9  Circles

A circle is a round, flat shape where every point on the edge is the same distance from the center. That distance is the radius, labeled \(r\). The diameter \(d\) is the distance across the circle through the center: \(d = 2r.\)

A labeled circle showing the radius r from the center to the upper-right edge, and the diameter d as a horizontal arrow spanning the full width through the center

We have two ways to measure the size of a circle.

Circumference is measured in units of length (feet, meters). Area is measured in square units (square feet, square meters).

Both formulas involve a special number called \(\pi\) (the Greek letter “pi”).

The number \(\pi\)

The ratio of circumference to diameter is the same for every circle, no matter the size. That ratio is \(\pi\): \[\pi = \dfrac{C}{d} \approx 3.14159\ldots\]

\(\pi\) is not an exact fraction — its decimal expansion continues forever without repeating. For calculations, use the \(\pi\) button on your calculator. For rough estimates by hand, \(\pi \approx 3.14\) is sufficient.

9.1 Circumference

Since \(\pi = C/d\) and \(d = 2r\), we can rearrange to get the circumference formula: \[C = \pi d = 2\pi r.\]

Example 9.1 The fountain at Taos Plaza is circular with a radius of 5 feet. What is the circumference of the fountain?

\[C = 2\pi(5) = 10\pi \approx 31.4 \text{ feet.}\]

Example 9.2 (Activity: Edging two garden beds) The UNM-Taos community garden has two circular planting beds.

  • Bed A has a radius of 4 feet.
  • Bed B has a diameter of 14 feet.

Each bed will be bordered with a metal edging strip.

  1. How many feet of edging does Bed A require?
  1. What is the radius of Bed B?
  1. How many feet of edging does Bed B require?

  1. \(C = 2\pi(4) = 8\pi \approx 25.1\) feet.

  2. \(r = 14 \div 2 = 7\) feet.

  3. \(C = 2\pi(7) = 14\pi \approx 44.0\) feet.

9.2 Area

The area of a circle is: \[A = \pi r^2.\]

Example 9.3 The county is considering installing a fountain in Taos Plaza with a radius of 5 feet. What will be the area of the fountain’s surface?

\[A = \pi(5)^2 = 25\pi \approx 78.5 \text{ square feet.}\]

Example 9.4 (Activity: Filling two garden beds) Recall the two planting beds from Example 9.2.

  1. How many square feet of soil are needed for Bed A?
  1. How many square feet of soil are needed for Bed B?
  1. What is the total area of both beds combined?

  1. \(A = \pi(4)^2 = 16\pi \approx 50.3\) square feet.

  2. \(A = \pi(7)^2 = 49\pi \approx 153.9\) square feet.

  3. \(50.3 + 153.9 = 204.2\) square feet total.

9.3 Practice: circumference and area

Example 9.5 (Activity: Computing circumference and area) Complete the table. Round all answers to two decimal places. For the last two rows, find \(r\) from the given diameter first.

\(r\) \(d = 2r\) \(C = 2\pi r\) \(A = \pi r^2\)
3 ft
5 ft
8 ft
12 ft
9 ft
\(r\) \(d\) \(C\) \(A\)
3 ft 6 ft 18.85 ft 28.27 sq ft
5 ft 10 ft 31.42 ft 78.54 sq ft
8 ft 16 ft 50.27 ft 201.06 sq ft
6 ft 12 ft 37.70 ft 113.10 sq ft
4.5 ft 9 ft 28.27 ft 63.62 sq ft

9.4 Finding a missing dimension

So far we have computed \(C\) and \(A\) when the radius is known. We can also reverse this: if the circumference or area is given, we can solve for the radius.

Finding \(r\) from circumference. Start from \(C = 2\pi r\) and solve for \(r\).

Example 9.6 A center-pivot irrigation system near Taos traces a circular path with a circumference of approximately 628 feet. How long is the pivot arm (the radius of the circle)?

\[628 = 2\pi r \Rightarrow r = \dfrac{628}{2\pi} \approx 100 \text{ feet.}\]

Finding \(r\) from area. Start from \(A = \pi r^2\) and solve for \(r\). The first step is to divide both sides by \(\pi\), leaving \(r^2\) alone. Then take the square root of both sides.

Example 9.7 A circular rug at a Taos gallery has an area of approximately 28.3 square feet. What is the radius of the rug?

\[28.3 = \pi r^2 \Rightarrow r^2 = \dfrac{28.3}{\pi} \approx 9 \Rightarrow r = \sqrt{9} = 3 \text{ feet.}\]

Example 9.8 (Activity: Circle problems) For each problem, set up an equation and solve. Round to two decimal places where needed.

  1. A circular hot tub at a Taos resort has a radius of 3.5 feet. What is the area of the cover needed?
  1. A decorative porthole window at a Santa Fe inn is circular with a diameter of 2 feet. What is the circumference of the window frame?
  1. A circular reflecting pool at the Millicent Rogers Museum has a circumference of approximately 47.1 feet. What is the radius of the pool?
  1. A circular dining table at a Taos restaurant has a surface area of approximately 12.57 square feet. What is the radius of the table?

  1. \(A = \pi(3.5)^2 = 12.25\pi \approx 38.48\) square feet.

  2. \(r = 2 \div 2 = 1\) foot. \(C = 2\pi(1) = 2\pi \approx 6.28\) feet.

  3. \(47.1 = 2\pi r \Rightarrow r = \dfrac{47.1}{2\pi} \approx 7.5\) feet.

  4. \(12.57 = \pi r^2 \Rightarrow r^2 = \dfrac{12.57}{\pi} \approx 4 \Rightarrow r = \sqrt{4} = 2\) feet.

Example 9.9 (Activity: Missing dimensions) Complete the table. Round all answers to two decimal places.

\(r\) \(d = 2r\) \(C = 2\pi r\) \(A = \pi r^2\)
25.13 ft
113.10 sq ft
62.83 ft
78.54 sq ft

Row 1: \(r = 25.13 \div (2\pi) \approx 4\) ft. Then \(d = 8\) ft, \(A = \pi(4)^2 \approx 50.27\) sq ft.

Row 2: \(r^2 = 113.10 \div \pi \approx 36\), so \(r = 6\) ft. Then \(d = 12\) ft, \(C = 2\pi(6) \approx 37.70\) ft.

Row 3: \(r = 62.83 \div (2\pi) \approx 10\) ft. Then \(d = 20\) ft, \(A = \pi(10)^2 \approx 314.16\) sq ft.

Row 4: \(r^2 = 78.54 \div \pi \approx 25\), so \(r = 5\) ft. Then \(d = 10\) ft, \(C = 2\pi(5) \approx 31.42\) ft.

9.5 Homework exercises

Exercise 9.1 For each problem, set up an equation and solve. Round to two decimal places where needed.

  1. A circular swimming pool at a Taos resort has a diameter of 18 feet. What is the circumference of the pool? What is the area of the pool’s surface?
  1. The observation area at a Taos Mountain overlook is a circular deck with a radius of 12 feet. What is the area of the deck?
  1. A circular hay bale has a circumference of approximately 94.2 inches. What is the diameter of the bale?
  1. A circular stained-glass window at a church in Ranchos de Taos has an area of approximately 176.7 square inches. What is the radius of the window? What is the diameter?

  1. \(r = 18 \div 2 = 9\) ft. \(C = \pi(18) = 18\pi \approx 56.55\) feet. \(A = \pi(9)^2 = 81\pi \approx 254.47\) square feet.

  2. \(A = \pi(12)^2 = 144\pi \approx 452.39\) square feet.

  3. \(94.2 = 2\pi r \Rightarrow r = \dfrac{94.2}{2\pi} \approx 15\) inches, so \(d = 2(15) = 30\) inches.

  4. \(r^2 = 176.7 \div \pi \approx 56.25\), so \(r = \sqrt{56.25} = 7.5\) inches. Diameter \(d = 15\) inches.

Exercise 9.2 Complete the table. Round all answers to two decimal places. For rows where the diameter is given, find \(r\) first.

\(r\) \(d = 2r\) \(C = 2\pi r\) \(A = \pi r^2\)
4 ft
16 ft
43.98 ft
254.47 sq ft
11 ft
18.85 ft

Row 1: \(d = 8\) ft, \(C = 2\pi(4) \approx 25.13\) ft, \(A = \pi(4)^2 \approx 50.27\) sq ft.

Row 2: \(r = 8\) ft, \(C = 2\pi(8) \approx 50.27\) ft, \(A = \pi(8)^2 \approx 201.06\) sq ft.

Row 3: \(r = 43.98 \div (2\pi) \approx 7\) ft, \(d = 14\) ft, \(A = \pi(7)^2 \approx 153.94\) sq ft.

Row 4: \(r^2 = 254.47 \div \pi \approx 81\), so \(r = 9\) ft, \(d = 18\) ft, \(C = 2\pi(9) \approx 56.55\) ft.

Row 5: \(d = 22\) ft, \(C = 2\pi(11) \approx 69.12\) ft, \(A = \pi(11)^2 \approx 380.13\) sq ft.

Row 6: \(r = 18.85 \div (2\pi) \approx 3\) ft, \(d = 6\) ft, \(A = \pi(3)^2 \approx 28.27\) sq ft.