Circles – SAT Math

📖 What You Need to Know About Circles

Circumference

\(C = 2\pi r\)

Distance around the circle

Area

\(A = \pi r^2\)

Space inside the circle

Diameter

\(d = 2r\)

Distance across the circle

Arc Length

\(s = \frac{\theta}{360^\circ} \times 2\pi r\)

Portion of circumference

Sector Area

\(A = \frac{\theta}{360^\circ} \times \pi r^2\)

Portion of area

Circle Equation

\((x-h)^2 + (y-k)^2 = r^2\)

Center \((h,k)\), radius \(r\)

1. Basic Circle Properties

📌 RULE: Circumference and Area

\[ \text{Circumference} = 2\pi r \]
\[ \text{Area} = \pi r^2 \]
\[ \text{Diameter} = 2r \]

Remember: \(\pi \approx 3.14\) or \(\pi \approx \frac{22}{7}\)

📝 SOLVED EXAMPLE 1 — Circumference and Area

Find the circumference and area of a circle with radius 7 cm. (Use \(\pi \approx 3.14\))

Step 1: \(C = 2\pi r = 2 \times 3.14 \times 7 = \color{var(--math)}{43.96}\) cm

Step 2: \(A = \pi r^2 = 3.14 \times 7^2 = 3.14 \times 49 = \color{var(--math)}{153.86}\) cm²

✅ Circumference = 43.96 cm, Area = 153.86 cm²

💡 Tip: Circumference is linear (cm), area is square (cm²).

2. Arcs and Sectors

📌 RULE: Arc Length

The arc length is the portion of the circumference corresponding to a central angle \(\theta\).

\[ \text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r \]
where \(\theta\) is the central angle in degrees.

📌 RULE: Sector Area

The sector area is the portion of the circle's area corresponding to a central angle \(\theta\).

\[ \text{Sector Area} = \frac{\theta}{360^\circ} \times \pi r^2 \]

💡 Strategy — Arcs and Sectors

1. Identify the central angle \(\theta\).
2. Use the fraction \(\frac{\theta}{360^\circ}\).
3. Multiply by the total circumference or area.

📝 SOLVED EXAMPLE 2 — Arc Length

Find the arc length of a circle with radius 8 cm and central angle \(45^\circ\). (Use \(\pi \approx 3.14\))

Step 1: Arc length = \(\frac{45}{360} \times 2\pi(8)\)

Step 2: \(\frac{1}{8} \times 2 \times 3.14 \times 8 = \frac{1}{8} \times 50.24 = \color{var(--math)}{6.28}\)

✅ Arc length = 6.28 cm

💡 Tip: \(45^\circ\) is \(\frac{1}{8}\) of a full circle (360°).

📝 SOLVED EXAMPLE 3 — Sector Area

Find the sector area of a circle with radius 6 cm and central angle \(60^\circ\). (Use \(\pi \approx 3.14\))

Step 1: Sector area = \(\frac{60}{360} \times \pi(6)^2\)

Step 2: \(\frac{1}{6} \times 3.14 \times 36 = \frac{1}{6} \times 113.04 = \color{var(--math)}{18.84}\)

✅ Sector area = 18.84 cm²

💡 Tip: \(60^\circ\) is \(\frac{1}{6}\) of a full circle.

3. Circle Angles

📌 RULE: Central Angle

A central angle has its vertex at the center of the circle. Its measure equals the measure of its intercepted arc.

📌 RULE: Inscribed Angle

An inscribed angle has its vertex on the circle. Its measure is half the measure of its intercepted arc.

\[ \text{Inscribed Angle} = \frac{1}{2} \times \text{Intercepted Arc} \]

📌 Key Concept

• Central angle = intercepted arc
• Inscribed angle = half of intercepted arc

📝 SOLVED EXAMPLE 4 — Inscribed Angle

An inscribed angle intercepts an arc of \(80^\circ\). What is the measure of the inscribed angle?

Step 1: Inscribed angle = \(\frac{1}{2} \times \text{arc}\)

Step 2: \(\frac{1}{2} \times 80^\circ = \color{var(--math)}{40^\circ}\)

✅ \(40^\circ\)

💡 Tip: Inscribed angle is always half the intercepted arc.

4. Circle Equation (Coordinate Geometry)

\((x - h)^2 + (y - k)^2 = r^2\)

📌 RULE: Circle Equation

The equation of a circle with center \((h, k)\) and radius \(r\) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Example: Center \((2, -3)\), radius \(4\):
\[ (x - 2)^2 + (y + 3)^2 = 16 \]

📝 SOLVED EXAMPLE 5 — Circle Equation

Find the center and radius of the circle: \((x - 3)^2 + (y + 2)^2 = 25\)

Step 1: Compare to \((x - h)^2 + (y - k)^2 = r^2\)

Step 2: \(h = 3\), \(k = -2\) → Center = \(\color{var(--math)}{(3, -2)}\)

Step 3: \(r^2 = 25\) → \(r = \color{var(--math)}{5}\)

✅ Center: \((3, -2)\), Radius: 5

💡 Tip: Remember: \((x - h)^2 + (y - k)^2 = r^2\)

🧪 Practice Questions

Solve each problem using the rules above. Click "Show Answer" to see the full solution.

Question 1

Find the circumference of a circle with radius 5 cm. (Use \(\pi \approx 3.14\))

A) 15.7 cm

B) 25.4 cm

C) 31.4 cm

D) 78.5 cm

✓ Answer: C

\(C = 2\pi r = 2 \times 3.14 \times 5 = 31.4\) cm

📝 Solution: \(2 \times 3.14 \times 5 = 31.4\)

Question 2

Find the area of a circle with radius 6 cm. (Use \(\pi \approx 3.14\))

A) 37.68 cm²

B) 75.36 cm²

C) 100.48 cm²

D) 113.04 cm²

✓ Answer: D

\(A = \pi r^2 = 3.14 \times 6^2 = 3.14 \times 36 = 113.04\) cm²

📝 Solution: \(3.14 \times 36 = 113.04\)

Question 3

Find the diameter of a circle with radius 9 cm.

A) 9 cm

B) 12 cm

C) 18 cm

D) 27 cm

✓ Answer: C

\(d = 2r = 2 \times 9 = 18\) cm

📝 Solution: \(2 \times 9 = 18\)

Question 4

Find the arc length of a circle with radius 10 cm and central angle \(90^\circ\). (Use \(\pi \approx 3.14\))

A) 10.7 cm

B) 15.7 cm

C) 20.4 cm

D) 31.4 cm

✓ Answer: B

Arc length = \(\frac{90}{360} \times 2\pi(10) = \frac{1}{4} \times 62.8 = 15.7\) cm

📝 Solution: \(0.25 \times 62.8 = 15.7\)

Question 5

Find the sector area of a circle with radius 8 cm and central angle \(45^\circ\). (Use \(\pi \approx 3.14\))

A) 12.56 cm²

B) 20.12 cm²

C) 25.12 cm²

D) 30.15 cm²

✓ Answer: C

Sector area = \(\frac{45}{360} \times \pi(8)^2 = \frac{1}{8} \times 3.14 \times 64 = \frac{1}{8} \times 200.96 = 25.12\) cm²

📝 Solution: \(0.125 \times 200.96 = 25.12\)

Question 6

Find the center and radius of the circle: \((x - 1)^2 + (y + 4)^2 = 36\)

A) Center: (1, 4), Radius: 6

B) Center: (-1, -4), Radius: 36

C) Center: (-1, 4), Radius: 6

D) Center: (1, -4), Radius: 6

✓ Answer: D

\((x - 1)^2 + (y + 4)^2 = 36\) → Center: \((1, -4)\), Radius: 6

📝 Solution: \(h = 1\), \(k = -4\), \(r = 6\)

Question 7

An inscribed angle intercepts an arc of \(120^\circ\). What is the measure of the inscribed angle?

A) \(30^\circ\)

B) \(45^\circ\)

C) \(60^\circ\)

D) \(120^\circ\)

✓ Answer: C

Inscribed angle = \(\frac{1}{2} \times 120^\circ = 60^\circ\)

📝 Solution: \(120/2 = 60\)

Question 8

A circle has circumference \(62.8\) cm. Find the radius. (Use \(\pi \approx 3.14\))

A) 5 cm

B) 8 cm

C) 9 cm

D) 10 cm

✓ Answer: D

\(C = 2\pi r\) → \(62.8 = 2 \times 3.14 \times r\) → \(62.8 = 6.28r\) → \(r = 10\) cm

📝 Solution: \(62.8/6.28 = 10\)

Question 9

Find the area of a circle with circumference \(50.24\) cm. (Use \(\pi \approx 3.14\))

A) 100.48 cm²

B) 150.72 cm²

C) 200.96 cm²

D) 301.44 cm²

✓ Answer: C

\(C = 2\pi r\) → \(50.24 = 6.28r\) → \(r = 8\) cm. \(A = \pi r^2 = 3.14 \times 64 = 200.96\) cm²

📝 Solution: \(r = 8\), \(A = 200.96\)

Question 10

Find the sector area of a circle with radius 12 cm and central angle \(30^\circ\). (Use \(\pi \approx 3.14\))

A) 25.12 cm²

B) 37.68 cm²

C) 50.24 cm²

D) 37.68 cm²

✓ Answer: D

Sector area = \(\frac{30}{360} \times \pi(12)^2 = \frac{1}{12} \times 3.14 \times 144 = \frac{1}{12} \times 452.16 = 37.68\) cm²

📝 Solution: \(0.0833 \times 452.16 = 37.68\)

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