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Unit 8: Geometry and Trigonometry

Topic 6

Circles

A circle is the set of all points in a plane that are a fixed distance (the radius, rr) from a fixed point (the center). The diameter is twice the radius: d=2rd = 2r. The circumference (perimeter) of a circle is:
C=2πr=πdC = 2\pi r = \pi d
The area of a circle is:
A=πr2A = \pi r^2
An arc is a portion of the circumference. A sector is the ``pie slice'' region bounded by two radii and the arc between them. The central angle is the angle formed at the center by those two radii. The key idea is that arcs and sectors are proportional to their central angles. If a central angle measures θ\theta degrees, then:
Arc length=θ360⋅2πr,Sector area=θ360⋅πr2\text{Arc length} = \frac{\theta}{360} \cdot 2\pi r, \qquad \text{Sector area} = \frac{\theta}{360} \cdot \pi r^2
Think of θ360\dfrac{\theta}{360} as the fraction of the full circle that the arc or sector represents. A 90∘90^\circ central angle gives 14\dfrac{1}{4} of the circumference and 14\dfrac{1}{4} of the area, a 180∘180^\circ angle gives half, and so on.
An inscribed angle is an angle formed by two chords that meet at a point on the circle. The Inscribed Angle Theorem states that an inscribed angle is exactly half the central angle that subtends (intercepts) the same arc:
Inscribed angle=12⋅central angle\text{Inscribed angle} = \frac{1}{2} \cdot \text{central angle}
A useful special case: any inscribed angle that intercepts a semicircle (an arc of 180∘180^\circ) is a right angle. In other words, if a triangle is inscribed in a circle with one side being a diameter, the angle opposite that diameter is 90∘90^\circ.
The standard form of a circle's equation in the coordinate plane is:
(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2
where (h,k)(h, k) is the center and rr is the radius. Be careful with signs: (x−3)2+(y+5)2=16(x - 3)^2 + (y + 5)^2 = 16 has center (3,−5)(3, -5) and radius 44. The SAT sometimes gives the equation in 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.

Core Skills

  • Calculate arc length and sector area using the central angle as a fraction of 360∘360^\circ.
  • Use the Inscribed Angle Theorem to relate inscribed angles to central angles.
  • Write, interpret, and convert between standard form and general form of a circle equation.
  • Find the center and radius of a circle by completing the square.
  • Solve problems that combine circle properties with other geometry or algebra skills.

Example 1: Arc Length

A circle has a radius of 10 cm. What is the length of an arc intercepted by a central angle of 72∘72^\circ? Step 1: Find the fraction of the circle.
72360=15\frac{72}{360} = \frac{1}{5}
Step 2: Multiply by the full circumference.
Arc length=15⋅2π(10)=20π5=4π\text{Arc length} = \frac{1}{5} \cdot 2\pi(10) = \frac{20\pi}{5} = 4\pi
The arc length is 4π\boxed{4\pi} cm.

Example 2: Sector Area

45°
A pizza has a diameter of 16 inches. A slice is cut with a central angle of 45∘45^\circ. What is the area of the slice? Step 1: The radius is 162=8\dfrac{16}{2} = 8 inches. The fraction of the circle is:
45360=18\frac{45}{360} = \frac{1}{8}
Step 2: Multiply by the full area.
Sector area=18⋅π(8)2=64π8=8π\text{Sector area} = \frac{1}{8} \cdot \pi(8)^2 = \frac{64\pi}{8} = 8\pi
The area of the slice is 8π\boxed{8\pi} square inches.

Example 3: Inscribed Angle

110°55°
A central angle in a circle measures 110∘110^\circ. What is the measure of the inscribed angle that intercepts the same arc? Step 1: The inscribed angle is half the central angle.
Inscribed angle=1102=55∘\text{Inscribed angle} = \frac{110}{2} = 55^\circ
The inscribed angle measures 55\boxed{55} degrees.

Example 4: Standard Form of a Circle

What are the center and radius of the circle (x+2)2+(y−7)2=36(x + 2)^2 + (y - 7)^2 = 36? Step 1: Compare to (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
(x−(−2))2+(y−7)2=62(x - (-2))^2 + (y - 7)^2 = 6^2
The center is (−2,7)(-2, 7) and the radius is 66. Center: (−2,7)\boxed{(-2, 7)}, Radius: 6\boxed{6}

Example 5: Completing the Square

Write the equation x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0 in standard form and identify the center and radius. Step 1: Group xx-terms and yy-terms and move the constant.
(x2−6x)+(y2+4y)=12(x^2 - 6x) + (y^2 + 4y) = 12
Step 2: Complete the square for each group. For xx: half of −6-6 is −3-3, and (−3)2=9(-3)^2 = 9. For yy: half of 44 is 22, and 22=42^2 = 4.
(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
The center is (3,−2)(3, -2) and the radius is 55. Center: (3,−2)\boxed{(3, -2)}, Radius: 5\boxed{5}

Example 6: Finding Arc Length from Inscribed Angle

In a circle with radius 12, an inscribed angle of 30∘30^\circ intercepts an arc. What is the length of that arc? Step 1: The central angle is twice the inscribed angle: 2×30=60∘2 \times 30 = 60^\circ. Step 2: Calculate the arc length.
Arc length=60360⋅2π(12)=16⋅24π=4π\text{Arc length} = \frac{60}{360} \cdot 2\pi(12) = \frac{1}{6} \cdot 24\pi = 4\pi
The arc length is 4π\boxed{4\pi}.

Key Takeaways

  • Arc length and sector area both use the same fraction: θ360\dfrac{\theta}{360}. Multiply that fraction by the circumference for arc length, or by the total area for sector area.
  • An inscribed angle is always half the central angle that intercepts the same arc. If the SAT gives you an inscribed angle and asks for an arc, double the inscribed angle first to get the central angle, then use the central angle in the arc/sector formulas.
  • When completing the square for a circle equation, remember to add the same values to both sides. The most common error is forgetting to add the completing-the-square constant to the right side.
  • In the standard equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, the right side is r2r^2, not rr. If the equation says =49= 49, the radius is 7, not 49.