Chapter 10 — Front Matter
Purpose and Overview
This chapter introduces circles as geometric objects with a rich internal structure. Unlike polygons, which are built from straight sides and angles, the circle is defined by a single relationship: every point on the circle is the same distance from a fixed center. That one idea generates an entire system of relationships — among radii, chords, arcs, central angles, inscribed angles, tangents, and secants — that you will study here.
This chapter also connects circle geometry to algebra. You will write and solve equations involving angle measures, segment lengths, and arc lengths. In the final sections, you will write equations that describe circles on the coordinate plane and graph circles from equations. These skills directly prepare you for area, sectors, coordinate geometry, trigonometry, engineering applications, and later analytic geometry.
Circles appear constantly in the world around you: wheels, clocks, lenses, orbits, pipes, pulleys, gears, stadiums, compasses, and radar screens. The mathematics of circles is not abstract decoration. It is the foundation of circular motion, wave behavior, design, architecture, and much of engineering and physics.
Opening Question
A surveyor needs to mark the exact center of a circular fountain that is already built. She can only access the outer rim. How can she find the center without demolishing the structure or estimating by eye? This kind of problem — finding a hidden center from information on the circle's edge — is exactly what circle geometry makes possible. By the end of this chapter, you will be able to answer this question and many like it.
14Sections
12Objectives
20+Key Terms
15+Worked Examples
Learning Objectives
- 1Define a circle and identify its parts: center, radius, diameter, chord, arc, tangent, and secant.
- 2Find central angle measures and arc measures, and apply the Arc Addition Postulate.
- 3Calculate arc length using proportional reasoning.
- 4Apply the Inscribed Angle Theorem and its corollaries.
- 5Use chord relationships to find segment lengths and distances from the center.
- 6Identify and apply properties of tangents, including two tangents from an external point.
- 7Apply angle relationships for chords, secants, and tangents intersecting inside and outside a circle.
- 8Apply segment relationships for intersecting chords, secants, and tangent-secant combinations.
- 9Write the equation of a circle given its center and radius.
- 10Graph a circle and convert expanded form to standard form by completing the square.
- 11Apply circle relationships to real-world measurement, design, and motion problems.
- 12Identify and correct common errors in circle reasoning.
Sections 10.1 – 10.4
Parts of a Circle, Arcs, and Arc Length
Definition · Radius · Diameter · Chords · Arcs · Arc Measure · Arc Length
Section 10.1
What a Circle Is
A circle is the set of all points in a plane that are the same distance from a fixed point called the center. The fixed distance from the center to any point on the circle is called the radius. Every radius of a given circle has the same length — this is guaranteed by the definition itself.
The circle is not a filled-in region. It is the boundary curve only. The filled-in region bounded by a circle is called a disk. In geometry, circle refers strictly to the curved edge.
A circle is named by its center. If the center is point P, the circle is called circle P, written ⊙P.
Diagram Instruction 10.1
Draw a point labeled P near the center of the page. Draw a smooth closed curve around P such that the distance from P to every point on the curve appears equal. Label several points on the curve A, B, C, and D. Draw segments PA, PB, PC, and PD. These are all radii and all have equal length. Mark them with the same tick mark to show congruence.
Why This Matters
Because every radius is equal, any two radii of the same circle are congruent. This seems obvious, but it is the starting point for nearly every circle proof. Many proofs begin with the statement "all radii of a circle are congruent" as a reason — keep this fact in mind throughout the chapter.
Section 10.2
Radius, Diameter, Chords, and Secants
| Part | What it is | Connects |
|---|
| Radius | Segment from center to any point on the circle | Center to circle |
| Diameter | Chord passing through the center; longest chord | Circle to circle, through center |
| Chord | Segment with both endpoints on the circle | Circle to circle (any chord) |
| Secant | Line intersecting the circle at exactly two points | Passes through circle at two points |
Diameter and Radius Relationship
d = 2r and r = d / 2
A diameter divides the circle into two equal halves, each called a semicircle. The diameter is the longest possible chord in any circle.
Example 10.2.1
Finding radius from diameter
- ProblemCircle M has a diameter of 26 cm. What is the radius?
- ThinkThe radius is half the diameter.
- Solve
r = d / 2 = 26 / 2 = 13 cm
Answer: The radius of circle M is 13 centimeters.
Example 10.2.2
Using algebra with radius and diameter
- ProblemThe radius of circle K is (3x + 4) and the diameter is (8x − 6). Find x and the actual radius.
- ThinkThe diameter equals twice the radius: d = 2r.
- Solve
8x − 6 = 2(3x + 4)
8x − 6 = 6x + 8
2x = 14
x = 7
r = 3(7) + 4 = 25
- Checkd = 8(7) − 6 = 50 and 2r = 2(25) = 50. ✓
Answer: x = 7 and the radius is 25 units.
Section 10.3
Central Angles and Arcs
A central angle is an angle whose vertex is at the center of the circle and whose two sides are radii. An arc is a connected portion of a circle. Every central angle intercepts two arcs — the arc inside the angle (the minor arc) and the arc on the other side (the major arc).
| Arc Type | Measure | Named with |
|---|
| Minor arc | Less than 180° | Two letters: AB⃗ (the shorter arc) |
| Semicircle | Exactly 180° | Three letters: AXB (any point X on the arc) |
| Major arc | More than 180° | Three letters: AXB (passes through a named midpoint) |
Arc Measure
The arc measure equals the measure of its central angle. A full circle = 360°. A semicircle = 180°.
Arc Addition Postulate: m⌢AC + m⌢CB = m⌢AB (when C is between A and B on the arc)
Example 10.3.1
Finding arc measures from a central angle
- ProblemIn circle P, central angle ∠APB = 74°. Find the minor arc and major arc.
- ThinkMinor arc = central angle. Major arc = 360° − minor arc.
- Solve
m⌢AB (minor) = 74°
m⌢AXB (major) = 360° − 74° = 286°
Answer: Minor arc = 74°; Major arc = 286°.
Example 10.3.2
Arc Addition Postulate
- ProblemIn circle Q, m⌢AB = 112° and m⌢BC = 85°. Find m⌢CA (the arc not passing through B).
- ThinkAll three arcs sum to 360°.
- Solve
112° + 85° + m⌢CA = 360°
m⌢CA = 360° − 197° = 163°
Answer: m⌢AC = 163°.
Section 10.4
Arc Measure and Arc Length
Arc measure tells you what portion of the circle the arc covers, expressed in degrees. Arc length tells you the actual physical distance along the arc in linear units. Two circles can both have a 90° arc — but if one circle is large and one is small, their arc lengths will be completely different. Both the central angle and the radius are needed to compute arc length.
Arc Length Formula
ℓ = (θ / 360) · 2πr
where θ is the arc measure in degrees and r is the radius
Circumference
C = 2πr = πd
Student Warning — Measure vs. Length
Arc measure is in degrees. Arc length is in centimeters, inches, or other linear units. If you are given only the arc measure and no radius, you cannot compute arc length. These are two separate pieces of information — never confuse them.
Example 10.4.1
Finding arc length from central angle
- ProblemCircle R has radius 10 cm. Find the arc length of a 72° arc.
- ThinkThe arc covers 72/360 = 1/5 of the full circle.
- Solve
ℓ = (72/360) · 2π(10)
= (1/5) · 20π
= 4π ≈ 12.57 cm
Answer: Arc length = 4π cm ≈ 12.57 cm.
Example 10.4.2
Finding arc measure from arc length
- ProblemA circle has circumference 48π inches. An arc has length 12π inches. Find the central angle.
- ThinkSet up the proportion: arc length / circumference = θ / 360.
- Solve
12π / 48π = θ / 360
1/4 = θ / 360
θ = 90°
Answer: The central angle and arc measure is 90°.
Sections 10.5 – 10.9
Inscribed Angles, Chords, and Angle Relationships
Inscribed Angles · Chord Properties · Tangents · Secants · Angle Formulas
Section 10.5
Inscribed Angles
An inscribed angle is an angle formed by two chords that share an endpoint on the circle. The vertex lies on the circle (not at the center), and the two sides are chords. The arc lying in the interior of the angle — the arc that does not contain the vertex — is called the intercepted arc.
Diagram Instruction 10.5
Draw circle P. Mark three points A, B, and C on the circle. Draw chords BA and BC. The angle ∠ABC is an inscribed angle with vertex B on the circle. The arc AC that does not contain B is the intercepted arc. Label it and distinguish it visually from the arc that does contain B.
Theorem 10.5
Inscribed Angle Theorem
The measure of an inscribed angle is half the measure of its intercepted arc.
m∠ABC = ½ · m⌢AC
An inscribed angle is always half the central angle that intercepts the same arc.
Corollaries to the Inscribed Angle Theorem
Two Important Consequences
Corollary 1: All inscribed angles intercepting the same arc are congruent.
Corollary 2: An inscribed angle that intercepts a semicircle is a right angle. If AC is a diameter and B is any point on the circle, then ∠ABC = 90°.
Diagram Instruction — Angle in a Semicircle
Draw circle P. Draw diameter AC. Mark any other point B on the circle. Draw chords BA and BC. Mark a right-angle symbol at B to show that ∠ABC = 90°. This right angle is guaranteed by Corollary 2, regardless of where B is placed on the circle.
Example 10.5.1
Using the Inscribed Angle Theorem
- ProblemInscribed angle ∠PQR = 38°. Find m⌢PR.
- ThinkThe inscribed angle is half the intercepted arc. So the arc is twice the angle.
- Solve
38° = ½ · m⌢PR
m⌢PR = 76°
Answer: The intercepted arc PR measures 76°.
Example 10.5.2
Inscribed angles intercepting the same arc
- Problem∠ACB and ∠ADB both intercept arc AB = 110°. Find both angle measures.
- ThinkBoth inscribed angles intercept the same arc, so both equal half of 110°.
- Solve
m∠ACB = ½(110°) = 55°
m∠ADB = ½(110°) = 55°
Answer: Both inscribed angles measure 55°. They are congruent, as Corollary 1 predicts.
Section 10.6
Chords and Distance from the Center
Theorem 10.6a
Perpendicular from Center to a Chord
If a diameter (or radius) is perpendicular to a chord, then it bisects the chord and its arc.
The two triangles formed on either side are congruent by HL (same hypotenuse = radius, shared perpendicular leg), so the half-chord lengths are equal.
Theorem 10.6b
Congruent Chords and Equal Distance
In the same circle, two chords are congruent if and only if they are equidistant from the center. Distance is measured perpendicularly from the center to the chord.
Example 10.6.1
Using the perpendicular bisector theorem
- ProblemDiameter RS is perpendicular to chord XY at point T. XT = 7 cm. Find XY.
- ThinkA diameter perpendicular to a chord bisects the chord.
- Solve
XY = 2 · XT = 2(7) = 14 cm
Answer: XY = 14 cm.
Example 10.6.2
Finding chord length with the Pythagorean Theorem
- ProblemIn circle O with radius 13, two chords are each at a distance of 5 from the center. Find the length of each chord.
- ThinkThe radius, half-chord, and perpendicular distance form a right triangle.
- DiagramDraw center O, chord AB, and perpendicular from O to midpoint M. Triangle OMA has OA = 13 (radius), OM = 5 (distance), MA = unknown.
- Solve
OA² = OM² + MA²
13² = 5² + MA²
169 = 25 + MA²
MA² = 144
MA = 12
AB = 2(12) = 24
- CheckBoth chords are equidistant from the center, so both have the same length.
Answer: Each chord is 24 units long.
Section 10.7
Tangents to Circles
A tangent to a circle is a line in the plane of the circle that intersects the circle at exactly one point. That single point of intersection is the point of tangency. A tangent touches the circle but does not pass through its interior.
Diagram Instruction 10.7
Draw circle P. Draw a line ℓ that touches the circle at exactly one point, labeled T. Draw the radius PT from the center to the point of tangency. Mark a right-angle symbol where the radius meets the tangent line — the radius and tangent are perpendicular at the point of tangency.
Theorem 10.7a
Tangent-Radius Perpendicularity
A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
PT ⊥ ℓ at point of tangency T
Converse: If a line is perpendicular to a radius at its outer endpoint, then the line is tangent to the circle.
Theorem 10.7b
Two Tangents from an External Point
If two tangent segments are drawn to a circle from the same external point, the two tangent segments are congruent.
Proof outline: Draw both radii and connect to the external point. The two triangles are congruent by HL (right angles at tangency, same hypotenuse, equal radii). Therefore the tangent segments are equal.
Example 10.7.1
Tangent segments from an external point
- ProblemFrom external point X, tangent segments touch circle C at A and B. XA = 3y − 4 and XB = y + 10. Find the tangent length.
- ThinkTangent segments from the same external point are equal: XA = XB.
- Solve
3y − 4 = y + 10
2y = 14
y = 7
XA = 3(7) − 4 = 17
- CheckXB = 7 + 10 = 17. ✓
Answer: Each tangent segment is 17 units long.
Internally and Externally Tangent Circles
Two circles can be tangent to each other. Externally tangent: the circles touch at one point and their interiors do not overlap; distance between centers = sum of radii. Internally tangent: one circle is inside the other and they touch at one point; distance between centers = difference of radii.
Section 10.8
Secants and Tangents
A secant of a circle is a line that intersects the circle in exactly two points. A chord is the segment between those two intersection points; the secant is the full line. When lines — whether tangents or secants — intersect from the same external point or cross inside the circle, they create predictable angle and segment relationships, organized in Sections 10.9 and 10.10.
An intercepted arc is the arc that lies in the interior of an angle formed by two lines or segments that intersect on or near a circle.
Section 10.9
Angle Relationships in Circles
The angle measure formed by two lines related to a circle depends entirely on where the vertex is located — at the center, on the circle, inside the circle, or outside the circle.
The Three Angle Cases — Summary
Vertex at center (central angle): angle = intercepted arc
Vertex on circle (inscribed angle): angle = ½ · intercepted arc
Vertex inside circle (two chords): angle = ½(sum of two intercepted arcs)
Vertex outside circle (secants/tangents): angle = ½(far arc − near arc)
| Vertex Location | Configuration | Formula |
|---|
| At center | Central angle | m∠ = m⌢arc |
| On circle | Inscribed angle (two chords) | m∠ = ½ · m⌢arc |
| Inside circle | Two intersecting chords | m∠ = ½(m⌢arc₁ + m⌢arc₂) |
| Outside circle | Two secants, secant-tangent, or two tangents | m∠ = ½(m⌢far − m⌢near) |
Diagram Instruction — Intersecting Chords Inside a Circle
Draw circle O. Draw two chords AB and CD crossing inside the circle at point E. Label the four arcs: ⌢AC, ⌢CB, ⌢BD, and ⌢DA. The angle ∠AEC at the intersection is formed by chords EA and EC. Its two intercepted arcs are ⌢AC (the arc inside the angle) and ⌢BD (the arc inside the vertical angle across from it).
Example 10.9.1
Chords intersecting inside a circle
- ProblemTwo chords intersect inside circle O. The two intercepted arcs measure 80° and 120°. Find the angle at intersection.
- ThinkVertex inside circle → use half the sum of the intercepted arcs.
- Solve
m∠ = ½(80° + 120°) = ½(200°) = 100°
Answer: The angle measures 100°.
Example 10.9.2
Two secants from an external point
- ProblemTwo secants from external point X intercept arcs of 160° (far) and 40° (near). Find the angle at X.
- ThinkVertex outside the circle → use half the difference of the arcs.
- Solve
m∠X = ½(160° − 40°) = ½(120°) = 60°
Answer: The angle at X is 60°.
Example 10.9.3
Using algebra with angle relationships
- ProblemTwo chords intersect inside circle O forming angle (5x + 10)°. The intercepted arcs are 74° and (4x + 12)°. Find x and the angle.
- ThinkSet up the chord-chord angle formula and solve for x.
- Solve
5x + 10 = ½(74 + 4x + 12)
5x + 10 = ½(86 + 4x)
5x + 10 = 43 + 2x
3x = 33
x = 11
m∠ = 5(11) + 10 = 65°
- Check½(74 + 4(11) + 12) = ½(130) = 65°. ✓
Answer: x = 11 and the angle measures 65°.
Sections 10.10 – 10.12
Segment Relationships and Equations of Circles
Intersecting Chords · Secant Products · Tangent² · Standard Form · Completing the Square
Section 10.10
Segment Relationships in Circles
Three Segment Theorems
Intersecting chords (inside): EA · EB = EC · ED
Two secants (outside): external × whole = external × whole
Tangent and secant (outside): tangent² = external × whole
Diagram Instruction — Intersecting Chords
Draw circle O. Draw two chords AB and CD meeting at interior point E. Label the four separate segments EA, EB, EC, and ED. These four lengths satisfy EA · EB = EC · ED. Note that each product uses the two segments of one chord — not the full chord lengths.
Example 10.10.1
Intersecting chords inside a circle
- ProblemChords AB and CD intersect at E inside circle O. EA = 6, EB = 8, EC = 4. Find ED.
- ThinkUse EA · EB = EC · ED.
- Solve
6 · 8 = 4 · ED
48 = 4 · ED
ED = 12
Answer: ED = 12.
Example 10.10.2
Two secants from an external point
- ProblemFrom external point P, one secant has external segment 4 and total length 10. The second has external segment x and total length (x + 12). Find x.
- Thinkexternal × whole = external × whole
- Solve
4 · 10 = x(x + 12)
40 = x² + 12x
x² + 12x − 40 = 0
(x + 20)(x − 2) = 0
x = 2 (reject x = −20)
Answer: x = 2.
Example 10.10.3
Tangent and secant from an external point
- ProblemFrom external point Q, tangent QT = 6 and secant passes through A and B with QA = 4. Find QB.
- ThinkUse tangent² = external × whole: QT² = QA · QB.
- Solve
6² = 4 · QB
36 = 4 · QB
QB = 9
Answer: QB = 9.
Section 10.11
Equations of Circles
Every point (x, y) on a circle with center (h, k) and radius r satisfies the distance formula: √[(x − h)² + (y − k)²] = r. Squaring both sides produces the standard form of the equation of a circle.
Standard Form — Equation of a Circle
(x − h)² + (y − k)² = r²
Center (h, k) · Radius r · Origin case: x² + y² = r²
Student Warning — Sign Convention
The equation (x + 3)² + (y − 5)² = 49 has center (−3, 5), not (3, 5). The term "(x + 3)" means x − (−3), so h = −3. Always reverse the signs of h and k as they appear in the equation to find the true center. And r² = 49 means r = 7 — not r = 49.
Example 10.11.1
Writing the equation of a circle
- ProblemWrite the equation of the circle with center (2, −7) and radius 5.
- ThinkSubstitute h = 2, k = −7, r = 5 into standard form.
- Solve
(x − 2)² + (y − (−7))² = 5²
(x − 2)² + (y + 7)² = 25
Answer: (x − 2)² + (y + 7)² = 25
Example 10.11.2
Finding center and radius from an equation
- ProblemIdentify the center and radius of (x + 4)² + (y − 1)² = 36.
- ThinkRewrite (x + 4)² as (x − (−4))², so h = −4. The y term gives k = 1. And r = √36 = 6.
- Solve
Center: (−4, 1) Radius: 6
Answer: Center (−4, 1), radius 6.
Example 10.11.3
Writing an equation from center and a point
- ProblemA circle has center (−1, 3) and passes through (5, 3). Write its equation.
- ThinkFind the radius using the distance formula, then substitute into standard form.
- Solve
r = √[(5−(−1))² + (3−3)²]
= √[6² + 0²] = √36 = 6
(x + 1)² + (y − 3)² = 36
Answer: (x + 1)² + (y − 3)² = 36
Section 10.12
Circles on the Coordinate Plane
Graphing a Circle from Its Equation
To graph the circle (x − h)² + (y − k)² = r²: (1) identify center (h, k), (2) identify radius r, (3) plot the center, (4) move r units in each cardinal direction (up, down, left, right) to find four guide points, (5) sketch the smooth circle through them.
Diagram Instruction — Graphing (x − 2)² + (y + 1)² = 9
Center: (2, −1). Radius: r = 3. Guide points: (2, 2) above, (2, −4) below, (5, −1) right, (−1, −1) left. Plot all five points on a coordinate grid and sketch the smooth circle through the four guide points, centered at (2, −1).
Completing the Square to Convert to Standard Form
Circle equations are sometimes given in expanded form such as x² + y² − 6x + 4y − 12 = 0. Use completing the square to rewrite in standard form: group x and y terms separately, complete the square for each variable (adding the same amounts to both sides), then factor each group.
Example 10.12.1
Completing the square to find center and radius
- ProblemRewrite x² + y² − 6x + 4y − 12 = 0 in standard form.
- ThinkGroup x and y terms, then complete the square for each.
- Solve
(x² − 6x) + (y² + 4y) = 12
For x: half of −6 = −3, (−3)² = 9. Add 9 to both sides.
For y: half of 4 = 2, (2)² = 4. Add 4 to both sides.
(x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x − 3)² + (y + 2)² = 25
Answer: Center (3, −2), radius 5.
Example 10.12.2
Does a point lie on, inside, or outside the circle?
- ProblemDoes point (7, 1) lie on, inside, or outside (x − 3)² + (y + 2)² = 25?
- ThinkSubstitute (7, 1) into the left side and compare to r² = 25.
- Solve
(7 − 3)² + (1 + 2)² = 4² + 3² = 16 + 9 = 25
- Check25 = 25, so the point satisfies the equation exactly.
Answer: The point (7, 1) lies ON the circle. (If result < 25: inside; if result > 25: outside.)
Sections 10.13 – 10.14
Applications and Common Mistakes
Architecture · Gears · Surveying · Radar · Six Recurring Errors
Section 10.13
Circles in Design, Motion, and Measurement
Circles appear throughout architecture, engineering, and daily life. Arches in bridges and doorways follow circular arcs. Knowing the arc length and radius allows builders to calculate how much material is needed for a curved edge. Gears are toothed wheels built around circles — the ratio of their radii determines how many times one gear rotates for each rotation of the other. Surveying uses circle theorems to locate hidden centers. Radar and sonar displays model objects as points on circular rings of fixed radius from the source.
Example 10.13.1
Arc length in architecture
- ProblemA garden path curves along a circle with radius 15 m. The arc spans a central angle of 120°. How long is the path?
- Solve
ℓ = (120/360) · 2π(15)
= (1/3) · 30π
= 10π ≈ 31.4 m
Answer: The path is 10π meters long, approximately 31.4 m.
Example 10.13.2
Gear ratio application
- ProblemA large gear (radius 12 cm) makes 5 rotations. A small gear (radius 4 cm) meshes with it. How many rotations does the small gear make?
- ThinkBoth gears travel the same arc distance at their contact point. Distance = rotations × circumference.
- Solve
Large gear distance = 5 × 2π(12) = 120π cm
Small gear circumference = 2π(4) = 8π cm
Rotations of small gear = 120π / 8π = 15
Answer: The small gear makes 15 full rotations.
Surveying Application — Finding a Hidden Center
Recall the opening question: how does a surveyor find the center of a circular fountain from the rim? Construct the perpendicular bisectors of two chords between known points on the rim. The perpendicular bisector of any chord passes through the center (Theorem 10.6a). Two perpendicular bisectors intersect at exactly one point — the center. This is a direct application of circle geometry to real construction.
Section 10.14
Common Circle Mistakes
Mistake 1 — Confusing Arc Measure with Arc Length
"Arc measure = 60°, so arc length = 60 cm."
Arc measure is in degrees; arc length is in linear units. You cannot find arc length without the radius. The correct formula is ℓ = (60/360) · 2πr — the radius is essential.
Mistake 2 — Confusing Inscribed and Central Angles
"The central angle is ½ the arc, just like the inscribed angle."
A central angle EQUALS its intercepted arc. An inscribed angle is HALF its intercepted arc. Check: where is the vertex? Center → central angle (equal). On the circle → inscribed angle (half).
Mistake 3 — Misreading the Circle Equation
"(x + 3)² + (y − 5)² = 16 has center (3, 5) and radius 16."
Correct: center is (−3, 5) — reverse the signs as written in the equation. Radius is √16 = 4, not 16. Never forget to take the square root of the right-hand side to find r.
Mistake 4 — Applying the Wrong Segment Theorem
Using the intersecting-chords formula for two secants from an external point.
Inside the circle (chords): each chord's two segments multiply equally. Outside the circle (secants): external segment × whole = external segment × whole. The setup differs — draw the diagram and label carefully before choosing a formula.
Mistake 5 — Incomplete Completing the Square
Adding to one side only when completing the square.
When you add (half the coefficient)² to complete the square for x, you must add the same amount to the right-hand side to keep the equation balanced. Forgetting this step produces the wrong center or radius.
Mistake 6 — Treating Any Circle Angle as Inscribed
Applying the ½ · arc formula to an angle whose vertex is inside or outside the circle.
The ½ · arc rule applies only when the vertex is ON the circle (inscribed angle). Inside the circle: ½(sum of arcs). Outside the circle: ½(difference of arcs). Identify the vertex location first, every time.
Chapter 10 — Back Matter
Chapter Summary
- Sec 10.1A circle is the set of all points in a plane equidistant from a fixed center. Every radius is congruent. The circle is the curved boundary, not the filled region.
- Sec 10.2The radius reaches from center to circle; the diameter passes through the center (d = 2r). A chord connects two points on the circle; a secant is the full line through two circle points.
- Sec 10.3A central angle has its vertex at the center. Arc measure equals the central angle. Minor arc < 180°; major arc > 180°; semicircle = 180°. Arc Addition Postulate: adjacent arcs add.
- Sec 10.4Arc length ℓ = (θ/360) · 2πr requires both angle and radius. Circumference C = 2πr. Arc measure and arc length are different quantities — one in degrees, one in linear units.
- Sec 10.5An inscribed angle has its vertex on the circle. Inscribed Angle Theorem: m∠ = ½ · intercepted arc. Same-arc corollary: all inscribed angles intercepting the same arc are equal. Semicircle corollary: inscribed angle in a semicircle = 90°.
- Sec 10.6A diameter perpendicular to a chord bisects the chord and its arc. Congruent chords are equidistant from the center. The Pythagorean Theorem connects radius, chord half-length, and perpendicular distance.
- Sec 10.7A tangent meets the circle at exactly one point. The tangent is perpendicular to the radius at the point of tangency. Two tangent segments from the same external point are congruent.
- Sec 10.8A secant intersects the circle at two points. The intercepted arc lies inside the angle formed by two intersecting lines related to the circle.
- Sec 10.9Angle formulas by vertex location: center = arc; on circle = ½ arc; inside = ½(sum); outside = ½(difference). Algebra can be used to write and solve equations with these formulas.
- Sec 10.10Segment products: intersecting chords EA·EB = EC·ED; two secants: ext×whole = ext×whole; tangent-secant: tangent² = ext×whole.
- Sec 10.11Standard form of a circle: (x − h)² + (y − k)² = r². Center (h, k), radius r. Reverse signs of h and k as written in the equation. r = √(right side).
- Sec 10.12Graph by plotting center and four guide points r units in each direction. Convert expanded form to standard form by completing the square for both x and y groups, adding equal amounts to both sides.
- Sec 10.13Arc length formulas apply to paths, gears, and architecture. Circle theorems solve real surveying problems. Gear ratios relate to circumference and radius. Radar and sonar models use the circle definition directly.
- Sec 10.14Six recurring errors: measure vs. length; inscribed vs. central; equation sign errors; r vs. r²; wrong segment theorem; assuming all circle angles are inscribed. Always identify vertex location before applying an angle formula.
Key Terms Review
arc
A connected portion of a circle.
arc length
The distance along an arc in linear units; requires both angle and radius.
arc measure
The degree measure of an arc, equal to its central angle.
central angle
An angle with vertex at the center and sides that are radii. Equals its intercepted arc.
chord
A segment with both endpoints on the circle. A diameter is a special chord through the center.
circle
The set of all points in a plane equidistant from a fixed center point.
circumference
The total distance around a circle: C = 2πr = πd.
diameter
A chord passing through the center; the longest chord; d = 2r.
inscribed angle
An angle with vertex on the circle and sides as chords. Equals half its intercepted arc.
intercepted arc
The arc lying in the interior of an angle formed by lines intersecting on or near a circle.
major arc
An arc greater than a semicircle; arc measure greater than 180°.
minor arc
An arc less than a semicircle; arc measure less than 180°.
point of tangency
The single point where a tangent line meets the circle.
radius
A segment from the center to any point on the circle; also its length. All radii of a circle are congruent.
secant
A line intersecting a circle at exactly two points.
semicircle
An arc equal to half the circle; arc measure = 180°. Created by a diameter.
standard form
The circle equation (x − h)² + (y − k)² = r² with center (h, k) and radius r.
tangent
A line intersecting a circle at exactly one point, perpendicular to the radius at that point.
Concept Review Questions
- What is the geometric definition of a circle? How does this definition generate the concept of radius?
- What is the difference between a chord and a diameter? Between a secant and a chord?
- What is the difference between arc measure and arc length? Why does arc length require knowing the radius?
- State the Inscribed Angle Theorem. Why does it say "half the intercepted arc" rather than "equal to the intercepted arc"?
- What is special about an inscribed angle that intercepts a semicircle?
- State the theorem about a radius perpendicular to a chord. Why does this follow from triangle congruence?
- State the tangent-radius relationship. Why does it produce a right angle?
- State the rule for angle measure when the vertex is inside the circle. When the vertex is outside the circle.
- State the intersecting-chords segment theorem. How is it different from the two-secants theorem?
- Write the standard form of the equation of a circle. What does each symbol represent?
- In circle Q with radius 11 cm, find the diameter.
- In circle R with diameter 30 in, find the radius.
- True or false: a chord can also be a diameter. Explain.
- The complete circle has arc measure ____. A semicircle has arc measure ____.
- True or false: two different circles with arcs both measuring 45° have the same arc length. Explain.
- Define an intercepted arc in your own words and draw an example.
- In circle O, central angle ∠AOB = 95°. Find m⌢AB.
- In circle P, m⌢XY = 140°. Find the major arc m⌢XZY.
- Three arcs: m⌢AB = 80°, m⌢BC = 115°. Find m⌢CA.
- Arc length of a 90° arc, radius 8 cm. Leave in terms of π.
- Arc length of a 45° arc, diameter 20 in. Leave in terms of π.
- Circumference 36π cm; arc length 9π cm. Find central angle.
- Inscribed angle ∠PQR = 52°. Find m⌢PR.
- m⌢ST = 126°. Find inscribed ∠SUT.
- ∠ACB and ∠ADB both intercept m⌢AB = 88°. Find both angles.
- AC is a diameter; ∠ABC is inscribed. What is m∠ABC? Why?
- Two chords intersect inside; arcs 60° and 100°. Find the angle.
- Two secants from external point; arcs 150° and 30°. Find external angle.
Algebra with angles:
- Chords intersect inside circle O forming angle (4x + 5)°; arcs 86° and (2x + 1)°. Find x and the angle.
- Inscribed angle (3x − 7)° intercepts arc (4x + 20)°. Find x and the angle.
- Tangent-radius: radius = 9, center-to-external = 15. Find tangent length.
- Two tangents from X: XA = 5t − 2, XB = 2t + 7. Find t and length.
- Explain why two tangent segments from external point are congruent. Name each theorem step.
- Chords AB and CD intersect at E: EA = 5, EB = 12, EC = 4. Find ED.
- Two secants from P: first has ext seg 3, total 12. Second has ext seg 4, total x. Find x.
- From Q: tangent QT = 8, secant with QA = 4. Find QB.
Write the equation of each circle:
- Center (0, 0), radius 5.
- Center (3, 4), radius 7.
- Center (−2, 5), radius √11.
- Center (0, −6), radius 3.
Identify center and radius from each equation:
- (x − 5)² + (y − 2)² = 64
- (x + 3)² + (y − 8)² = 100
- (x + 1)² + y² = 49
- x² + y² = 144
Write the equation from given information:
- Center (4, −1); passes through (4, 5).
- Center (−3, 0); passes through the origin.
Convert to standard form by completing the square. Then identify center and radius:
- x² + y² − 4x + 6y − 3 = 0
- x² + y² + 8x − 2y − 8 = 0
- x² + y² − 10x = 0
- x² + y² + 2x − 4y − 20 = 0
Does the point lie on, inside, or outside the circle?
- Point (1, 4) and circle (x − 1)² + (y − 1)² = 16.
- Point (6, 6) and circle (x − 3)² + (y − 2)² = 25.
Word Problems and Applications
For each problem: identify all given information, draw a diagram where helpful, select the correct formula or theorem, solve, and check.
- A circular clock has radius 15 cm. The minute hand sweeps through 240° in 40 minutes. How far does the tip of the minute hand travel? Leave in terms of π.
- A park path curves along a circle of radius 50 m. The arc spans a central angle of 72°. How long is the path to the nearest tenth of a meter?
- A wheel of radius 30 cm rolls along a flat surface. How far does it travel in 5 full rotations? Leave in terms of π.
- A circular pool has center at the origin. The pool has radius 6 m. Write its equation. Does the point (4, 4) lie inside, on, or outside the pool?
- An engineer draws a circular arch for a bridge. The arch is part of a circle passing through (10, 0), (−10, 0), and (0, 10). What is the radius? Write the equation. (Hint: substitute each point and solve the system.)
- Two gears mesh. Large gear radius = 18 cm; small gear radius = 6 cm. Large gear makes 3 rotations. How many does the small gear make?
- In circle O, diameter AC = 20. Point B is on the circle and AB = 12. Since AC is a diameter, ∠ABC = 90°. Find BC and the area of △ABC.
Error Analysis
Each problem contains an error. Find and correct it.
Problem 1
"Circle P has radius 4. Arc = 90°. Arc length = (90/360) · 4 = 1 unit."
What is the error? What is the correct calculation?
Problem 2
"Inscribed ∠AXB = 70°. The central angle ∠AOB is also 70° because they both intercept the same arc."
Identify the error. What is the correct central angle?
Problem 3
"(x + 2)² + (y − 3)² = 25 has center (+2, −3) and radius 25."
Correct both errors. State the true center and radius.
Problem 4
"Chords AB and CD intersect at E. EA = 3, EB = 4, EC = 2. ED = EA + EB − EC = 5."
Identify the error. Use the correct theorem to find ED.
Problem 5
"Two secants from external point X: external segments 5 and 6; total lengths 10 and 12. So 5 × 10 = 50 and 6 × 12 = 72. These aren't equal, so the problem is wrong."
What theorem should the student have used? Redo correctly and explain.
Chapter 10 Checkpoint
Chapter 10 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 10
Part A — Vocabulary and Concepts
- Define a circle in one precise sentence using the words "set of all points," "plane," and "distance."
- What is the difference between a major arc and a minor arc? How are they named differently?
- State the Inscribed Angle Theorem.
- State the tangent-radius perpendicularity theorem.
- What does the equation (x − h)² + (y − k)² = r² represent? Define h, k, and r.
Part B — Skill Problems
- Circle O has radius 9. Find the circumference and the arc length of a 40° arc. Leave in terms of π.
- Inscribed angle ∠ACB = 63°. Find m⌢AB.
- Two chords intersect inside circle Q forming a 75° angle. One intercepted arc is 90°. Find the other intercepted arc.
- From external point X, two tangent segments: XA = 4m − 3 and XB = 2m + 7. Find m and the tangent length.
- Chords PQ and RS intersect at T: PT = 6, QT = 9, RT = 4. Find ST.
- Write the equation of the circle with center (−3, 4) and radius 8.
- Find the center and radius of (x + 1)² + (y − 6)² = 49.
Part C — Extended Problems
- Convert x² + y² − 6x + 10y + 9 = 0 to standard form. State the center and radius. Then determine whether the point (6, −5) lies on, inside, or outside the circle.
- In circle O, diameter AC has length 20. Point B is on the circle and AB = 12. Since AC is a diameter, ∠ABC = 90°. Use the Pythagorean Theorem to find BC. Then find the area of △ABC.
- A clock has radius 20 cm. The minute hand sweeps from 12 to 4 (one-third of the circle). Find the arc length swept by the tip of the minute hand. Leave in terms of π.
Looking Ahead
Chapters 11 and 12: Area, Surface Area, and Volume
Chapter 11 continues from circles into area. You will compute the area enclosed by a circle (A = πr²) and the area of a sector — the "pie slice" cut by a central angle. Sectors connect arc length and area using the same proportional reasoning from this chapter.
Chapter 12 extends measurement into three dimensions: surface area of cylinders (built from circles) and volume of spheres. Every formula involves π and r, directly extending what you learned here.
Chapter 13 returns to coordinate geometry and uses the equation of a circle together with distance, midpoint, and slope to prove geometric relationships and solve problems algebraically.
In Trigonometry and Precalculus, the unit circle — radius 1, centered at the origin — becomes the foundation for extending sine and cosine to all angles. Every circle relationship from this chapter reappears in that context, especially central angles, arcs, and coordinates of points on the circle.
MGU Library Connection
- Circle Theorem Reference — Appendix I of this textbook
- Area Formula Sheet — Appendix J (includes sector area)
- Coordinate Geometry Formula Sheet — Appendix M (includes standard form of circle equation)
- Algebra I Readiness Review — Appendix A (completing the square)
- Chapter 11: Area and Surface Area — sectors, circle area, cylinder surface area
- Chapter 13: Coordinate Geometry — proving circle relationships with algebra
- Trigonometry and Precalculus: Unit Circle and Radian Measure
- Physics: Circular Motion, Orbital Paths, Angular Velocity
- Engineering: Gear Ratios, Pulley Systems, Pipe Cross-Sections