MathHub US Geometry Chapter 10

Chapter 10: Circles — Arcs, Chords, and Angles

High School Geometry · Unit 5: Circles · 3 interactive diagrams · 8 practice problems

Learning Objectives

10.1 Circle Basics and Central Angles

Definition: Circle and Key Parts

A circle is the set of all points in a plane equidistant from a fixed point called the center. The common distance is the radius $r$.

The arc measure (in degrees) equals the measure of its central angle. Arc length converts that degree measure to an actual distance along the circle:

$$\text{Arc Length} = \frac{\theta}{360^\circ} \cdot 2\pi r$$

Theorem: Central Angle and Arc Measure

The measure of a central angle equals the measure of its intercepted arc.

$$m\angle AOB = m\overset{\frown}{AB}$$

where $O$ is the center and $A$, $B$ are points on the circle.

Example 10.1 — Arc Length from a Central Angle

A circle has radius $r = 8$. The central angle $\angle AOB = 75°$. Find the arc length of $\overset{\frown}{AB}$.

$$\text{Arc Length} = \frac{75}{360} \cdot 2\pi(8) = \frac{75}{360} \cdot 16\pi \approx \frac{75}{360} \cdot 50.27 \approx \mathbf{10.47}$$

Example 10.2 — Central Angle from Arc Measure

Arc $RS = 140°$. Find the central angle $\angle ROS$.

By the Central Angle and Arc Measure Theorem: $\angle ROS = m\overset{\frown}{RS} = \mathbf{140°}$.

TRY IT

A circle has radius $r = 10$ and an arc with measure $60°$. Find the arc length.

Show Answer
Arc Length $= \dfrac{60}{360} \cdot 2\pi(10) = \dfrac{1}{6} \cdot 20\pi = \dfrac{20\pi}{6} = \dfrac{10\pi}{3} \approx \mathbf{10.47}$ units.

Circle with center, radius, and 75° central angle — the arc from A to B is highlighted.

Figure 10.1 — Central Angle and Arc on a Circle

10.2 Inscribed Angles and Intercepted Arcs

An inscribed angle has its vertex on the circle, with both sides as chords of the circle. This differs from a central angle, which sits at the center.

Inscribed Angle Theorem and Corollaries

Example 10.3 — Inscribed Angle from Arc Measure

Arc $BC = 100°$. Point $A$ is on the circle (not on arc $BC$). Find inscribed angle $\angle BAC$.

$$\angle BAC = \frac{1}{2}(100°) = \mathbf{50°}$$

Example 10.4 — Inscribed Quadrilateral

Quadrilateral $ABCD$ is inscribed in a circle. $\angle A = 75°$. Find $\angle C$.

Opposite angles of an inscribed quadrilateral are supplementary:

$$\angle C = 180° - \angle A = 180° - 75° = \mathbf{105°}$$

TRY IT

Two inscribed angles $\angle PQR$ and $\angle PSR$ both intercept arc $PR = 84°$. Find both angle measures.

Show Answer
By the Inscribed Angle Theorem: $\angle PQR = \frac{1}{2}(84°) = 42°$ and $\angle PSR = \frac{1}{2}(84°) = 42°$. By Corollary 1, they are equal since they intercept the same arc. Both $= \mathbf{42°}$.

Key Distinction: A central angle equals its arc. An inscribed angle is half its arc. When the inscribed angle subtends a diameter, the arc is 180° and the angle is 90° — always a right angle.

Circle with inscribed angle at A and central angle at O — both subtend arc BC. Notice the central angle is twice the inscribed angle.

Figure 10.2 — Inscribed Angle vs. Central Angle over the Same Arc

10.3 Chord and Secant Relationships

A chord is a segment whose two endpoints lie on the circle. When two chords intersect inside the circle, they create angle and length relationships.

Chord Intersection Power Theorem

If two chords $\overline{AB}$ and $\overline{CD}$ intersect at point $P$ inside a circle, then:

$$PA \cdot PB = PC \cdot PD$$

The products of the two segment lengths are always equal.

Chord-Chord Angle: When two chords intersect at point $P$ inside the circle, the angle formed equals half the sum of the intercepted arcs:

$$\angle P = \frac{1}{2}(m\overset{\frown}{AC} + m\overset{\frown}{BD})$$

Secant-Secant (from external point): The angle formed by two secants drawn from an external point equals half the positive difference of the intercepted arcs:

$$\angle P = \frac{1}{2}|m\overset{\frown}{\text{far}} - m\overset{\frown}{\text{near}}|$$

Example 10.5 — Chord-Chord Power Theorem

Chords $\overline{AB}$ and $\overline{CD}$ intersect at $P$ inside a circle. $PA = 3$, $PB = 8$, $PC = 4$. Find $PD$.

By the Power Theorem:

$$PA \cdot PB = PC \cdot PD \implies 3 \cdot 8 = 4 \cdot PD \implies PD = \frac{24}{4} = \mathbf{6}$$

Example 10.6 — Chord-Chord Angle

Two chords intersect inside a circle. The intercepted arcs are $80°$ and $120°$. Find the chord-chord angle.

$$\angle P = \frac{1}{2}(80° + 120°) = \frac{1}{2}(200°) = \mathbf{100°}$$
TRY IT

Chords $\overline{EF}$ and $\overline{GH}$ intersect at $K$ inside a circle. $EK = 5$, $KF = 6$, $GK = 3$. Find $KH$.

Show Answer
$EK \cdot KF = GK \cdot KH \implies 5 \cdot 6 = 3 \cdot KH \implies KH = \dfrac{30}{3} = \mathbf{10}$.

10.4 Tangent Lines and Tangent-Chord Angles

A tangent line touches the circle at exactly one point, called the point of tangency. The tangent is always perpendicular to the radius drawn to that point.

Tangent-Radius Perpendicularity and Tangent Segment Lengths

Example 10.7 — Tangent Length from External Point

External point $P$ is $13$ cm from the center $O$. The radius $r = 5$ cm. Find the length of the tangent from $P$ to the circle.

$$t^2 + r^2 = d^2 \implies t^2 + 5^2 = 13^2 \implies t^2 + 25 = 169 \implies t^2 = 144 \implies t = \mathbf{12 \text{ cm}}$$

Example 10.8 — Tangent-Chord Angle

A tangent and a chord form an angle of $65°$ at the point of tangency. Find the measure of the intercepted arc.

Tangent-Chord Angle $= \frac{1}{2}$ intercepted arc, so:

$$\text{Intercepted Arc} = 2 \times 65° = \mathbf{130°}$$
TRY IT

Two tangents from external point $P$ touch a circle at $A$ and $B$. $PA = 4x - 3$ and $PB = 2x + 7$. Find $PA$.

Show Answer
Equal tangent segments from the same external point: $PA = PB$, so $4x - 3 = 2x + 7 \implies 2x = 10 \implies x = 5$. $PA = 4(5) - 3 = \mathbf{17}$.

Circle with external point P, tangent line touching at T, dashed radius OT (perpendicular), and dashed hypotenuse OP.

Figure 10.3 — Tangent Line, Radius, and the Right Triangle Relationship

Practice Problems

1

A circle has radius $r = 6$. Find the arc length of an arc with central angle $120°$.

Show Solution
Arc Length $= \dfrac{120}{360} \cdot 2\pi(6) = \dfrac{1}{3} \cdot 12\pi = 4\pi \approx \mathbf{12.57}$ units.
2

Arc $PQ = 200°$. Find the central angle $\angle POQ$ and the inscribed angle $\angle PRQ$ where $R$ is on the major arc.

Show Solution
Central angle $\angle POQ = 200°$ (major arc). For inscribed angle intercepting arc $PQ = 200°$: $\angle PRQ = \frac{1}{2}(200°) = \mathbf{100°}$. If $R$ is on the minor arc, the intercepted arc is the major arc $PQ = 200°$ — same result.
3

Inscribed quadrilateral $ABCD$ has $\angle A = 82°$ and $\angle B = 67°$. Find $\angle C$ and $\angle D$.

Show Solution
Opposite angles supplement each other: $\angle C = 180° - 82° = \mathbf{98°}$ and $\angle D = 180° - 67° = \mathbf{113°}$.
4

Two chords intersect at $P$. $PA = 6$, $PB = 4$, $PC = 3$. Find $PD$ using the Power Theorem.

Show Solution
$PA \cdot PB = PC \cdot PD \implies 6 \cdot 4 = 3 \cdot PD \implies PD = \dfrac{24}{3} = \mathbf{8}$.
5

Two chords intersect inside a circle, creating arcs of $50°$ and $90°$ on opposite sides. Find the chord-chord angle.

Show Solution
Chord-Chord Angle $= \frac{1}{2}(50° + 90°) = \frac{1}{2}(140°) = \mathbf{70°}$.
6

External point $P$ is $17$ cm from the center. Radius $= 8$ cm. Find the tangent length from $P$.

Show Solution
$t^2 + 8^2 = 17^2 \implies t^2 = 289 - 64 = 225 \implies t = \mathbf{15}$ cm.
7

A tangent-chord angle measures $48°$. Find both arcs created by the chord (the intercepted arc and the other arc).

Show Solution
Tangent-Chord Angle $= \frac{1}{2} \cdot \text{intercepted arc} \implies$ intercepted arc $= 96°$. The chord divides the circle into two arcs; the other arc $= 360° - 96° = \mathbf{264°}$.
8 — AP Challenge

In a circle, chord $\overline{AB}$ has central angle $100°$. Inscribed angle $\angle ACB$ (with $C$ on the major arc) subtends arc $AB$. A tangent at $A$ forms angle $\theta$ with chord $\overline{AB}$. Find all: $m\overset{\frown}{AB}$, $\angle ACB$, and $\theta$.

Show Solution
$m\overset{\frown}{AB}$ (minor) $= 100°$ (equals central angle).
$\angle ACB = \frac{1}{2}(100°) = \mathbf{50°}$ (inscribed angle theorem, $C$ on major arc intercepts minor arc).
Tangent-Chord angle at $A$: $\theta = \frac{1}{2}(100°) = \mathbf{50°}$ (tangent-chord angle = half intercepted arc).
Note: $\angle ACB = \theta = 50°$ — an elegant result of circle geometry.

📋 Chapter Summary

Key Formulas

Arc Length

$L = \dfrac{\theta}{360°} \cdot 2\pi r$ where $\theta$ is the central angle in degrees. In radians: $L = r\theta$.

Sector Area

$A = \dfrac{\theta}{360°} \cdot \pi r^2$ — the "pie slice" area corresponding to central angle $\theta$.

Inscribed Angle Theorem

An inscribed angle $= \frac{1}{2}$ its intercepted arc. Inscribed angles intercepting the same arc are congruent.

Tangent-Radius

A tangent line is perpendicular to the radius at the point of tangency. $OT \perp$ tangent at $T$.

Angle-Arc Relationships

Central Angle

Central angle $=$ intercepted arc (in degrees). The arc measure equals the central angle that subtends it.

Inscribed Angle

Inscribed angle $= \frac{1}{2}$ intercepted arc. An angle inscribed in a semicircle is always $90°$.

Two Chords Inside

Angle $= \frac{1}{2}(\text{arc}_1 + \text{arc}_2)$. The angle formed by two chords inside a circle averages the two arcs.

Two Secants/Tangents Outside

Angle $= \frac{1}{2}(\text{larger arc} - \text{smaller arc})$. The angle formed outside is half the difference of the arcs.

📘 Key Terms

ChordA segment with both endpoints on the circle. The longest chord passes through the center (diameter).
TangentA line that touches the circle at exactly one point (the point of tangency). Perpendicular to the radius there.
SecantA line that intersects a circle at two points.
Inscribed AngleAn angle with its vertex on the circle and sides that are chords. Equals half the intercepted arc.
Arc LengthThe distance along the circle between two points: $L = \frac{\theta}{360°} \cdot 2\pi r$.
SectorA "pie slice" region bounded by two radii and an arc. Area $= \frac{\theta}{360°}\cdot\pi r^2$.
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