Segments of a Circle A chord of a circle divides the circle into two regions, which are called the segments of the circle.

The minor segment is the region bounded by the chord and the minor arc intercepted by the chord.

The major segment is the region bounded by the chord and the major arc intercepted by the chord.

Angles in Different Segments

Angles in the Same Segment

Theorem

Use the information given in the diagram to prove that the angles in the same segment of a circle areequal. That is, a = b.

Given:

To prove:

Construction:

Join O to A and B.

Proof:

In general:

Angles in the same segment of a circle are equal.

Practical applications

Danger Angle

If there are rocks near the shore, then boats are informed by the chart (map) to keep the angle subtended by two land marks, A and B, smaller than the given danger angle.

Angle for Scoring a Goal in Soccer

All positions on the same arc of a circle give the same angle for scoring a goal in soccer. Note that the distance of the shot changes but the angle of possible shots remains constant.

Example 25

Find the value of the pronumeral in the following circle centred at O.

Solution:

Example 26

Find the value of each of the pronumerals in the following circle centred at O.

Solution:

Example 27

Find the value of each of the pronumerals in the following circle centred at O:

Solution:

Key Terms

chord, segments of a circle, minor segment, major segment, angles in different segments, angles in the same segment, angles in the same segment theorem

All the "parts" of a circle, such as the radius, the diameter, etc., have a relationship with the circle or another "part" that can always be expressed as a theorem.  The two theorems that deal with chords and radii (plural of radius) are outlined below.

1.  If a radius of a circle is perpendicular to a chord, then the radius bisects the chord. Here's a graphical representation of this theorem:

2.  In a circle or in congruent circles, if two chords are the same distance from the center, then they are congruent.

Using these theorems in action is seen in the example below:

1. Problem: Find CD.

Given: Circle R is congruent to circle S. Chord AB = 8. RM = SN. Solution: By theorem number 2 above, segment AB is congruent to segment CD. Therefore, CD equals 8.

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Oh, the wonderfully confusing world of geometry!  :-)  The tangent being discussed here is not the trigonometric ratio.  This kind of tangent is a line or line segment that touches the perimeter of a circle at one point only and is perpendicular to the radius that contains the point. 

1. Problem: Find the value of x. Given: Segment AB is tangent to circle C at B.

Solution: x is a radius of the circle. Since x contains B, and AB is a tangent segment, x must be perpendicular to AB (the definition of a tangent tells us that). If it is perpendicular, the triangle formed by x, AB, and CA is a right triangle. Use the Pythagorean Theorem to solve for x. 152 + x2 = 172

x2 = 64 x = 8

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Congruent arcs are arcs that have the same degree measure and are in the same circle or in congruent circles.

Arcs are very important and let us find out a lot about circles.  Two theorems involving arcs and their central angles are outlined below.

1.  For a circle or for congruent circles, if two minor arcs are congruent, then their central angles are congruent.

2.  For a circle or for congruent circles, if two central angles are congruent, then their arcs are congruent.

Example:

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An inscribed angle is an angle with its vertex on a circle and with sides that contain chords of the circle.  The figure below shows an inscribed angle.

The most important theorem dealing with inscribed angles is stated below.

The measure of an inscribed angle is equal to one-half the degree measure of its intercepted arc.

1. Problem: Find the measure of each arc or angle listed below. arc QSR angle Q angle R

Solution: Arc QSR is 180o because it is twice the measure of its inscribed angle (angle QPR, which is 90o).

Angle Q is 60o because it is half of its intercepted arc, which is 120o. Angle R is 30o by the Triangle Sum Theorem which says a triangle has three angles which have measures that equal 180o when added together.

In the last problem's figure, you noticed that angle P is inscribed in semicircle QPR and angle P = 90o.  This leads us to our next theorem, which is stated below.

Any angle inscribed in a semicircle is a right angle.

The one last theorem dealing with inscribed angles is a bit more complicated because it deals with quadrilaterals, too.  It is stated below.

If a quadrilateral is inscribed in a circle, then both pairs of opposite angles are supplementary.

1. Problem: Find the measure of arc GDE.

Solution: By the theorem stated above, angle D and angle F are supplementary. Therefore, angle F equals 95o. The first theorem discussed in this section tells us the measure of an arc is twice that of its inscribed angle. With that theorem, arc GDE is 190o.

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When two secants intersect inside a circle, the measure of each angle formed is related to one-half the sum of the measures of the intercepted arcs.  The figure below shows this theorem in action.

In the figure, arc AB and arc CD are 60o and 50o, respectively.  By the above stated theorem, the measures of both angle 1 and angle 2 in the figure are 55o.

Sometimes, secants intersect outside of circles.  When this happens, the measure of the angle formed is equal to one-half the difference of the degree measures of the intercepted arcs.

1. Problem: Find the measure of angle 1.

Givens: Arc AB = 60o

Arc CD = 100o

Solution: By the theorem stated above, the measure of angle 1 = .5((arc CD) - (arc AB)) angle 1 = .5((100 - 60)) angle 1 = 20o

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Another way secants can intersect in circles is if they are only in line segments.  There is a theorem that tells us when two chords intersect inside a circle, the product of the measures of the two segments of one chord is equal to the product of the measures of the two segments of the other chord.  In the figure below, chords PR and QS intersect.  By the theorem stated above, PT * TR = ST * TQ.

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One last thing that has to be discussed when dealing with circles is circumference, or the distance around a circle.  The circumference of a circle equals 2 times PI times the measure of the radius.  That postulate is usually represented by the following equation (where C represents circumference and r stands for radius): C = 2(PI)r.

For example, if a circle has a radius of 3, the circumference of the circle is 6(PI).

Also, you can find the length of any arc when you know its degree measure and the measure of a radius with the following formula (L = length, n = degree measure of arc, r = radius): L = (n/360)(2(PI)r).

1. Problem: Find the length of a 24o arc of a circle with a 5 cm radius.

Solution: n 24 2(PI) L = ---(2(PI)r) = ---(2(PI))5 = ----- 360 360 3 The length of the arc is (2/3)(PI) cm.

Theorems for Segments and Circles

Tangent Segments

Given a point outside a circle, two lines can be drawn through that point that are tangent to the circle. The tangent segments whose endpoints are the points of tangency and the fixed point outside the circle are equal. In other words, tangent segments drawn to the same circle from the same point (there are two for every circle) are equal.

Figure %: Tangent segments that share an endpoint not on the circle are equal

Chords

Chords within a circle can be related many ways. Parallel chords in the same circle always cut congruent arcs. That is, the arcs whose endpoints include one endpoint from each chord have equal measures.

Figure %: Arcs AC and BD have equal measures

When congruent chords are in the same circle, they are equidistant from the center.

Figure %: Congruent chords in the same circle are equidistant from the center

In the figure above, chords WX and YZ are congruent. Therefore, their distances from the center, the lengths of segments LC and MC, are equal.

A final word on chords: Chords of the same length in the same circle cut congruent arcs. That is, if the endpoints of one chord are the endpoints of one arc, then the two arcs defined by the two congruent chords in the same circle are congruent.

Intersecting Chords, Tangents, and Secants

A number of interesting theorems arise from the relationships between chords, secant segments, and tangent segments that intersect. First of all, we must define a secant segment. A secant segment is a segment with one endpoint on a circle, one endpoint outside the circle, and one point between these points that intersects the circle. Three theorems exist concerning the above segments.

Theorem 1

PARGRAPH When two chords of the same circle intersect, each chord is divided into two segments by the other chord. The product of the segments of one chord is equal to the product of the segments of the other chord.

Figure %: Chords of the same circle that intersect

In the figure above, chords QR and ST intersect. The theorem states that the product of QB and BR is equal to the product of SB and BT.

Theorem 2

Every secant segment is divided into two segments by the circle it intersects. The internal segment is a chord, and the external segment is the segment with one endpoint at the intersection of the secant segment and the circle, and the other endpoint at the fixed point outside the circle. Given these conditions, a theorem states that when two secant segments share an endpoint not on the circle, the products of the lengths of each secant segment and its external segment are equal.

Figure %: Secant segments that share an endpoint not on the circle

In the figure above, the secant segments DE and FE share an endpoint, E, outside the circle. The theorem states that the product of the lengths of DE and ME is equal to the product of the lengths of FE and NE.

Theorem 3

A similar theorem exists when a secant segment and a tangent segment share an endpoint not on the circle. This theorem states that the length of the tangent segment squared is equal to the product of the secant segment and its external segment.

Figure %: A secant segment and a tangent segment that share an endpoint not on the circle

In the figure above, secant segment QR and tangent segment SR share an endpoint, R, not on the circle. The theorem states that the length of SR squared is equal to the product of the lengths of QR and KR.

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Problems

Problem : If two chords in the same circle cut two arcs of 75 degrees, what do we know about the chords?

Solution for Problem 1 >>If the arcs have an endpoint on each chord, then the chords are parallel. If the arcs have both endpoints on the same chord, then the chords are congruent. Close

Problem : Find the value of x .

Solution for Problem 2 >>x = 2 Close

Problem : Find the value of x .

Solution for Problem 3 >>x = 5 Close

Problem : Find the value of x .

Solution for Problem 4 >>x = 5 Close

Problem : Find the value of x .

Solution for Problem 5 >>x = 4

Segments of Chords, Secants, Tangents

In Figure 1 , chords QS and RT intersect at P. By drawing QT and RS, it can be proven that Δ QPT ∼ Δ RPS. Because the ratios of corresponding sides of similar triangles are equal, a/ c = d/ b. The Cross Products Property produces ( a) ( b) = ( c) ( d). This is stated as a theorem.

Figure 1 Two chords intersecting inside a circle.

Theorem 83: If two chords intersect inside a circle, then the product of the segments of one chord equals the product of the segments of the other chord.

Example 1: Find x in each of the following figures in Figure 2 .

Figure 2 Two chords intersecting inside a circle.

In Figure 3 , secant segments AB and CD intersect outside the circle at E. By drawing BC and AO, it can be proven that Δ EBC ∼ Δ EDA. This makes

Figure 3 Two secant segments intersecting outside a circle.

By using the Cross-Products Property,

(EB)(EA) = (ED)(EC)

This is stated as a theorem.

Theorem 84: If two secant segments intersect outside a circle, then the product of the secant segment with its external portion equals the product of the other secant segment with its external portion.

Example 2: Find x in each of the following figures in 4 .

Figure 4 More secant segments intersecting outside a circle.

In Figure 5 , tangent segment AB and secant segment BD intersect outside the circle at B. By drawing AC and AD, it can be proven that Δ ADB ∼ Δ CAB. Therefore,

Figure 5 A tangent segment and a secant segment intersecting outside a circle.

This is stated as a theorem.

Theorem 85: If a tangent segment and a secant segment intersect outside a circle, then the square of the measure of the tangent segment equals the product of the measures of the secant segment and its external portion.

Also,

Theorem 86: If two tangent segments intersect outside a circle, then the tangent segments have equal measures.

Example 3: Find x in the following figures in 6 .

Figure 6 A tangent segment and a secant segment (or another tangent segment) intersecting outside a circle.