Skip to content

Circles

Understand circle equations, arc length, sector area, central and inscribed angles, tangent lines, and circle theorems tested on the SAT.

Written and reviewed by the Study Mondo Education TeamLast updated
🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Circles on the SAT

Essential Circle Formulas

PropertyFormula
AreaA=πr2A = \pi r^2
CircumferenceC=2πr=πdC = 2\pi r = \pi d
Diameterd=2rd = 2r
Arc lengthL=θ360×2πrL = \frac{\theta}{360} \times 2\pi r
Sector areaA=θ360×πr2A = \frac{\theta}{360} \times \pi r^2

Equation of a Circle

Standard Form

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

  • Center: (h,k)(h, k)
  • Radius: rr

General Form

x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0

To convert to standard form: complete the square for both xx and yy.

Example: x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0 (x2−6x+9)+(y2+4y+4)=12+9+4(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25

Center: (3,−2)(3, -2), Radius: 55

Arc Length and Sector Area

An arc is a portion of the circumference. A sector is the "pie slice" region.

For a central angle of θ\theta degrees:

Arc length=θ360×2πr\text{Arc length} = \frac{\theta}{360} \times 2\pi r

Sector area=θ360×πr2\text{Sector area} = \frac{\theta}{360} \times \pi r^2

Key insight: The fraction θ360\frac{\theta}{360} represents what fraction of the full circle is captured.

Central and Inscribed Angles

  • Central angle: vertex at the center of the circle
  • Inscribed angle: vertex on the circle

An inscribed angle is half the central angle that subtends the same arc.

If a central angle is 80°, the inscribed angle on the same arc is 40°.

Special case: Inscribed angle on a semicircle

If an inscribed angle subtends a diameter (semicircle), the angle is always 90°.

Tangent Lines

A tangent line touches the circle at exactly one point and is perpendicular to the radius at that point.

If a tangent and a radius meet at the point of tangency, they form a 90° angle.

SAT Question Types

Type 1: Find the Area or Circumference

Plug into the formulas. Watch for diameter vs. radius!

Type 2: Equation of a Circle

Given center and radius → write equation, or given equation → find center and radius.

Type 3: Complete the Square

Convert general form to standard form.

Type 4: Arc Length / Sector Area

Use the fraction of the circle based on the central angle.

Type 5: Inscribed/Central Angles

Use the 2:1 relationship between central and inscribed angles.

Common SAT Mistakes

  1. Confusing radius and diameter — the formula uses radius, but the problem may give diameter
  2. Forgetting to square rr in the circle equation — it's r2r^2, not rr
  3. Sign errors in the circle equation — (x−3)2(x-3)^2 means center x=3x = 3 (positive), not −3-3
  4. Using 360 instead of 2π2\pi when the angle is in radians
  5. Not completing the square properly when converting circle equations

📚 Practice Problems

1Problem 1easy

❓ Question:

A circle has a diameter of 12. What is its area?

💡 Show Solution

Solution:

Given: Diameter = 12 Find radius: r=d2=122=6r = \frac{d}{2} = \frac{12}{2} = 6

Area formula: A=πr2A = \pi r^2 A=π(6)2=36πA = \pi(6)^2 = 36\pi

Answer: 36π36\pi

SAT Tip: Always convert diameter to radius first! Area uses radius, not diameter.

2Problem 2medium

❓ Question:

What is the center and radius of the circle (x−4)2+(y+1)2=9(x - 4)^2 + (y + 1)^2 = 9?

💡 Show Solution

Solution:

Standard form: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Compare to given equation: (x−4)2+(y−(−1))2=32(x - 4)^2 + (y - (-1))^2 = 3^2

Center: (h,k)=(4,−1)(h, k) = (4, -1) Radius: r=9=3r = \sqrt{9} = 3

Answer: Center: (4,−1)(4, -1), Radius: 33

SAT Tip: Watch the signs! (y+1)(y + 1) means k=−1k = -1, not +1+1

3Problem 3hard

❓ Question:

A circle has radius 9. A sector of this circle has a central angle of 40°40°. What is the area of the sector?

💡 Show Solution

Solution:

Sector area = (fraction of circle) × (total area)

Fraction: 40°360°=40360=19\frac{40°}{360°} = \frac{40}{360} = \frac{1}{9}

Total area: A=πr2=π(9)2=81πA = \pi r^2 = \pi(9)^2 = 81\pi

Sector area: 19×81π=9π\frac{1}{9} \times 81\pi = 9\pi

Answer: 9π9\pi

SAT Tip: Sector = "pizza slice." Find what fraction of the whole circle it is!

4Problem 4easy

❓ Question:

A circle has a diameter of 14 cm. What is its area?

💡 Show Solution

Step 1: Find the radius: r=d2=142=7r = \frac{d}{2} = \frac{14}{2} = 7 cm

Step 2: Calculate area: A=πr2=π(7)2=49π≈153.94 cm2A = \pi r^2 = \pi(7)^2 = 49\pi \approx 153.94 \text{ cm}^2

Answer: 49π49\pi cm² (approximately 153.94 cm²)

Common mistake: Using the diameter (14) instead of the radius (7) in the formula.

5Problem 5easy

❓ Question:

A circle has a diameter of 14 cm. What is its area?

💡 Show Solution

Step 1: Find the radius: r=d2=142=7r = \frac{d}{2} = \frac{14}{2} = 7 cm

Step 2: Calculate area: A=πr2=π(7)2=49π≈153.94 cm2A = \pi r^2 = \pi(7)^2 = 49\pi \approx 153.94 \text{ cm}^2

Answer: 49π49\pi cm² (approximately 153.94 cm²)

Common mistake: Using the diameter (14) instead of the radius (7) in the formula.

6Problem 6medium

❓ Question:

What is the center and radius of the circle (x+3)2+(y−5)2=36(x + 3)^2 + (y - 5)^2 = 36?

💡 Show Solution

Standard form: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Compare with (x+3)2+(y−5)2=36(x + 3)^2 + (y - 5)^2 = 36:

(x+3)2=(x−(−3))2(x + 3)^2 = (x - (-3))^2, so h=−3h = -3 (y−5)2(y - 5)^2, so k=5k = 5 r2=36r^2 = 36, so r=6r = 6

Center: (−3,5)(-3, 5) Radius: 66

Key: Watch the signs! (x+3)2(x + 3)^2 means the center xx-coordinate is −3-3, not +3+3.

7Problem 7medium

❓ Question:

What is the center and radius of the circle (x+3)2+(y−5)2=36(x + 3)^2 + (y - 5)^2 = 36?

💡 Show Solution

Standard form: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Compare with (x+3)2+(y−5)2=36(x + 3)^2 + (y - 5)^2 = 36:

(x+3)2=(x−(−3))2(x + 3)^2 = (x - (-3))^2, so h=−3h = -3 (y−5)2(y - 5)^2, so k=5k = 5 r2=36r^2 = 36, so r=6r = 6

Center: (−3,5)(-3, 5) Radius: 66

Key: Watch the signs! (x+3)2(x + 3)^2 means the center xx-coordinate is −3-3, not +3+3.

8Problem 8medium

❓ Question:

A sector of a circle with radius 10 has a central angle of 72°. What is the area of the sector?

💡 Show Solution

Sector area formula: A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2

A=72360×π(10)2=15×100π=20π≈62.83A = \frac{72}{360} \times \pi(10)^2 = \frac{1}{5} \times 100\pi = 20\pi \approx 62.83

Answer: 20π20\pi square units (approximately 62.83)

Check: 72° is 15\frac{1}{5} of 360°, so the sector is 15\frac{1}{5} of the full circle area. 15×100π=20π\frac{1}{5} \times 100\pi = 20\pi ✓

9Problem 9medium

❓ Question:

A sector of a circle with radius 10 has a central angle of 72°. What is the area of the sector?

💡 Show Solution

Sector area formula: A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2

A=72360×π(10)2=15×100π=20π≈62.83A = \frac{72}{360} \times \pi(10)^2 = \frac{1}{5} \times 100\pi = 20\pi \approx 62.83

Answer: 20π20\pi square units (approximately 62.83)

Check: 72° is 15\frac{1}{5} of 360°, so the sector is 15\frac{1}{5} of the full circle area. 15×100π=20π\frac{1}{5} \times 100\pi = 20\pi ✓

10Problem 10hard

❓ Question:

Convert to standard form and find the center and radius: x2+y2+8x−10y+16=0x^2 + y^2 + 8x - 10y + 16 = 0

💡 Show Solution

Step 1: Group xx terms and yy terms, move constant: (x2+8x)+(y2−10y)=−16(x^2 + 8x) + (y^2 - 10y) = -16

Step 2: Complete the square for xx: Half of 8 = 4, square it = 16 (x2+8x+16)(x^2 + 8x + 16)

Step 3: Complete the square for yy: Half of −10-10 = −5-5, square it = 25 (y2−10y+25)(y^2 - 10y + 25)

Step 4: Add the same values to the right side: (x2+8x+16)+(y2−10y+25)=−16+16+25(x^2 + 8x + 16) + (y^2 - 10y + 25) = -16 + 16 + 25 (x+4)2+(y−5)2=25(x + 4)^2 + (y - 5)^2 = 25

Answer: Center (−4,5)(-4, 5), Radius =25=5= \sqrt{25} = 5

11Problem 11hard

❓ Question:

Convert to standard form and find the center and radius: x2+y2+8x−10y+16=0x^2 + y^2 + 8x - 10y + 16 = 0

💡 Show Solution

Step 1: Group xx terms and yy terms, move constant: (x2+8x)+(y2−10y)=−16(x^2 + 8x) + (y^2 - 10y) = -16

Step 2: Complete the square for xx: Half of 8 = 4, square it = 16 (x2+8x+16)(x^2 + 8x + 16)

Step 3: Complete the square for yy: Half of −10-10 = −5-5, square it = 25 (y2−10y+25)(y^2 - 10y + 25)

Step 4: Add the same values to the right side: (x2+8x+16)+(y2−10y+25)=−16+16+25(x^2 + 8x + 16) + (y^2 - 10y + 25) = -16 + 16 + 25 (x+4)2+(y−5)2=25(x + 4)^2 + (y - 5)^2 = 25

Answer: Center (−4,5)(-4, 5), Radius =25=5= \sqrt{25} = 5

12Problem 12expert

❓ Question:

A circle with center (2,3)(2, 3) is tangent to the line y=−1y = -1. What is the equation of the circle?

💡 Show Solution

Step 1: "Tangent to the line y=−1y = -1" means the circle touches this horizontal line at exactly one point. The distance from the center to the line equals the radius.

Step 2: The distance from center (2,3)(2, 3) to the line y=−1y = -1 is: r=∣3−(−1)∣=∣4∣=4r = |3 - (-1)| = |4| = 4

Step 3: Write the equation: (x−2)2+(y−3)2=16(x - 2)^2 + (y - 3)^2 = 16

Check: The closest point on the circle to y=−1y = -1 is directly below the center: (2,3−4)=(2,−1)(2, 3-4) = (2, -1). This point is on the line y=−1y = -1 ✓

Answer: (x−2)2+(y−3)2=16(x - 2)^2 + (y - 3)^2 = 16

13Problem 13expert

❓ Question:

A circle with center (2,3)(2, 3) is tangent to the line y=−1y = -1. What is the equation of the circle?

💡 Show Solution

Step 1: "Tangent to the line y=−1y = -1" means the circle touches this horizontal line at exactly one point. The distance from the center to the line equals the radius.

Step 2: The distance from center (2,3)(2, 3) to the line y=−1y = -1 is: r=∣3−(−1)∣=∣4∣=4r = |3 - (-1)| = |4| = 4

Step 3: Write the equation: (x−2)2+(y−3)2=16(x - 2)^2 + (y - 3)^2 = 16

Check: The closest point on the circle to y=−1y = -1 is directly below the center: (2,3−4)=(2,−1)(2, 3-4) = (2, -1). This point is on the line y=−1y = -1 ✓

Answer: (x−2)2+(y−3)2=16(x - 2)^2 + (y - 3)^2 = 16

Explain using:
← Previous Topic
Geometry Basics
🎴

Practice with Flashcards

Rate this topic's cards with spaced repetition. Cards join your deck when you finish a topic's lesson and take its exit quiz.

📖

Browse All Topics

Explore more SAT Prep topics

📌 Related Topics in Geometry and Trigonometry

Geometry and TrigonometryGeometry Basics

❓ Frequently Asked Questions

What is Circles?▾
Understand circle equations, arc length, sector area, central and inscribed angles, tangent lines, and circle theorems tested on the SAT.
How can I study Circles effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 13 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Circles study guide free?▾
Yes — all study notes, flashcards, and practice problems for Circles on Study Mondo are free to access. No account is needed.
What course covers Circles?▾
Circles is part of the SAT Prep course on Study Mondo, specifically in the Geometry and Trigonometry section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Circles?▾
Yes, this page includes 13 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.