Circles

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

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Circles on the SAT

Essential Circle Formulas

| Property | Formula | |---|---| | Area | A=πr2A = \pi r^2 | | Circumference | C=2πr=πdC = 2\pi r = \pi d | | Diameter | d=2rd = 2r | | Arc length | L=θ360×2πrL = \frac{\theta}{360} \times 2\pi r | | Sector area | A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2 |


Equation of a Circle

Standard Form

(xh)2+(yk)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+y26x+4y12=0x^2 + y^2 - 6x + 4y - 12 = 0 (x26x+9)+(y2+4y+4)=12+9+4(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 (x3)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 — (x3)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 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=49π153.94 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.

2Problem 2medium

Question:

What is the center and radius of the circle (x+3)2+(y5)2=36(x + 3)^2 + (y - 5)^2 = 36?

💡 Show Solution

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

Compare with (x+3)2+(y5)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 (y5)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.

3Problem 3medium

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=20π62.83A = \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

4Problem 4hard

Question:

Convert to standard form and find the center and radius: x2+y2+8x10y+16=0x^2 + y^2 + 8x - 10y + 16 = 0

💡 Show Solution

Step 1: Group xx terms and yy terms, move constant: (x2+8x)+(y210y)=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 (y210y+25)(y^2 - 10y + 25)

Step 4: Add the same values to the right side: (x2+8x+16)+(y210y+25)=16+16+25(x^2 + 8x + 16) + (y^2 - 10y + 25) = -16 + 16 + 25 (x+4)2+(y5)2=25(x + 4)^2 + (y - 5)^2 = 25

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

5Problem 5expert

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: (x2)2+(y3)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: (2,34)=(2,1)(2, 3-4) = (2, -1). This point is on the line y=1y = -1

Answer: (x2)2+(y3)2=16(x - 2)^2 + (y - 3)^2 = 16