Circle Theorems and Arc Relationships

⭕ Circle Theorems & Arc Relationships

Part 1 of 5 — Arcs, Central Angles & the Vocabulary of Circles

Topics in This Part

🔑 Key Concept: Almost every circle theorem connects an angle to an arc. Before we can use those theorems, we need to measure arcs — and that begins with the central angle.

Parts of a Circle

Every circle theorem is built from a few key parts:

💡 A diameter is a special chord, and a secant is the line that a chord lies on. Knowing which is which decides which theorem applies.

Arcs and Their Measures

Two points on a circle split it into two arcs:

• A minor arc is the shorter arc (measure $< 180°$). Named with two letters, e.g. $\overgroup{AB}$.
• A major arc is the longer arc (measure $> 180°$). Named with three letters, e.g. $\overgroup{ACB}$, to make the path clear.
• A semicircle is exactly half the circle (measure $= 180°$), cut off by a diameter.

The measure of an arc is not its length — it is the number of degrees of the central angle that opens onto it. A full circle is always $360°$:

$\text{minor arc} + \text{major arc} = 360°$

🔑 Key Idea: Arc measure is in degrees and tells you "how much turn." Arc length is an actual distance and depends on the radius. This lesson is about measure.

Name That Part 🔽

Match each description to the correct term.

The Central Angle = Its Arc

A central angle has its vertex at the center of the circle. This gives us our very first theorem — and the simplest one:

$\boxed{\text{central angle} = \text{intercepted arc}}$

If central angle $\angle AOB = 70°$, then the minor arc $\overgroup{AB} = 70°$ as well.

Example

A central angle measures $110°$. Then:

• minor arc $\overgroup{AB} = 110°$
• major arc $\overgroup{ACB} = 360° - 110° = 250°$

💡 This is the only angle that equals its arc exactly. Every other angle in this lesson is some fraction or combination of arcs — so always ask first: "Is the vertex at the center?"

Concept Check 🎯

Arc Addition

Adjacent arcs add, just like adjacent angles:

$\overgroup{AB} + \overgroup{BC} = \overgroup{ABC}$

Worked Example

Points $A$, $B$, $C$ lie on a circle in order. If $\overgroup{AB} = 75°$ and $\overgroup{BC} = 60°$, then:

$\overgroup{ABC} = 75° + 60° = 135°$

And the remaining arc back to $A$ is $360° - 135° = 225°$.

⚠️ Common Mistake: Don't confuse a semicircle ($180°$) with a major arc ($>180°$). A diameter creates two semicircles, each exactly $180°$.

Arc Arithmetic 🧮

Points $P$, $Q$, $R$ lie on a circle in that order. Use arc addition and the $360°$ rule.

1) If $\overgroup{PQ} = 40°$ and $\overgroup{QR} = 95°$, then $\overgroup{PQR} = \,?$ degrees 2) Using part 1, the remaining major arc $\overgroup{RP}$ (back to $P$) $= \,?$ degrees 3) A central angle intercepts an arc of $148°$. The central angle $= \,?$ degrees

⭕ Circle Theorems & Arc Relationships

Part 2 of 5 — The Inscribed Angle Theorem

🔑 The Big One: An inscribed angle has its vertex on the circle. The Inscribed Angle Theorem says it is half of its intercepted arc. This single theorem powers most circle problems.

The Inscribed Angle Theorem

An inscribed angle is formed by two chords that share an endpoint on the circle. The arc "inside" the angle is the intercepted arc.

$\boxed{\text{inscribed angle} = \frac{1}{2}\,(\text{intercepted arc})}$

Equivalently, the arc is twice the inscribed angle.

Compare to the Central Angle

Worked Example

An inscribed angle intercepts an arc of $100°$. Then:

$\text{inscribed angle} = \frac{1}{2}(100°) = 50°$

Going the other way: if the inscribed angle is $35°$, its intercepted arc is $2(35°) = 70°$.

Half the Arc 🧮

Use inscribed angle $= \frac{1}{2}\,(\text{arc})$, and arc $= 2 \times (\text{inscribed angle})$.

1) An inscribed angle intercepts a $146°$ arc. The angle $= \,?$ degrees 2) An inscribed angle measures $52°$. Its intercepted arc $= \,?$ degrees 3) A central angle of $90°$ and an inscribed angle intercept the same arc. The inscribed angle $= \,?$ degrees

Powerful Corollary: Same Arc → Same Angle

Because every inscribed angle is half its arc, two inscribed angles that intercept the same arc must be equal — no matter where their vertices sit on the circle.

$\angle ACB = \angle ADB \quad \text{(both intercept arc } \overgroup{AB}\text{)}$

Why this is so useful

In many figures you'll see several angles all "looking at" the same chord. They're all equal. This turns scary diagrams into simple ones.

💡 Pro tip: When two angles open onto the same chord from the same side, set them equal. When you see a central and inscribed angle on the same arc, the central is twice the inscribed.

Concept Check 🎯

Putting It Together

Two theorems, side by side, decided entirely by where the vertex sits:

And the headline corollary: angles on the same arc are equal, while a central angle is twice an inscribed angle on that same arc.

💡 Keep this picture in mind for the next dropdown — every item is just "central vs. inscribed."

Pick the Relationship 🔽

For each pair, choose how the angle relates to the arc it intercepts.

⭕ Circle Theorems & Arc Relationships

Part 3 of 5 — Special Inscribed Angles: Semicircles, Cyclic Quadrilaterals & Tangent–Chord

🔑 From one theorem, three classics: The Inscribed Angle Theorem gives us (1) the right angle in a semicircle, (2) supplementary opposite angles in a cyclic quadrilateral, and (3) the tangent–chord angle.

Angle in a Semicircle (Thales' Theorem)

If the intercepted arc is a semicircle ($180°$), the inscribed angle is:

$\frac{1}{2}(180°) = 90°$

So an angle inscribed in a semicircle is always a right angle. Equivalently: if a triangle's longest side is a diameter, the angle opposite it is $90°$.

Worked Example

$\overline{AB}$ is a diameter and $C$ is any other point on the circle. Then $\angle ACB = 90°$ — guaranteed, wherever $C$ is.

If $\angle CAB = 28°$, then in right triangle $ACB$:

$\angle ABC = 180° - 90° - 28° = 62°$

💡 Spotting a diameter in a problem is a free $90°$. Always look for it.

Right Angle in a Semicircle 🧮

$\overline{AB}$ is a diameter of a circle and $C$ is another point on the circle, so $\angle ACB = 90°$.

1) $\angle ACB = \,?$ degrees 2) If $\angle CAB = 35°$, then $\angle ABC = \,?$ degrees 3) The arc $\overgroup{ACB}$ cut off by the diameter measures $\,?$ degrees

Cyclic Quadrilaterals

A cyclic quadrilateral has all four vertices on a circle. Its opposite angles are supplementary (add to $180°$):

$\angle A + \angle C = 180° \qquad \angle B + \angle D = 180°$

Why? Opposite angles intercept arcs that together make the whole circle. Each angle is half its arc, so together they are half of $360° = 180°$.

Worked Example

In cyclic quadrilateral $ABCD$, $\angle A = 95°$. Then the opposite angle:

$\angle C = 180° - 95° = 85°$

⚠️ Only opposite angles are supplementary. Adjacent angles ($\angle A$ and $\angle B$) are not generally supplementary.

Concept Check 🎯

Tangent–Chord Angle

When a tangent and a chord meet at the point of tangency, the angle they form is half the intercepted arc — the same "half the arc" rule, just with a tangent as one side:

$\text{tangent–chord angle} = \frac{1}{2}\,(\text{intercepted arc})$

Worked Example

A chord meets a tangent at the point of tangency, cutting off an arc of $130°$ on the side of the angle. Then:

$\text{angle} = \frac{1}{2}(130°) = 65°$

🔑 Pattern so far: Central angle $=$ arc. Everything else with its vertex ON the circle (inscribed angle, tangent–chord angle) $= \frac{1}{2}$ arc.

Classify & Compute 🔽

Choose the correct value for each special case.

⭕ Circle Theorems & Arc Relationships

Part 4 of 5 — Angles from Chords, Secants & Tangents

🔑 Where is the vertex? That one question selects the formula. Inside the circle → average of two arcs. Outside the circle → half the difference of two arcs.

Two Chords Meeting Inside the Circle

When two chords cross inside a circle, each angle is the average of the two arcs it (and its vertical partner) intercept:

$\text{angle} = \frac{1}{2}\,(\text{arc}_1 + \text{arc}_2)$

Here $\text{arc}_1$ is the arc "inside" the angle and $\text{arc}_2$ is the arc inside its vertical angle.

Worked Example

Two chords cross inside a circle. The arc inside the angle is $70°$ and the arc inside its vertical angle is $50°$:

$\text{angle} = \frac{1}{2}(70° + 50°) = \frac{1}{2}(120°) = 60°$

💡 An inscribed angle ($\frac{1}{2}$ of one arc) is just the special case where the vertex slides out to the circle and one of the two arcs shrinks to $0°$.

Average the Arcs (Vertex Inside) 🧮

Two chords intersect inside a circle. Use angle $= \frac{1}{2}(\text{arc}_1 + \text{arc}_2)$.

1) Arcs of $80°$ and $40°$. Angle $= \,?$ degrees 2) Arcs of $130°$ and $90°$. Angle $= \,?$ degrees 3) The angle is $75°$ and one intercepted arc is $60°$. The other arc $= \,?$ degrees

Vertex Outside the Circle

When two secants, two tangents, or a secant and a tangent meet outside the circle, the angle is half the difference of the intercepted arcs (far arc minus near arc):

$\text{angle} = \frac{1}{2}\,(\text{far arc} - \text{near arc})$

This works for all three exterior cases:

Worked Example

Two secants from an external point intercept a far arc of $120°$ and a near arc of $40°$:

$\text{angle} = \frac{1}{2}(120° - 40°) = \frac{1}{2}(80°) = 40°$

⚠️ Inside vs. outside: Inside you add the arcs (average); outside you subtract them (half the difference). Mixing these up is the #1 error on circle angle problems.

Concept Check 🎯

One Decision, Four Formulas

Everything in this lesson collapses to a single question — where is the vertex?

As the vertex moves from the center, outward to the circle, then outside, the angle for the same arcs gets smaller. That's why outside angles use a difference: they "see" less of the circle.

💡 Memory hook: Inside → add (the angle is "full of" circle); outside → subtract (the circle is "far away").

Inside or Outside? 🔽

Decide which operation to use, then read off the result.

⭕ Circle Theorems & Arc Relationships

Part 5 of 5 — Segment Lengths & Mastery Check

We've measured arcs and angles. Now we measure lengths of the chords, secants, and tangents themselves — then put everything together.

Segment Length Relationships

These "power of a point" rules relate the lengths of segments — not their arcs.

1) Two chords intersecting inside: the products of the two pieces are equal.

$a \cdot b = c \cdot d$

2) Two secants from an external point: (whole) $\times$ (external part) is equal for both.

$(\text{whole}_1)(\text{external}_1) = (\text{whole}_2)(\text{external}_2)$

3) Tangent and secant from an external point: the tangent squared equals (whole secant)(external part).

$t^2 = (\text{whole})(\text{external})$

🔑 Theme: Every rule multiplies a segment by its matching partner. Inside the circle it's piece $\times$ piece; outside it's whole $\times$ external.

Power of a Point 🧮

1) Two chords cross inside a circle. One chord is split into $4$ and $6$; the other into $3$ and $x$. Using $4 \cdot 6 = 3 \cdot x$, find $x = \,?$ 2) A tangent of length $t$ and a secant from the same external point: the secant's whole length is $16$ and its external part is $4$. Using $t^2 = 16 \cdot 4$, find $t = \,?$ 3) Two chords cross: one chord splits into $8$ and $3$; the other into $6$ and $y$. Find $y = \,?$ (decimal is fine)

Quick Reference — The Whole Lesson

⚠️ Always ask first: Where is the vertex? That decides the angle formula every single time.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.