Video Transcript
In this video, we will learn how to
identify the relationship between chords that are equal or different in length and
the center of the circle and use the properties of chords in congruent circles to
solve problems. We begin by recalling that
perpendicular bisectors of chords go through the center of the circle as shown. Let’s now consider this in more
detail and how it leads to definitions and theorems we will use in this video.
In the diagram shown, the line
segment 𝑂𝐶 is the perpendicular bisector of the chord 𝐴𝐵. This leads us to the following
definition. 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. Adding the radius 𝑂𝐴, we see that
triangle 𝑂𝐶𝐴 is a right triangle. Using the Pythagorean theorem, 𝐴𝐶
squared plus 𝑂𝐶 squared is equal to 𝑂𝐴 squared. Since 𝐶 is the midpoint of the
chord 𝐴𝐵, we also know that 𝐴𝐵 is equal to two multiplied by 𝐴𝐶. This means that if we are given the
radius of the circle 𝑂𝐴 together with the distance of a chord from the center of
the circle 𝑂𝐶, then we can calculate the length of the chord 𝐴𝐵.
Let’s now consider a second chord
𝐷𝐸 on the same diagram. Adding the perpendicular bisector
of the chord 𝑂𝐹 together with the radius 𝑂𝐷, our diagram is as shown. Since 𝑂𝐴 and 𝑂𝐷 are radii of
the circle, they have the same length. We wish to consider the
relationship between the lengths of the chords 𝐴𝐵 and 𝐷𝐸. If we know that 𝐷𝐸 is further
from the center than 𝐴𝐵, we can assume that 𝑂𝐶 is less than 𝑂𝐹. Comparing the half chords 𝐴𝐶 and
𝐷𝐹 and using the Pythagorean theorem once again, we have 𝐴𝐶 squared plus 𝑂𝐶
squared is equal to 𝑂𝐴 squared and 𝐷𝐹 squared plus 𝑂𝐹 squared is equal to 𝑂𝐷
squared.
We know that 𝑂𝐴 is equal to
𝑂𝐷. This means that the left-hand side
of both equations must also be equal. 𝐴𝐶 squared plus 𝑂𝐶 squared is
equal to 𝐷𝐹 squared plus 𝑂𝐹 squared. We can rearrange this equation such
that 𝐴𝐶 squared minus 𝐷𝐹 squared is equal to 𝑂𝐹 squared minus 𝑂𝐶
squared. Since 𝑂𝐹 is greater than 𝑂𝐶,
the right-hand side of our equation must be greater than zero. This means that the left-hand side
must also be greater than zero or positive. Adding 𝐷𝐹 squared to both sides
of this inequality, we have 𝐴𝐶 squared is greater than 𝐷𝐹 squared. And since 𝐴𝐶 and 𝐷𝐹 are both
positive lengths, square rooting both sides gives us 𝐴𝐶 is greater than 𝐷𝐹.
This leads us to the following
theorem. If we consider two chords in the
same circle whose distances from the center are different, then the chord which is
closer to the center of the circle has a greater length than the other. We will now apply this theorem to a
specific example.
Suppose that 𝐵𝐶 equals eight
centimeters and 𝐵𝐴 equals seven centimeters. Which of the following is true? Is it (A) 𝐷𝑀 is equal to 𝑋𝑌,
(B) 𝐷𝑀 is greater than 𝑋𝑌, or (C) 𝐷𝑀 is less than 𝑋𝑌?
Let’s begin by adding the lengths
𝐵𝐶 and 𝐵𝐴 to our diagram. These are the distances from the
chords 𝐷𝑀 and 𝑋𝑌, respectively, to the center of the circle 𝐵. We recall that the chord that is
closer to the center of the circle has a greater length. From the diagram, we see that the
chord 𝑋𝑌 is seven centimeters from the center. This is the length of 𝐵𝐴. The chord 𝐷𝑀, on the other hand,
is eight centimeters from the center. This means that 𝑋𝑌 is closer to
the center of the circle than 𝐷𝑀. As this chord will have a greater
length, we can conclude that 𝑋𝑌 is greater than 𝐷𝑀. From the three options listed, the
correct answer is option (C) 𝐷𝑀 is less than 𝑋𝑌.
So far, we have only considered the
situation where the distances of two chords from the center of a circle are not
equal. Let’s now consider what happens
when the two chords are equidistant from the center.
In the diagram drawn, we assume
that the chords 𝐴𝐵 and 𝐷𝐸 are equidistant from the center. This means that 𝑂𝐶 is equal to
𝑂𝐹. The two radii 𝑂𝐴 and 𝑂𝐷 will
also be equal in length. This means that two sides of our
right triangles 𝑂𝐶𝐴 and 𝑂𝐹𝐷 are equal. Using our knowledge of the
Pythagorean theorem, this means that the length of the third sides of our triangle
must also be equal. The half chord 𝐴𝐶 is equal to the
half chord 𝐷𝐹. This in turn leads to the fact that
the chords 𝐴𝐵 and 𝐷𝐸 are equal in length. This can be summarized as
follows. If we consider two chords in the
same circle and if they are equidistant from the center of the circle, then their
lengths are equal.
Whilst we will not see an example
in this video, it is also important to note that this theorem holds for congruent
circles. In that case, if the chords are
equidistant from the respective centers of the circles, then the lengths are
equal. Let’s now consider an example where
we need to find the missing length of a chord in a given diagram.
Given that 𝑀𝐶 is equal to 𝑀𝐹,
which is equal to three centimeters, 𝐴𝐶 is equal to four centimeters, line segment
𝑀𝐶 is perpendicular to line segment 𝐴𝐵, and line segment 𝑀𝐹 is perpendicular
to line segment 𝐷𝐸, find the length of the line segment 𝐷𝐸.
We are told in the question that
𝑀𝐶 is equal to 𝑀𝐹. And this means that the two chords
𝐴𝐵 and 𝐷𝐸 are equidistant from the center 𝑀. They are both three centimeters
away from the center. We recall the theorem that states
that if two chords are equidistant from the center, they are also equal in
length. This means that in this question,
the length 𝐴𝐵 is equal to the length 𝐷𝐸. We also know that 𝑀𝐶 is the
perpendicular bisector of 𝐴𝐵. And this means that 𝐴𝐵 is equal
to two multiplied by 𝐴𝐶. Since 𝐴𝐶 is equal to four
centimeters, 𝐴𝐵 is equal to eight centimeters. We can therefore conclude that the
length of the chord 𝐷𝐸 is eight centimeters.
So far in this video, we have
discussed the lengths of chords depending on their distance from the center of the
circle. We will now consider the converse
relationship. In the diagram shown, we have two
congruent circles. We will consider the case where the
chords 𝐴𝐵 and 𝐶𝐷 are equal in length and, more importantly, what this says about
the distance of the chords from their respective centers. In this example, these are the
lengths 𝑂𝐸 and 𝑃𝐹, respectively, Adding the radii 𝑂𝐴 and 𝑃𝐶, we know that
these lengths must be equal, as the circles are congruent. As the chords are equal in length,
the half chords must also be equal in length such that 𝐴𝐸 is equal to 𝐶𝐹. This is because 𝐴𝐸 is equal to a
half of 𝐴𝐵 and 𝐶𝐹 is equal to a half of 𝐶𝐷.
Using our knowledge of the
Pythagorean theorem, the third sides of our right triangles must also be equal in
length. The length 𝑂𝐸 is equal to the
length 𝑃𝐹. In other words, the distance of the
chords from their respective centers are equal. This can be summarized as
follows. Two chords of equal lengths in the
same circle or congruent circles are equidistant from the center of the circle or
the respective centers.
In our final example, we will use
this statement to find a missing length.
Given that 𝐴𝐵 is equal to 𝐶𝐷,
𝑀𝐶 is equal to 10 centimeters, and 𝐷𝐹 is equal to eight centimeters, find the
length of line segment 𝑀𝐸.
In this question, we are trying to
calculate the length of 𝑀𝐸, which is the distance from the chord 𝐴𝐵 to the
center of the circle 𝑀. We begin by recalling that two
chords of equal lengths are equidistant from the center. And in this question, we are told
that the two chords 𝐴𝐵 and 𝐶𝐷 are equal in length. This means that the length 𝑀𝐹
must be equal to 𝑀𝐸. The line segment 𝑀𝐸
perpendicularly bisects the chord 𝐴𝐵. Likewise, 𝑀𝐹 is the perpendicular
bisector of 𝐶𝐷. Since we are told 𝐷𝐹 is equal to
eight centimeters, 𝐶𝐹, 𝐴𝐸, and 𝐵𝐸 are all also equal to eight centimeters.
Our next step is to consider the
right triangle 𝑀𝐹𝐶. Using the Pythagorean theorem, 𝑀𝐹
squared plus 𝐶𝐹 squared is equal to 𝑀𝐶 squared. By subtracting 𝐶𝐹 squared from
both sides and substituting in the values of 𝐶𝐹 and 𝑀𝐶, we have 𝑀𝐹 squared is
equal to 10 squared minus eight squared. This is equal to 36. Square rooting both sides of this
equation and knowing that 𝑀𝐹 must be a positive answer, we have 𝑀𝐹 is equal to
six. Since 𝑀𝐹 is equal to six
centimeters, 𝑀𝐸 must also be equal to six centimeters. The perpendicular distance from the
center of the chord 𝐴𝐵 to the center of the circle 𝑀, which is the line segment
𝑀𝐸, is equal to six centimeters.
We will now finish this video by
summarizing the key points from it. 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. When two chords whose distances
from the center of a circle are different, then the chord which is closer to the
center has a greater length. If two chords are equidistant from
the center of a circle, their lengths are equal. The converse of this is also
true. Two chords of equal lengths in the
same circle are equidistant from the center. It is important to note that the
last three statements are also true for two congruent circles.
