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Geometry and Trigonometry - Circles

Grade 12IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The length of an arc ll and the area of a sector AA are proportional to the central angle θ\theta (measured in radians). For a circle with radius rr, the arc length is l=rθl = r\theta and the area is A=12r2θA = \frac{1}{2}r^2\theta.

Diagram showing a circle with a sector highlighted, labeling the radius r, the central angle theta, and the arc length l.
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A chord is a straight line segment whose endpoints both lie on a circle. A chord divides the circle into two segments: the major segment and the minor segment. The area of the minor segment is calculated by subtracting the area of the triangle formed by the chord and the radii from the area of the sector: A=12r2(θ−sin⁡θ)A = \frac{1}{2}r^2(\theta - \sin \theta).

Diagram of a circle with a chord drawn, illustrating the area of the minor segment between the chord and the arc.
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The standard equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) represents the center of the circle and rr is the radius. If the equation is given in expanded form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0, completing the square for xx and yy terms is required to find the center and radius.

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Radians are a measure of angles based on the radius of a circle. One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius. Conversion: 180∘=π radians180^{\circ} = \pi \text{ radians}.

📐Formulae

Angle in radians=Angle in degrees×π180\text{Angle in radians} = \text{Angle in degrees} \times \frac{\pi}{180}

l=rθl = r\theta

A=12r2θA = \frac{1}{2}r^2\theta

Asegment=12r2(θ−sin⁡θ)A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin \theta)

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

💡Examples

Problem 1:

A sector of a circle has a radius of 66 cm and a central angle of 1.51.5 radians. Calculate the arc length and the area of the sector.

Solution:

Using the formula for arc length: l=rθ=6×1.5=9 cml = r\theta = 6 \times 1.5 = 9 \text{ cm} Using the formula for sector area: A=12r2θ=12×62×1.5=12×36×1.5=27 cm2A = \frac{1}{2}r^2\theta = \frac{1}{2} \times 6^2 \times 1.5 = \frac{1}{2} \times 36 \times 1.5 = 27 \text{ cm}^2

Explanation:

Substitute the given values r=6r = 6 and θ=1.5\theta = 1.5 directly into the radian-based formulae for arc length and sector area.

Problem 2:

Find the area of the segment cut off by a chord in a circle of radius 88 cm, where the chord subtends an angle of π3\frac{\pi}{3} at the center.

Solution:

Area of sector: Asector=12×82×π3=64π6=32π3≈33.51 cm2A_{\text{sector}} = \frac{1}{2} \times 8^2 \times \frac{\pi}{3} = \frac{64\pi}{6} = \frac{32\pi}{3} \approx 33.51 \text{ cm}^2 Area of triangle: Atriangle=12×82×sin⁡(π3)=32×32=163≈27.71 cm2A_{\text{triangle}} = \frac{1}{2} \times 8^2 \times \sin\left(\frac{\pi}{3}\right) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.71 \text{ cm}^2 Area of segment: Asegment=33.51−27.71=5.80 cm2A_{\text{segment}} = 33.51 - 27.71 = 5.80 \text{ cm}^2

Explanation:

The segment area is the difference between the sector area and the area of the triangle formed by the radius and the chord. Ensure the calculator is in Radian mode when calculating sin⁡(π3)\sin\left(\frac{\pi}{3}\right).

Problem 3:

A circle is defined by the equation x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0. Find the coordinates of the center and the length of the radius.

Solution:

Rearrange and complete the square for xx and yy: (x2−4x)+(y2+6y)=12(x^2 - 4x) + (y^2 + 6y) = 12 (x2−4x+4)+(y2+6y+9)=12+4+9(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9 (x−2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25 Comparing to (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2: Center (h,k)=(2,−3)(h, k) = (2, -3), Radius r=25=5r = \sqrt{25} = 5.

Explanation:

To find the center and radius from a general quadratic form, complete the square for both the xx and yy terms to put the equation into standard form.

Problem 4:

A windshield wiper of length 4040 cm rotates through an angle of 120∘120^{\circ}. Calculate the area of the windshield cleaned by the wiper, giving your answer in terms of π\pi.

A sector representing the sweep of a windshield wiper with a radius of 40 cm and a central angle of 120 degrees.

Solution:

  1. Convert the angle to radians: θ=120∘×π180∘=2π3 rad\theta = 120^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{2\pi}{3} \text{ rad}
  2. Identify the radius: r=40 cmr = 40 \text{ cm}
  3. Use the sector area formula: A=12r2θ=12(40)2(2π3)A = \frac{1}{2}r^2\theta = \frac{1}{2}(40)^2(\frac{2\pi}{3})
  4. Simplify: A=12(1600)(2π3)=800×2π3=1600π3 cm2A = \frac{1}{2}(1600)(\frac{2\pi}{3}) = 800 \times \frac{2\pi}{3} = \frac{1600\pi}{3} \text{ cm}^2

Explanation:

The area cleaned by a wiper is a sector of a circle. We convert the degree measurement to radians first because the standard sector formula A=12r2θA = \frac{1}{2}r^2\theta requires θ\theta to be in radians.

Problem 5:

A circle passes through the origin (0,0)(0, 0) and has its center at (3,4)(3, 4). Determine the equation of the circle in the form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.

A coordinate plane showing a circle with center at (3, 4) passing through the origin (0, 0).

Solution:

  1. Identify the center (h,k)(h, k): (h,k)=(3,4)(h, k) = (3, 4)
  2. Calculate the radius rr using the distance formula between the center and (0,0)(0, 0): r=(3−0)2+(4−0)2=32+42=9+16=25=5r = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
  3. Substitute h,k,h, k, and rr into the standard equation: (x−3)2+(y−4)2=52(x - 3)^2 + (y - 4)^2 = 5^2 (x−3)2+(y−4)2=25(x - 3)^2 + (y - 4)^2 = 25

Explanation:

The radius is the distance from the center of the circle to any point on its circumference. Since the origin is on the circle, we find the distance between (3,4)(3, 4) and (0,0)(0, 0) to determine rr.