krit.club logo

Geometry and Trigonometry - Circle geometry

Grade 9IB

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

🔑Concepts

•

The angle at the center of a circle is twice the angle at the circumference when both angles are subtended by the same arc. This fundamental property leads to several other circle theorems.

Diagram showing the angle at the center 2x is twice the angle at the circumference x.
•

A tangent to a circle is a straight line that touches the circle at exactly one point. The radius of the circle is perpendicular (90∘90^\circ) to the tangent at the point of contact.

Diagram of a tangent line perpendicular to the radius at the point of contact.
•

Angles subtended by the same arc (or segment) in the same part of the circle are equal. This is often called 'angles in the same segment'.

•

In a cyclic quadrilateral, where all four vertices lie on the circumference of a circle, the opposite angles add up to 180∘180^\circ (supplementary).

•

The angle in a semi-circle is always a right angle (90∘90^\circ). This occurs when the chord subtending the angle is the diameter of the circle.

📐Formulae

C=2πrC = 2\pi r

A=πr2A = \pi r^2

Arc Length=θ360∘×2πr\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r

Area of Sector=θ360∘×πr2\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2

Area of Segment=θ360∘πr2−12r2sin⁡(θ)\text{Area of Segment} = \frac{\theta}{360^\circ} \pi r^2 - \frac{1}{2} r^2 \sin(\theta)

💡Examples

Problem 1:

Calculate the area of a sector with a radius of 12 cm12\text{ cm} and a central angle of 60∘60^\circ. Leave your answer in terms of π\pi.

Solution:

Area=60∘360∘×π×122\text{Area} = \frac{60^\circ}{360^\circ} \times \pi \times 12^2 Area=16×144π\text{Area} = \frac{1}{6} \times 144\pi Area=24π cm2\text{Area} = 24\pi \text{ cm}^2

Explanation:

Substitute the given radius r=12r = 12 and angle θ=60∘\theta = 60^\circ into the sector area formula. Simplify the fraction and calculate the final value.

Problem 2:

In a circle with center OO, points AA, BB, and CC lie on the circumference. If ∠BOC=130∘\angle BOC = 130^\circ (where OO is the center), find the size of ∠BAC\angle BAC.

Solution:

∠BAC=12×∠BOC\angle BAC = \frac{1}{2} \times \angle BOC ∠BAC=130∘2\angle BAC = \frac{130^\circ}{2} ∠BAC=65∘\angle BAC = 65^\circ

Explanation:

Using the 'Angle at the Center' theorem, the angle subtended at the circumference (∠BAC\angle BAC) is half the angle subtended at the center (∠BOC\angle BOC) by the same arc BCBC.

Problem 3:

A circle has a circumference of 20π cm20\pi \text{ cm}. Find its area.

Solution:

First, find rr: 2πr=20π  ⟹  r=10 cm2\pi r = 20\pi \implies r = 10\text{ cm} Now, find the area: A=π(10)2=100π cm2A = \pi (10)^2 = 100\pi \text{ cm}^2

Explanation:

Use the circumference formula to solve for the radius rr. Once rr is known, substitute it into the area formula A=πr2A = \pi r^2.

Problem 4:

In the given circle with center OO, segment ABAB is a diameter. Point CC lies on the circumference. If ∠BAC=35∘\angle BAC = 35^\circ, find the value of ∠ABC\angle ABC.

A circle with diameter AB and point C on the circumference forming triangle ABC.

Solution:

  1. Identify that △ABC\triangle ABC is inscribed in a semi-circle because ABAB is the diameter.
  2. According to the circle theorem, the angle in a semi-circle is 90∘90^\circ, so ∠ACB=90∘\angle ACB = 90^\circ.
  3. The sum of angles in a triangle is 180∘180^\circ. Therefore: ∠ABC=180∘−(∠ACB+∠BAC)\angle ABC = 180^\circ - (\angle ACB + \angle BAC) ∠ABC=180∘−(90∘+35∘)\angle ABC = 180^\circ - (90^\circ + 35^\circ) ∠ABC=180∘−125∘=55∘\angle ABC = 180^\circ - 125^\circ = 55^\circ

Explanation:

This problem uses the theorem that any angle subtended by a diameter at the circumference is a right angle, combined with the basic triangle angle sum property.

Problem 5:

Calculate the length of an arc that subtends an angle of 120∘120^\circ at the center of a circle with a radius of 9 cm9\text{ cm}. Give your answer in terms of π\pi.

A circle sector with a central angle of 120 degrees and radius 9 cm.

Solution:

  1. Use the arc length formula: Arc Length=θ360∘×2πr\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r
  2. Substitute the given values θ=120∘\theta = 120^\circ and r=9 cmr = 9\text{ cm}: Arc Length=120360×2×π×9\text{Arc Length} = \frac{120}{360} \times 2 \times \pi \times 9
  3. Simplify the fraction: Arc Length=13×18π\text{Arc Length} = \frac{1}{3} \times 18\pi
  4. Calculate the final value: Arc Length=6π cm\text{Arc Length} = 6\pi \text{ cm}

Explanation:

Arc length is a fraction of the total circumference. Since 120∘120^\circ is one-third of 360∘360^\circ, the arc length is one-third of the circumference.