Chord

Line Segments with Both Endpoints on a Circle

Each chord determines a minor arc and a major arc, and it divides the disk into a minor segment and a major segment.

Among chords of the same circle, a chord farther from the center is shorter. The longest chord passes through the center and is called a diameter.

Useful Chord Configurations

The following configurations help organize chord problems:

1. Non-diameter chord: A chord that does not pass through the center. It is shorter than the diameter of the same circle.

2. Diameter: A chord that passes through the center. It is the longest chord in the circle and divides the disk into two semicircles.

3. Parallel chords: A relationship between two or more chords with parallel supporting lines. Their lengths are equal exactly when their perpendicular distances from the center are equal.

Relationship Between Chord and Distance to Center

The relevant distance is the perpendicular distance from the center to the chord. The perpendicular from the center bisects the chord, creating a right triangle with legs $d$ and $l/2$ and hypotenuse $r$.

If $d$ is that perpendicular distance, $r$ is the radius, and $l$ is the chord length, the Pythagorean theorem gives:

This formula shows that when $d = 0$ (chord passes through the center), then $l = 2r$ which is the diameter.

Properties of Equal Length Chords

Equal chords in the same circle have equal perpendicular distances from the center. Conversely, chords at equal perpendicular distances from the center have equal lengths.

Equal chords determine equal central angles. Therefore, if two chords have the same length, then:

1. The distance of both chords to the center of the circle is the same
2. Their corresponding minor arcs have the same length
3. Their subtended minor central angles have the same measure

Intersecting Chords Theorem

When two chords intersect inside a circle, there is a special relationship between the segments formed. If chords $AB$ and $CD$ intersect at point $P$, then:

This relationship is called the intersecting chords theorem. It determines an unknown segment length when three of the four segment lengths are known.

Practice Problems

Each problem gives parts of a circle and asks for the chord length, so use the radius and the perpendicular distance from the centre to the chord.

1. A circle has a radius of $13 \text{ cm}$. If the distance of a chord to the center of the circle is $5 \text{ cm}$, determine the length of the chord.

2. In a circle with radius $10 \text{ cm}$, there is a chord with length $16 \text{ cm}$. Determine the distance of the chord to the center of the circle.

3. Two chords $AB$ and $CD$ in the same circle have lengths of $24 \text{ cm}$ and $18 \text{ cm}$ respectively. If the radius of the circle is $15 \text{ cm}$, determine the difference in distance between the two chords to the center of the circle.

4. Chords $PQ$ and $RS$ intersect at point $T$ in a circle. If $PT = 6 \text{ cm}$, $TQ = 8 \text{ cm}$, and $RT = 4 \text{ cm}$, determine the length of $TS$.

5. A circle has a diameter of $26 \text{ cm}$. Determine the length of a chord that is $12 \text{ cm}$ away from the center of the circle.

Worked Solutions

1. Answer: $24 \text{ cm}$

   Given: $r = 13 \text{ cm}$, $d = 5 \text{ cm}$

   Using the chord length formula:

   $$l = 2\sqrt{r^2 - d^2}$$

   $$l = 2\sqrt{13^2 - 5^2}$$

   $$l = 2\sqrt{169 - 25}$$

   $$l = 2\sqrt{144} = 2 \times 12 = 24 \text{ cm}$$

2. Answer: $6 \text{ cm}$

   Given: $r = 10 \text{ cm}$, $l = 16 \text{ cm}$

   Using the chord length formula:

   $$16 = 2\sqrt{10^2 - d^2}$$

   $$8 = \sqrt{100 - d^2}$$

   $$64 = 100 - d^2$$

   $$d^2 = 100 - 64 = 36$$

   $$d = 6 \text{ cm}$$

3. Answer: $3 \text{ cm}$

   Given: $r = 15 \text{ cm}$, $l_1 = 24 \text{ cm}$, $l_2 = 18 \text{ cm}$

   Finding the distance of the first chord:

   $$24 = 2\sqrt{15^2 - d_1^2}$$

   $$12 = \sqrt{225 - d_1^2}$$

   $$144 = 225 - d_1^2$$

   $$d_1^2 = 81, \text{ so } d_1 = 9 \text{ cm}$$

   Finding the distance of the second chord:

   $$18 = 2\sqrt{15^2 - d_2^2}$$

   $$9 = \sqrt{225 - d_2^2}$$

   $$81 = 225 - d_2^2$$

   $$d_2^2 = 144, \text{ so } d_2 = 12 \text{ cm}$$

   Difference in distance: $d_2 - d_1 = 12 - 9 = 3 \text{ cm}$

4. Answer: $12 \text{ cm}$

   Given: PT = $6 \text{ cm}$, TQ = $8 \text{ cm}$, RT = $4 \text{ cm}$

   Using the intersecting chords theorem:

   $$PT \times TQ = RT \times TS$$

   $$6 \times 8 = 4 \times TS$$

   $$48 = 4 \times TS$$

   $$TS = 12 \text{ cm}$$

5. Answer: $10 \text{ cm}$

   Given: diameter = $26 \text{ cm}$, so $r = 13 \text{ cm}$, $d = 12 \text{ cm}$

   Using the chord length formula:

   $$l = 2\sqrt{r^2 - d^2}$$

   $$l = 2\sqrt{13^2 - 12^2}$$

   $$l = 2\sqrt{169 - 144}$$

   $$l = 2\sqrt{25} = 2 \times 5 = 10 \text{ cm}$$