Contents / Geometry / Circles & Polygons

Chapter 3

Circles & Polygons

Circles, radii and diameters, polygons, inscribed angles, and area.

Introduction

Circles, radii and diameters, polygons, inscribed angles, and area.

A circle is the set of all points in a plane that are at a fixed distance (the radius) from a given center point.

It is the most symmetric of all plane figures and appears everywhere from wheels to orbits to the definition of π.

3.1Definition and Basic Terminology

Definition 3.1 (Circle). A circle with center $O$ and radius $r$ is the set of points of the plane at distance $r$ from $O$ . Its diameter $d = 2 r$ is the longest chord, the one passing through the center. A chord is a segment joining two points on the circle, and an arc is a portion of the circle between two points.

Theorem 3.2 (Circumference and area). A circle of radius $r$ has

C = 2 π r, A = π r^{2} .

Intuition. A circle is every point exactly r steps away from the center, like a perfect round fence with the center as the post.

The diameter is the widest straight cut through the middle; twice as wide as from center to edge.

Example 3.3. A circle with center $(0, 0)$ and radius $5$ consists of all $(x, y)$ satisfying $x^{2} + y^{2} = 25$ .

Solution.

test a point: substitute $(3, 4)$ : $3^{2} + 4^{2} = 9 + 16 = 25$ . It satisfies the equation, so $(3, 4)$ lies on the circle.
locate the diameter: the diameter along the $x$ -axis runs from $(- 5, 0)$ to $(5, 0)$ , length $10 = 2 r$ .
Sanity check: the distance from the center to $(3, 4)$ is $3^{2} + 4^{2} = 25 = 5 = r$ , confirming the point is exactly one radius away. Good.

□

Exercises for §3.1 →

3.2Chords, Arcs, and Central Angles

Definition 3.4 (Central angle). A central angle is formed by two radii. The measure of the arc it cuts off equals the measure of the central angle.

Proposition 3.5 (Chords and the angles they subtend). In a given circle, longer chords subtend larger central angles, and equal chords subtend equal arcs and equal central angles.

Intuition. The "slice" of pie from the center (central angle) tells you how big the crust arc is.

A bigger slice has a longer crust piece and a longer straight chord across the top of the slice.

Example 3.6. In a circle of radius $10$ , a central angle of $60°$ subtends an arc of $60°$ , which is $\frac{60}{360} = \frac{1}{6}$ of the full circle.

Solution.

arc length: the circumference is $C = 2 π r = 2 π (10) = 20 π$ , so the arc has length $\frac{1}{6} (20 π) = \frac{10 π}{3} \approx 10.47$ .
chord length: drop a perpendicular from the center $O$ to the chord. This splits the isosceles triangle (two radii of length $10$ plus the chord) into two right triangles, each with a half-angle of $\frac{60°}{2} = 30°$ at $O$ . The half-chord is the opposite side, so $half-chord = 10 sin (30°) = 10 \cdot \frac{1}{2} = 5$ , giving a full chord of $2 (5) = 10$ .
Sanity check: the two radii ( $10$ and $10$ ) and the chord ( $10$ ) form an equilateral triangle, which matches the $60°$ vertex angle. The chord ( $10$ ) is a little shorter than its arc ( $\approx 10.47$ ), as it must be, since the straight path is shorter than the curved one. Both good.

□

Exercises for §3.2 →

3.3Inscribed Angles and Angles in Semicircles

Definition 3.7 (Inscribed angle). An inscribed angle has its vertex on the circle and its sides passing through two other points of the circle.

Theorem 3.8 (Inscribed angle theorem). An inscribed angle measures half the intercepted arc, and therefore half the central angle standing on the same arc.

Corollary 3.9 (Angle in a semicircle). An angle inscribed in a semicircle is a right angle (Thales' theorem).

Intuition. An angle sitting on the edge of the circle "sees" only half as much of the arc as the center does.

If the arc is exactly a semicircle (180°), the inscribed angle is 90° — that's why a diameter makes a right triangle with any point on the circle.

Example 3.10. Arc $A B$ measures $80°$ (that is, the central angle $∠ A OB = 80°$ ).

Solution.

An inscribed angle $∠ A PB$ , with vertex $P$ anywhere on the major arc (the part of the circle not between $A$ and $B$ ), intercepts arc $A B$ .
By the inscribed angle theorem, $∠ A PB = \frac{1}{2} (arc A B) = \frac{1}{2} (80°) = 40°$ .
Sanity check: the inscribed angle ( $40°$ ) is exactly half the central angle ( $80°$ ) for the same arc, as expected. As a second case, if arc $A B$ were $180°$ (a semicircle), the inscribed angle would be $\frac{1}{2} (180°) = 90°$ — Thales' theorem falls right out.

□

Exercises for §3.3 →

Common Mistakes

Confusing central and inscribed angles: Inscribed is half the central for the same arc.
Thinking the diameter is any chord through the center: Only the one passing through center is diameter; others are just long chords.
Forgetting that all radii are equal: This is what makes the circle so symmetric and allows so many isosceles triangles inside it.
Mixing arc measure (degrees) with arc length (units of length): Arc measure is angle; length is (theta/360)*2πr.

The radius of a circle or sphere is the distance from its center to any point on its circumference (or surface). It is usually denoted r and is the fundamental length parameter of the circle.

All radii of a given circle are equal, and the radius determines every other measurement of the circle.

3.4Definition and Properties

Definition 3.11 (Radius). The radius $r$ is half the diameter: the set of points at distance $r$ from the center $O$ is the circle, and any segment from the center to the circumference is a radius, of length $r$ .

Because all radii are equal, triangles formed by two radii and a chord are isosceles.

Intuition. The radius is the "spoke" length from the axle (center) to the rim. Every spoke is the same.

Once you fix r, the whole circle is fixed; you cannot have a bigger or smaller circle with the same r.

Example 3.12. Circle with center $(2, 3)$ and radius $4$ .

Solution.

an axis-aligned point: $(2 + 4, 3) = (6, 3)$ is a distance $(6 - 2)^{2} + (3 - 3)^{2} = 16 = 4$ from the center, so it is on the circle. Likewise $(2, 3 + 4) = (2, 7)$ is on it.
an off-axis point: is $(2, 3) + (\frac{12}{5}, \frac{16}{5}) = (\frac{22}{5}, \frac{31}{5})$ on the circle? Its distance from the center is $(\frac{12}{5})^{2} + (\frac{16}{5})^{2} = \frac{144 + 256}{25} = \frac{400}{25} = 16 = 4$ . Yes.
Sanity check: every point we found sits exactly $r = 4$ from $(2, 3)$ , which is the defining property — any point at distance exactly $4$ from the center lies on the circle, and no others do. Good.

□

Exercises for §3.4 →

3.5Radius in Formulas

Circumference $C = 2 π r$ . Area $A = π r^{2}$ . In a unit circle $r = 1$ , many trig identities simplify.

Theorem 3.13 (Chord length). The chord subtending a central angle $2 θ$ has length

2 r sin θ .

Proof. Drop a perpendicular from the center to the chord. It bisects both the chord and the central angle, creating a right triangle in which the half-chord is the side opposite the half-angle $θ$ , of length $r sin θ$ . The full chord is twice that.∎

Intuition. Circumference is "unrolled" length around the circle: like the string you wrap exactly once.

Area is "how much carpet" inside the circle; it grows with the square of the radius (bigger r means much more area).

Example 3.14. Take $r = 5$ .

Solution.

circumference: $C = 2 π r = 2 π (5) = 10 π \approx 31.4$ .
area: $A = π r^{2} = π (5)^{2} = 25 π \approx 78.5$ .
chord for a $60°$ central angle: here $2 θ = 60°$ , so $θ = 30°$ , and the chord is $2 r sin θ = 2 (5) sin (30°) = 10 \cdot \frac{1}{2} = 5$ .
Sanity check: the chord ( $5$ ) equals the radius ( $5$ ), exactly what we expect for a $60°$ central angle (the equilateral-triangle case). Good.

□

Exercises for §3.5 →

3.6Radius vs Other Lengths

For a circle, every point on the boundary is the same distance $r$ from the center, so "the distance from center to boundary" is unambiguous — it is always $r$ . For a polygon, though, the boundary is not equidistant from the center, and two different lengths matter.

Definition 3.15 (Circumradius and apothem). The circumradius of a polygon is the distance from its center to a vertex — the farthest boundary points. The apothem is the perpendicular distance from the center to a side — the nearest boundary points, equal to the radius of the inscribed circle.

In 3D the radius of a sphere is the direct analogue of the circle's radius.

Intuition. The radius is the shortest "ruler" from middle to edge. Any other line from center to edge is also a radius, same length.

For a square, the radius of the inscribed circle touches the middle of each side (apothem); the circumradius goes to the corners.

Example 3.16. For a regular hexagon with side length $s$ , find both the circumradius and the apothem.

Solution.

circumradius: connect the center to two adjacent vertices. The central angle is $\frac{360°}{6} = 60°$ , and the two connecting segments are equal (both circumradii), so the triangle is isosceles with a $60°$ apex — hence equilateral. Its third side is a hexagon side of length $s$ , so the circumradius $R = s$ .
apothem: the apothem $a$ drops from the center perpendicular to the midpoint of a side, splitting that equilateral triangle in half. It is the height of an equilateral triangle of side $s$ , so $a = \frac{3}{2} s \approx 0.866 s$ .
Sanity check: the apothem ( $\approx 0.866 s$ ) is shorter than the circumradius ( $s$ ), as it must be — the nearest boundary point (side midpoint) is closer than the farthest (a vertex). Good.

□

Exercises for §3.6 →

Common Mistakes

Using diameter when radius is needed in formulas: Many formulas (area, $C = 2 π r$ ) explicitly use $r$ ; doubling by mistake is common.
Thinking radius can vary around the circle: In a true circle it cannot; varying distance means it is not a circle (ellipse, etc.).
Confusing radius of circle with radius of curvature: For more complex curves the local radius of curvature can change.
Forgetting r=0 is a degenerate point-circle: A circle of radius 0 is just its center point.

The diameter of a circle is a chord that passes through the center. It is the longest possible chord and its length is twice the radius (d = 2r).

Any angle inscribed in a semicircle (with the diameter as one side) is a right angle.

3.7Definition and Properties

Definition 3.17 (Diameter). The diameter is a chord through the center,

d = 2 r,

and it is the maximum distance between any two points of the circle. The center is the midpoint of every diameter.

A diameter divides the circle into two equal semicircles.

Intuition. The diameter is the straight-line distance from one side of the circle, through the middle, to the opposite side — the widest you can go.

Cut a round pizza through the center; the cut is a diameter and you get two equal halves.

Example 3.18. Circle center (0,0), r=5. One diameter runs from (−5,0) to (5,0), length 10. Another from (0,−5) to (0,5). The center (0,0) is exactly in the middle of each.

Exercises for §3.7 →

3.8Diameter and Right Angles (Thales)

Theorem 3.19 (Thales' theorem). The angle subtended by a diameter at any point of the circle is $9 0^{\circ}$ .

This gives a quick way to construct right angles or recognize right triangles inscribed in circles.

Intuition. If you take the diameter as the base of a triangle and put the third corner anywhere on the circle, the corner at the circumference will always be a perfect 90° corner.

This is why you can draw a right triangle by picking any point on a circle with a given hypotenuse as diameter.

Example 3.20. Diameter $A B$ with center $O$ , and a point $C$ anywhere on the circle. Show $∠ A CB = 90°$ .

Solution.

$O A$ , $OB$ , and $OC$ are all radii, so $O A = OB = OC$ . That makes $△ O A C$ and $△ OBC$ both isosceles.
Let $∠ O A C = ∠ OC A = α$ (base angles of isosceles $△ O A C$ ) and $∠ OBC = ∠ OCB = β$ (base angles of isosceles $△ OBC$ ).
The three angles of $△ A BC$ sum to $180°$ . Angle $A = α$ , angle $B = β$ , and angle $C = α + β$ . So $α + β + (α + β) = 180°$ , giving $2 (α + β) = 180°$ and $α + β = 90°$ .
Therefore $∠ A CB = α + β = 90°$ .
Sanity check: this matches the inscribed-angle view — the central angle for the semicircular arc $A B$ is $180°$ , and the inscribed angle at $C$ is half of it, $90°$ . Both routes agree.

□

Exercises for §3.8 →

3.9Diameter in Calculations

Proposition 3.21 (Circumference and area from the diameter).

C = π d, A = \frac{π}{4} d^{2} .

Proof. Substitute $r = \frac{d}{2}$ into $C = 2 π r$ and $A = π r^{2}$ .∎

The longest chord is the diameter; any other chord is shorter.

Intuition. Using d instead of r is convenient when you measure across the whole circle (e.g., with a ruler on a physical wheel).

The formulas are just rewritten with the factor of $2$ absorbed into $π$ or $\frac{1}{4}$ .

Example 3.22. Circle with $d = 10$ (so $r = 5$ ).

Solution.

circumference from $d$ : $C = π d = π (10) = 10 π \approx 31.4$ .
area from $d$ : $A = \frac{π}{4} d^{2} = \frac{π}{4} (10)^{2} = \frac{π}{4} (100) = 25 π \approx 78.5$ .
Sanity check: using $r = 5$ directly, $C = 2 π (5) = 10 π$ and $A = π (5)^{2} = 25 π$ — the $d$ -based and $r$ -based formulas agree. Good.

□

Exercises for §3.9 →

Common Mistakes

Calling any long chord a diameter: Only those passing exactly through the center qualify.
Forgetting Thales gives a right angle only when the hypotenuse is the diameter: If the hypotenuse is a shorter chord the inscribed angle is acute or obtuse accordingly.
Mixing radius and diameter in formulas: $C = 2 π r = π d$ ; using $2 π d$ by mistake doubles the circumference.
Thinking the center is not part of the diameter segment: The diameter segment includes the center as its midpoint.

A polygon is a closed plane figure bounded by a finite chain of straight line segments (sides) meeting at vertices.

Polygons are the basic building blocks for tiling, computer graphics, and much of discrete geometry.

3.10Definition and Classification

Definition 3.23 (Polygon). A polygon with $n$ sides is an $n$ -gon. Simple polygons do not cross themselves; complex ones may. Convex polygons have all interior angles less than $180°$ and no sides that bend back inward. Regular polygons have all sides and all angles equal.

Intuition. A polygon is a closed shape made of straight sides, like a stop sign (octagon) or a yield sign (triangle).

Convex means you can draw a rubber band around it and it touches all vertices; no dents.

Example 3.24. Quadrilateral (4-gon), pentagon (5), hexagon (6), heptagon (7), octagon (8), nonagon (9), decagon (10). A star pentagram is a complex (self-intersecting) polygon.

Exercises for §3.10 →

3.11Interior and Exterior Angles

Theorem 3.25 (Interior angle sum). The interior angles of a simple $n$ -gon sum to

(n - 2) \times 180°.

Proof. Pick one vertex and draw the diagonals to all the others. This cuts the polygon into $n - 2$ triangles, each contributing $180°$ , and every interior angle of the polygon is accounted for exactly once.∎

Theorem 3.26 (Exterior angle sum). The exterior angles of any simple polygon — one at each vertex, all turning the same way — sum to $360°$ , regular or not. For a regular $n$ -gon each of them is therefore

\frac{360°}{n} .

Intuition. As you walk around the polygon, the total turn you make is a full circle (360°). That is the sum of the exterior angles.

Interior angles are what is inside at each corner; for regular they are all equal.

Example 3.27. Find every angle of a regular pentagon ( $n = 5$ ).

Solution.

exterior angle: the five exterior angles are equal and sum to $360°$ , so each is $\frac{360°}{5} = 72°$ .
interior angle: an interior angle and its exterior angle are supplementary (they form a straight line along the extended side), so each interior angle is $180° - 72° = 108°$ .
check the interior sum: $5 \times 108° = 540°$ .
Sanity check: the formula gives $(n - 2) \times 180° = (5 - 2) \times 180° = 3 \times 180° = 540°$ — a match. Good.

□

Exercises for §3.11 →

3.12Convexity and Simple vs Complex

Definition 3.28 (Convexity). A polygon is convex when any line segment between two points inside it stays inside.

Theorem 3.29 (Jordan curve theorem, polygonal case). The boundary of a simple polygon divides the plane into an interior and an exterior region.

Intuition. Convex = no "caves"; you can see every point from every other point inside without the wall getting in the way.

A star is simple? No, self-intersecting; a simple dented shape is concave but still simple (no crossings).

Example 3.30. A rectangle is convex. An arrowhead shape (chevron) is concave (one interior angle >180°). A pentagram crosses itself and is complex.

Exercises for §3.12 →

Common Mistakes

Counting sides vs vertices: An n-gon has n sides and n vertices.
Assuming all polygons are convex or regular: Many are irregular or concave.
Forgetting the (n-2) formula: Quadrilateral 360°, pentagon 540°, etc.
Including self-intersecting when talking 'interior angle sum': The simple formula assumes simple polygons.

A regular polygon is both equilateral (all sides equal) and equiangular (all interior angles equal). It has rotational symmetry of order n and reflection symmetries.

Regular polygons tile the plane in only three cases (triangles, squares, hexagons) and are the faces of the Platonic solids.

3.13Properties

Theorem 3.31 (Measurements of a regular polygon). For a regular $n$ -gon with side length $s$ and circumradius $r$ (center to vertex):

each central angle is $\frac{360°}{n}$ ;
each interior angle is $\frac{( n - 2 )}{n} \times 180°$ ;
the area is

Area = \frac{1}{2} n r^{2} sin (\frac{360°}{n}) = \frac{n s ^{2}}{4 tan ( 180°/ n )} .

Proof. The $n$ radii to the vertices cut the polygon into $n$ congruent isosceles triangles, each with two sides $r$ and included angle $\frac{360°}{n}$ , so each has area $\frac{1}{2} r^{2} sin (\frac{360°}{n})$ . Summing the $n$ of them gives the first expression; rewriting $r$ in terms of the side $s$ gives the second.∎

Intuition. A regular polygon is perfectly "fair"; every corner and every side is identical. You can spin it and it looks the same every 360/n degrees.

The more sides, the closer it looks to a circle.

Example 3.32. Analyze a regular hexagon ( $n = 6$ ) with side length $s$ .

Solution.

central angle: $\frac{360°}{6} = 60°$ .
interior angle: $\frac{( 6 - 2 )}{6} \times 180° = \frac{4}{6} \times 180° = 120°$ .
circumradius: since the central angle is $60°$ and the two bounding segments are equal circumradii, each of the $6$ central triangles is equilateral, so the circumradius $r = s$ .
area via the general formula: $Area = \frac{1}{2} n r^{2} sin (\frac{360°}{n}) = \frac{1}{2} (6) (s^{2}) sin (60°) = 3 s^{2} \cdot \frac{3}{2} = \frac{3 3}{2} s^{2}$ .
Sanity check: six equilateral triangles of side $s$ each have area $\frac{3}{4} s^{2}$ , so the total is $6 \cdot \frac{3}{4} s^{2} = \frac{3 3}{2} s^{2}$ — matches. Good.

□

Exercises for §3.13 →

3.14Construction and Symmetry

Regular polygons with 3, 4, 5, 6 sides are classically constructible with ruler and compass. Gauss showed the 17-gon is constructible; not all $n$ are.

Theorem 3.33 (Symmetry group of a regular polygon). The symmetries of a regular $n$ -gon are the $n$ rotations about its center and the $n$ reflections in its axes: the symmetry group is the dihedral group $D_{n}$ , with $2 n$ elements.

Intuition. You can draw some regular polygons exactly with just a compass and straightedge (equilateral triangle, square, regular hexagon, and some others like pentagon). Others require more advanced tools or are impossible exactly.

The symmetries are all the ways you can turn or flip it and have it land on itself.

Example 3.34. Count the symmetries of a regular pentagon ( $n = 5$ ) and check constructibility.

Solution.

rotations: rotating by any multiple of $\frac{360°}{5} = 72°$ maps the pentagon onto itself. That gives $5$ rotations: $0°, 72°, 144°, 216°, 288°$ (the $0°$ rotation is the identity).
reflections: a regular polygon with an odd number of sides has one reflection axis through each vertex and the midpoint of the opposite side — that is $5$ reflection axes, hence $5$ reflections.
total: $5 + 5 = 10 = 2 n$ symmetries, so the symmetry group is $D_{5}$ with $10$ elements.
constructibility: by the Gauss–Wantzel theorem a regular $n$ -gon is ruler-and-compass constructible exactly when $n$ is a power of $2$ times a product of distinct Fermat primes ( $3, 5, 17, 257, \dots$ ). Since $5$ is a Fermat prime, the regular pentagon is constructible; its classical construction (Euclid, Book IV) relies on building the golden ratio via isosceles triangles with $36°$ – $72°$ – $72°$ angles, then stepping equal chords around a circle.
Sanity check: the pentagon's exterior angle is $72°$ , so five turns of $72°$ close the figure ( $5 \times 72° = 360°$ ) — consistent with the $72°$ rotation being a symmetry. Good.

□

Exercises for §3.14 →

3.15Limits and Approximation to Circle

Theorem 3.35 (Regular polygons exhaust the circle). As $n \to \infty$ , a regular $n$ -gon inscribed in a circle of radius $r$ approaches the circle:

P_{n} \to 2 π r, A_{n} \to π r^{2} .

This is one historical way to define and compute $π$ (Archimedes' method).

Intuition. With more and more sides the polygon gets "rounder" until you cannot see the difference from a true circle.

Archimedes used $96$ -sided polygons to bound $π$ between $3 \frac{10}{71} \approx 3.1408$ and $3 \frac{1}{7} \approx 3.1429$ .

Example 3.36. Watch the perimeter of an inscribed regular $n$ -gon close in on $2 π r$ (take $r = 1$ , so the target is $2 π \approx 6.2832$ ).

Solution. side length: a side subtends a central angle $\frac{360°}{n}$ , so by the chord formula each side has length $2 r sin (\frac{180°}{n})$ , and the perimeter is $P_{n} = 2 n r sin (\frac{180°}{n})$ .

plug in values (with $r = 1$ ):

$n = 6$ : $P_{6} = 12 sin (30°) = 12 (0.5) = 6.0000$ . $n = 12$ : $P_{12} = 24 sin (15°) \approx 24 (0.2588) \approx 6.2117$ . $n = 96$ : $P_{96} = 192 sin (1.875°) \approx 192 (0.03272) \approx 6.2821$ .

area by the "half perimeter times apothem" rule: the apothem is $a_{n} = r cos (\frac{180°}{n})$ , which $\to r$ as $n \to \infty$ . Since $P_{n} \to 2 π r$ , the area $\frac{1}{2} P_{n} a_{n} \to \frac{1}{2} (2 π r) (r) = π r^{2}$ .

Sanity check: the perimeters rise $6.0000 < 6.2117 < 6.2821 < 6.2832 \approx 2 π$ , staying just below the circle (an inscribed polygon is shorter than the curve it sits inside) and converging to it. Good.□

Exercises for §3.15 →

Common Mistakes

Calling a regular polygon 'equilateral' only: It must also have equal angles; a rhombus is equilateral but not regular unless it is a square.
Confusing radius and apothem: Radius goes to a vertex; apothem is perpendicular distance to a side (inradius).
Assuming all regular polygons tile the plane: Only 3,4,6 do; others leave gaps or overlap.
Forgetting that constructibility is rare: Most regular n-gons are not ruler-and-compass constructible.

An inscribed angle is an angle whose vertex lies on the circle and whose sides (chords) pass through two other points on the circle.

Its measure is half the measure of the intercepted arc.

3.16Definition and Measure

Definition 3.37 (Intercepted arc). An inscribed angle intercepts the arc lying between its two sides.

Theorem 3.38 (Inscribed angle theorem). An inscribed angle is half its intercepted arc,

∠ A PB = \frac{1}{2} (arc A B),

which is half the central angle subtending the same arc.

Intuition. The angle sitting on the rim of the circle sees only half the arc that the center sees.

The farther from the center, the "smaller" the angle looks for the same arc.

Example 3.39. Arc $A B = 120°$ (central angle $∠ A OB = 120°$ ). Find the inscribed angle and confirm it does not depend on where the vertex sits.

Solution.

Place the vertex $P$ anywhere on the major arc. The inscribed angle $∠ A PB$ intercepts the same arc $A B$ .
Applying the inscribed angle theorem, $∠ A PB = \frac{1}{2} (arc A B) = \frac{1}{2} (120°) = 60°$ .
Move the vertex to a different point $Q$ on the major arc. It still intercepts arc $A B$ , so $∠ A QB = \frac{1}{2} (120°) = 60°$ as well — the measure is fixed by the arc, not the vertex position.
Sanity check: $60°$ is exactly half the central angle $120°$ for the same arc, as the theorem requires. Good.

□

Exercises for §3.16 →

3.17Inscribed Angle in Semicircle

Corollary 3.40 (Thales' theorem). When the intercepted arc is a semicircle ( $180°$ ), the inscribed angle is $9 0^{\circ}$ .

This gives a construction for right angles and shows that any triangle with its hypotenuse as diameter is right-angled at the circumference.

Intuition. Put the hypotenuse as the diameter; no matter where you put the third corner on the circle, the angle at that corner is a perfect right angle.

This is a quick way to draw or recognize right triangles.

Example 3.41. Diameter $A B$ , point $C$ on the circle. Show $∠ A CB = 90°$ , then use it.

Solution.

The diameter splits the circle into two semicircular arcs. The arc $A B$ not containing $C$ measures $180°$ .
$∠ A CB$ is inscribed and intercepts that $180°$ arc, so $∠ A CB = \frac{1}{2} (180°) = 90°$ .
application: if $A B = 10$ and $A C = 6$ , then since $∠ A CB = 90°$ , triangle $A CB$ is right-angled at $C$ , and $BC = A B^{2} - A C^{2} = 100 - 36 = 64 = 8$ .
Sanity check: $6$ – $8$ – $10$ is a scaled $3$ – $4$ – $5$ right triangle, and $6^{2} + 8^{2} = 36 + 64 = 100 = 1 0^{2}$ confirms the right angle at $C$ . Good.

□

Exercises for §3.17 →

Common Mistakes

Using the wrong arc: The inscribed angle intercepts the arc that lies in its interior (the one "seen" by the angle), not the reflex arc.
Confusing with central: Always half for the same arc.
Forgetting the vertex must be on the circle: If the vertex is inside or outside, the angle/arc relations differ (power of point, etc.).
Assuming all inscribed angles on the same side are equal: They are, for the same arc; different arcs give different measures.

The circumscribed circle (or circumcircle) of a polygon is the unique circle that passes through all the vertices of the polygon.

Its center is the circumcenter; its radius is the circumradius.

3.18Definition and Existence

Definition 3.42 (Circumcircle, circumcenter, circumradius). The circumcircle of a polygon is a circle through all of its vertices; its center is the circumcenter and its radius the circumradius.

Theorem 3.43 (Every triangle has a circumcircle). For a triangle the circumcircle always exists and is unique, and its center is the point where the perpendicular bisectors of the three sides meet.

For a general polygon, a circumcircle exists only if the polygon is cyclic — that is, only if its vertices lie on a common circle.

Intuition. The circumcircle is the "round fence" that just touches all the corners of your polygon.

For a triangle you can always find such a circle; for a random quadrilateral the four corners may not lie on one circle.

Example 3.44. Find the circumcircle of $A = (0, 0)$ , $B = (4, 0)$ , $C = (0, 6)$ .

Solution.

perpendicular bisector of $A B$ : $A B$ runs along the $x$ -axis with midpoint $(2, 0)$ ; its perpendicular bisector is the vertical line $x = 2$ .
perpendicular bisector of $A C$ : $A C$ runs along the $y$ -axis with midpoint $(0, 3)$ ; its perpendicular bisector is the horizontal line $y = 3$ .
circumcenter: the two bisectors meet at $(2, 3)$ .
circumradius: distance from $(2, 3)$ to $A = (0, 0)$ is $2^{2} + 3^{2} = 13$ .
Sanity check: distance to $B = (4, 0)$ is $(4 - 2)^{2} + (0 - 3)^{2} = 4 + 9 = 13$ , and to $C = (0, 6)$ is $(0 - 2)^{2} + (6 - 3)^{2} = 4 + 9 = 13$ — all three vertices are equidistant, so $(2, 3)$ is indeed the circumcenter. (Note $∠ A = 90°$ , so by Thales the hypotenuse $BC$ is a diameter; its midpoint $(\frac{4 + 0}{2}, \frac{0 + 6}{2}) = (2, 3)$ is the center, confirming the result.)

□

Exercises for §3.18 →

3.19Properties

Proposition 3.45 (Where the circumcenter sits). In a right triangle the circumcenter is the midpoint of the hypotenuse, so the circumradius is half the hypotenuse. In an acute triangle the circumcenter lies inside the triangle; in an obtuse triangle, outside it.

Theorem 3.46 (Extended law of sines). For every side–angle pair of a triangle,

\frac{a}{sin A} = 2 R,

where $R$ is the circumradius and $a$ is the side opposite the angle $A$ .

Intuition. In a right triangle the hypotenuse is the diameter of the circumcircle (Thales again).

The bigger the triangle relative to its angles, the larger R.

Example 3.47. Right triangle with legs $3$ and $4$ , hypotenuse $5$ .

Solution.

circumradius: for a right triangle the hypotenuse is a diameter (Thales), so $R = \frac{hypotenuse}{2} = \frac{5}{2} = 2.5$ , and the circumcenter is the midpoint of the hypotenuse.
cross-check with the law of sines: the right angle ( $90°$ ) is opposite the hypotenuse ( $5$ ), so $2 R = \frac{5}{s i n 90°} = \frac{5}{1} = 5$ , giving $R = 2.5$ .
check another side–angle pair: the side of length $3$ is opposite an angle $A$ with $sin A = \frac{3}{5} = 0.6$ (opposite over hypotenuse). Then $2 R = \frac{3}{0.6} = 5$ , so $R = 2.5$ again.
Sanity check: all three methods give $R = 2.5$ , and each vertex is exactly $2.5$ from the hypotenuse's midpoint. Good.

□

Exercises for §3.19 →

3.20For Regular Polygons and Cyclic Quads

Every regular polygon has a circumcircle, by symmetry.

Theorem 3.48 (Cyclic quadrilaterals). A quadrilateral has a circumcircle if and only if its opposite angles sum to $180°$ . Such a quadrilateral is called cyclic.

Intuition. Regular polygons are perfectly symmetric so their vertices lie on a circle centered at the polygon's center.

For a quad to fit on a circle, the opposite corners must "add up" to a straight angle in a certain way.

Example 3.49. Test which quadrilaterals are cyclic using the "opposite angles sum to $180°$ " rule.

Solution.

square: every angle is $90°$ , so each pair of opposite angles sums to $90° + 90° = 180°$ . Cyclic — its circumcircle has the diagonal as diameter.
non-square rectangle: again every angle is $90°$ , so opposite angles sum to $180°$ . Cyclic — the equal diagonals meet at the center, which is the circumcenter.
a general (non-rectangular) parallelogram: opposite angles are equal, say $70°$ and $70°$ . Their sum is $70° + 70° = 140° \neq = 180°$ , so it is not cyclic. A parallelogram is cyclic only when each angle is $90°$ — i.e., when it is a rectangle.
Sanity check: in the square/rectangle cases the two adjacent angles at the ends of a diagonal already show the four vertices lie on one circle (Thales, since each angle is $90°$ subtending the diagonal). Consistent. Good.

□

Exercises for §3.20 →

Common Mistakes

Confusing circumcircle with incircle: Circum passes through vertices (outside or through); in touches the sides (inside).
Assuming every quadrilateral has a circumcircle: Only cyclic ones do.
Forgetting the circumcenter may lie outside the triangle: In obtuse triangles it does.
Mixing R with inradius r: R is usually larger.

The area of a circle of radius r is $A = π r^{2}$ .

This formula can be derived by dividing the circle into sectors and rearranging them into a rectangle-like shape whose dimensions involve $π r$ and $r$ .

3.21Derivation by Sectors

Theorem 3.50 (Area of a circle). A circle of radius $r$ has area

A = π r^{2} .

Proof. Divide the circle into $n$ thin sectors, each with angle $\frac{360°}{n}$ , arc length $\frac{2 π r}{n}$ and "height" $r$ . Rearranged by alternating them point-up and point-down, the sectors interlock into a shape that approaches a rectangle of width $π r$ and height $r$ as $n \to \infty$ : the width is $π r$ because half the sectors contribute their arcs to the top edge and half to the bottom, so each long edge totals half the circumference, $\frac{1}{2} (2 π r) = π r$ . Hence the area is $π r \times r = π r^{2}$ .∎

Intuition. Cut the pizza into many skinny slices and rearrange them into a rectangle that is half the circumference wide and radius tall.

The wiggly edges become straight in the limit, giving the clean rectangle area.

Example 3.51. Apply the rearrangement to $r = 5$ .

Solution.

rectangle width: half the circumference is $\frac{1}{2} (2 π r) = π r = 5 π$ .
rectangle height: the radius, $r = 5$ .
area: $width \times height = 5 π \times 5 = 25 π \approx 78.5$ .
Sanity check: the direct formula gives $A = π r^{2} = π (5)^{2} = 25 π$ — a match. (With only $8$ sectors the rearranged edge is still visibly wavy, so the estimate is rough; the equality becomes exact only in the limit $n \to \infty$ .)

□

Exercises for §3.21 →

3.22Other Derivations

Proof. By integral calculus, slicing the disc into vertical strips of height $2 r^{2} - x^{2}$ at each $x$ ,

A = \int_{- r}^{r} 2 r^{2} - x^{2} d x = π r^{2} .

∎

Using Green's theorem or polar coordinates also yields the formula.

Intuition. You can add up the areas of infinitely many thin rings or use trig to integrate the circle equation.

All methods agree on $π r^{2}$ .

Example 3.52. Derive the area in polar coordinates.

Solution.

set up the sector-sweep integral: the area swept by a radius as $θ$ goes from $0$ to $2 π$ is $A = \frac{1}{2} \int_{0}^{2 π} r^{2} d θ$ , where $r$ is constant for a circle.
integrate: since $r^{2}$ is constant, $A = \frac{1}{2} r^{2} \int_{0}^{2 π} d θ = \frac{1}{2} r^{2} (2 π) = π r^{2}$ .
Sanity check: this matches both the sector-rearrangement result and $\int_{- r}^{r} 2 r^{2} - x^{2} d x = π r^{2}$ . All three methods agree. Good.

□

Exercises for §3.22 →

3.23Applications and Scaling

Proposition 3.53 (Quadratic scaling of area). Area scales with the square of the linear dimensions:

A (k r) = π (k r)^{2} = k^{2} A (r),

so doubling the radius quadruples the area.

Used in physics (pressure on a circular piston, flux through circular aperture), probability (normal distribution in 2D), and design.

Intuition. A circle twice as wide has four times as much 'stuff' inside (area).

This quadratic scaling is why large pizzas are better value per crust (less crust per area).

Example 3.54. Show how area responds to scaling the radius.

Solution.

compute a few areas: $r = 1 \Rightarrow A = π (1)^{2} = π$ ; $r = 2 \Rightarrow A = π (2)^{2} = 4 π$ ; $r = 10 \Rightarrow A = π (10)^{2} = 100 π$ .
read off the factors: going from $r = 1$ to $r = 2$ (radius $\times 2$ ) multiplies area by $\frac{4 π}{π} = 4 = 2^{2}$ ; going from $r = 1$ to $r = 10$ (radius $\times 10$ ) multiplies area by $\frac{100 π}{π} = 100 = 1 0^{2}$ .
general rule: scaling the radius by a factor $k$ scales area by $k^{2}$ , since $A (k r) = π (k r)^{2} = k^{2} π r^{2} = k^{2} A (r)$ .
Sanity check: the observed factors $4$ and $100$ equal $2^{2}$ and $1 0^{2}$ , matching the $k^{2}$ law. Good.

□

Exercises for §3.23 →

Common Mistakes

Using $2 π r$ for area: That is circumference. Area is $π r^{2}$ .
Forgetting the square: $π r$ is not area.
Using diameter instead of radius without adjusting: $A = \frac{π}{4} d^{2}$ is correct but easy to forget the $\frac{1}{4}$ .
Confusing with sphere surface/volume: Sphere surface $4 π r^{2}$ , volume $\frac{4}{3} π r^{3}$ .

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter