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Geometry - Properties of Circles

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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The angle at the center of a circle is twice the angle at the circumference subtended by the same arc. This is known as the Central Angle Theorem.

Diagram showing the angle at the center being twice the angle at the circumference.
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The angle in a semicircle is always a right angle (90∘90^\circ). Any triangle drawn with the diameter as its base and the third vertex on the circumference is a right-angled triangle.

A right-angled triangle inscribed in a semicircle.
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Angles in the same segment of a circle are equal. This means that two angles subtended by the same arc at any point on the circumference are equivalent.

Angles in the same segment showing equality.
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Opposite angles in a cyclic quadrilateral sum to 180∘180^\circ. A cyclic quadrilateral is a four-sided figure where all vertices lie on the circumference of a circle.

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A tangent to a circle is perpendicular to the radius at the point of contact. This means the angle between the tangent and the radius is exactly 90∘90^\circ.

📐Formulae

Circumference = 2πr2\pi r or πd\pi d

Area = πr2\pi r^2

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

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

💡Examples

Problem 1:

A circle has a radius of 7 cm. Calculate the length of an arc that subtends an angle of 60∘60^\circ at the center. (Use π≈3.142\pi \approx 3.142)

Solution:

Arc Length = 60360×2×3.142×7≈7.33\frac{60}{360} \times 2 \times 3.142 \times 7 \approx 7.33 cm

Explanation:

Apply the arc length formula by substituting θ=60\theta = 60 and r=7r = 7. Simplify the fraction 60360\frac{60}{360} to 16\frac{1}{6} and multiply.

Problem 2:

In a cyclic quadrilateral ABCDABCD, angle A=85∘A = 85^\circ. Find the size of the opposite angle CC.

Solution:

180∘−85∘=95∘180^\circ - 85^\circ = 95^\circ

Explanation:

According to the property of cyclic quadrilaterals, opposite angles are supplementary, meaning they add up to 180∘180^\circ.

Problem 3:

A triangle is drawn inside a circle where one side is the diameter. If one of the other angles is 35∘35^\circ, find the third angle.

Solution:

180∘−(90∘+35∘)=55∘180^\circ - (90^\circ + 35^\circ) = 55^\circ

Explanation:

The property 'angle in a semi-circle' states the angle opposite the diameter is 90∘90^\circ. Since the sum of angles in a triangle is 180∘180^\circ, we subtract the known angles from 180∘180^\circ.

Problem 4:

In the given diagram, OO is the center of the circle. If the angle at the center ∠AOB=110∘\angle AOB = 110^\circ, find the value of angle xx at the circumference.

Circle with center angle 110 degrees and circumference angle x.

Solution:

  1. Identify the relationship: The angle at the center is twice the angle at the circumference subtended by the same arc.
  2. Apply the theorem: ∠AOB=2×∠ACB\angle AOB = 2 \times \angle ACB.
  3. Substitute values: 110∘=2x110^\circ = 2x.
  4. Solve for xx: x=110∘2=55∘x = \frac{110^\circ}{2} = 55^\circ.

Explanation:

Using the Central Angle Theorem, we determine that the angle subtended by arc ABAB at the circumference is half the angle subtended by the same arc at the center.

Problem 5:

A tangent PTPT touches a circle at point TT. The center of the circle is OO. If the radius OT=5OT = 5 cm and the distance from the center to point PP is OP=13OP = 13 cm, calculate the length of the tangent segment PTPT.

Right-angled triangle formed by radius, tangent and line to center.

Solution:

  1. Recognize the property: A tangent is perpendicular to the radius at the point of contact, so ∠OTP=90∘\angle OTP = 90^\circ.
  2. Apply Pythagoras' Theorem to the right-angled triangle △OTP\triangle OTP: OT2+PT2=OP2OT^2 + PT^2 = OP^2
  3. Substitute the known values: 52+PT2=1325^2 + PT^2 = 13^2
  4. Calculate: 25+PT2=16925 + PT^2 = 169
  5. Solve for PTPT: PT2=169−25=144PT^2 = 169 - 25 = 144
  6. PT=144=12 cmPT = \sqrt{144} = 12 \text{ cm}

Explanation:

Because the tangent is perpendicular to the radius, we can use the Pythagorean theorem to find the missing side of the right triangle formed by the radius, the tangent, and the line from the center to the external point.