Its Third Is the Circle’s Center

Find 𝑚𝑀𝑋𝑌.

Answer

We recall that two chords of equal lengths in the same circle are equidistant from the center of the circle. We also know that the distance of a chord from the center of the circle is measured by the length of the line segment from the center intersecting perpendicularly with the chord.

In this example, we have two chords 𝐴𝐵 and 𝐴𝐶 that have equal lengths. We recall that the perpendicular bisector of a chord goes through the center of the circle. Since 𝑋 and 𝑌 are midpoints of the two chords and 𝑀 is the center of the circle, line segments 𝑀𝑋 and 𝑀𝑌 must be the perpendicular bisectors of the two chords. In particular, these lines intersect perpendicularly with the respective chords. This tells us that 𝑀𝑋 and 𝑀𝑌 are the respective distances of chords 𝐴𝐵 and 𝐴𝐶 from the center of the circle.

Since the two chords have equal lengths, they must be equidistant from the center. This tells us 𝑀𝑋=𝑀𝑌.

This also tells us that two sides of triangle 𝑀𝑋𝑌 have equal lengths. In other words, 𝑀𝑋𝑌 is an isosceles triangle. Hence, 𝑚𝑀𝑋𝑌=𝑚𝑀𝑌𝑋.

We also know that the sum of the interior angles of a triangle is equal to 180. We can write 𝑚𝑋𝑀𝑌+𝑚𝑀𝑋𝑌+𝑚𝑀𝑌𝑋=180.

We know that 𝑚𝑋𝑀𝑌=102 and also 𝑚𝑀𝑋𝑌=𝑚𝑀𝑌𝑋. Substituting these expressions into the equation above, 102+2𝑚𝑀𝑋𝑌=180,2𝑚𝑀𝑋𝑌=180102=78.whichleadsto

Therefore, 𝑚𝑀𝑋𝑌=782=39.

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

Key Points

  • The distance of a chord from the center of the circle is measured by the length of the line segment from the center intersecting perpendicularly with the chord.
  • Consider two chords in the same circle, or in two congruent circles, whose distances from the center, or the respective centers, are different. The chord that is closer to the respective center is of greater length than the other.
  • Consider two chords in the same circle, or in congruent circles. If they are equidistant from the center of the circle, or from the respective centers of the circles, their lengths are equal.
  • Two chords of equal lengths in the same circle, or in congruent circles, are equidistant from the center of the circle, or the respective centers of the circles.

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