Lesson 1: Symmetrical Properties of Circles

Video Lesson:

Competency:

At the end of this lesson, you will be able to:

• Discover the symmetrical properties of circles.
• Use the symmetrical properties of circles to solve related problems.
• Explain angle properties of circles in their own words.
• Apply angle properties of circles to solve related problems.

Key terms:

• Circle
• Centre
• Line of symmetry
• Chord 

Brainstorming Questions:

1. What is a circle?
2. Draw a circle and indicate its centre, radius and diameter.
3. What is line of symmetry?
4. How many lines of symmetry does an equilateral triangle have?

Answer: A circle is set of points in a plane each of which is equidistant from a fixed point in the plane. The fixed point is called the center of the circle and the constant distance is called its radius.

1. 𝑂 is the centre, 𝑅 is the radius and 𝐷 is the diameter.
2.  The line of symmetry can be defined as the axis or imaginary line that passes. Through the centre of the shape or object and divides it into identical halves.
3. Equilateral triangle has three lines of symmetry.

1.1. Introduction to circles

Definition:

A circle is the locus of points (set of points) in a plane each of which is equidistant from a fixed point in the plane.

• The fixed point is called the centre of the circle and
• The constant distance is called its radius.
• Thus, the circle is defined by its centre O and radius r.
• Recall that area of a circle, 𝐴 = 𝜋𝑟² and perimeter of the circle, 𝑃 = 2𝜋𝑟.
• Any diameter of a circle can be considered as a line of symmetry for the circle.
• An object can have zero lines of symmetry or it can have infinite lines of symmetry.

Example 1:

Determine the number of lines of symmetry for the figure below.

Solution:

We know that a circle has infinite lines of symmetry but as per the given figure above, a circle has been inscribed in a square. A square has 4 lines of symmetry. Therefore, the given figure above has 4 lines of symmetry as shown in figure.

Theorem:
The line segment joining the center of a circle to the midpoint of a chord is perpendicular to the chord.
Proof:
Given A circle with center 𝑂 and a chord (PQ) whose midpoint is 𝑀. We want to prove that ∠𝑂MP is a right angle. Draw the diameter (ST) through point M. Then, the circle is symmetric about (ST) and (PM) ≡ (QM) . So, (ST) is perpendicular bisector of and hence ∠𝑂MP is a right angle.

Example 2:

1. 𝛥𝐴𝐵𝐶 is an equilateral triangle and circle O is its circumcircle. How many lines of symmetry does figure below have?

2. In figure show that ∆𝑂𝑃M ≡ ∆𝑂𝑄M.

Solution:
An equilateral triangle has three lines of symmetry
Proof:
Join (OP) and(OQ) .
i. (OP) ≡ (OQ) … radii of circle
ii. (PM) ≡ (QM) … M is midpoint of (PQ)
iii. (OM) ≡ (OM) … common side
iv. ∆𝑂𝑃𝑀 ≡ ∆𝑂𝑄𝑀 … by SSS- criteria of congruency
Characteristics of Chord (1)
Theorem:
The line segment drawn from the centre of a circle perpendicular to a chord
bisects the chord.
Proof:
Given: A circle with centre 𝑂 and (ON) is drawn from centre 𝑂 perpendicular to the chord 𝐴𝐵 as shown in figure below.
We want to prove that (AN) ≡ (NB).
Join (OA) and (OB).

1. (OA) ≡ (OB) …radii of circle
2. m(∠𝐴𝑁𝑂) = 𝑚(∠𝐵𝑁𝑂) …both equal to 90°
3. (ON) ≡ (ON) …common side
4. ∆𝐴𝑂𝑁 ≡ ∆𝐵𝑂𝑁 …by RHS-criteria of congruency
5. (AN) ≡ (BN) …by step 4

Theorem:
Equal chords of a circle are equidistant from the center of the circle.
Proof:
Given: Chords (AB) and (CD) are equal in length. Construction: Join points 𝐴 and 𝐶 with center 𝑂 and drop perpendiculars from 𝑂 to the chords (AB) and (CD) (see figure). We want to prove: (OP) ≡ (OQ).

Characteristics of Chord

Theorem:
If the angles subtended by the chords of a circle are equal in measure, then the length of the chords are equal.

Proof:
From figure below, consider ∆AOB and ∆POQ.

Theorem:
Chords which are equal in length subtend equal angles at the Centre of the circle.

Proof:
From figure below, consider ∆AOB and ∆POQ
We want to prove ∠AOB ≡ ∠POQ.

Steps	Statement	Reason
1	AB ≡ PQ	Given
2	OA ≡ OB ≡ OP≡ OQ	Radii of the same circle
3	∆𝐴𝑂𝐵 ≡ ∆𝑃𝑂𝑄	SSS postulate of Congruence
4	m(∠AOB) = 𝑚(∠𝑃𝑂𝑄)	From step 3

Example 3

If the chord of a circle of radius 10 cm is 16 cm long. Find the distance of the chord from the centre.
Solution:

$$\begin{gathered} \text{Given:}AB=16\text{cm and}OB=10\text{cm} \\ \text{Thus}MB=8\text{cm; (}\frac{1}{2}\text{AB)} \\ O{M}^2+M{B}^2=O{B}^2 \\ OM=\sqrt{\left(OB\right){}^2–\left(MB\right){}^2} \\ =\sqrt{\left(10\right){}^2–8{}^2} \\ =\sqrt{(100–64)} \\ =\sqrt{\left(36\right)} \\ =6\text{cm} \end{gathered}$$

1.2 Angle Properties of Circles

1.2.1. Central Angles and Inscribed Angles

Activity

Define the following terms: chord, diameter, radius, tangent, secant, arc, major and minor arc.

Answer

1. The chord of a circle can be defined as the line segment joining any two points on the circumference of the circle.
2. The diameter is the longest chord in a circle.
3. The distance from the centre point to any endpoint on the circle is called the radius of a circle.
4. A tangent to a circle is a straight line which touches the circle at only one point.
5. A straight line that intersects a circle at two points is called a secant line.
6. An arc is a smooth curve joining two points. Consequently, on a circle, every pair of distinct points determines two arcs:

Two points lying on a circle actually define two arcs. The shorter is called the ‘minor arc‘ and the longer one is called the ‘major arc‘.

Remark:

1. A major arc is an arc connecting two endpoints on a circle and its measure is greater than180°or π. (See figure to the right). Arc ACB (is major arc).

2. Minor arc is an arc connecting two endpoints on a circle and its measure is less than 180° or π. (See figure below) Arc ACB ( is minor arc. Usually denoted by two letters. So we call minor arc AC.

3. A major arc is usually referred to with three letters and a minor arc is usually referred to with only two letters.

4. A central angle is an angle formed by two radii with vertex at the center of the circle. (See figure below) ∠AOB is central angle

5. An inscribed angle is an angle with vertex on the circle formed by two intersecting chords. (See figure below) ∠APB is inscribed angle.

Example 4:

In figure below, m(∠𝐴𝑂𝐵) is a central angle with an intercepted minor arc 𝐴𝐵 whose measure is 82°, i.e., 𝑥 = 82°. Arc AB is minor arc. (Why?)

Theorem:
If an inscribed and a central angle intercept the same arc, then the measure of an inscribed angle is half of the measure of a central angle.

Proof:
Let’s prove 𝜃 = 2𝑎 for all 𝜃 and 𝑎, where 𝜃 is the central angle. These three cases account for all possible situations where an inscribed angle and a central angle intercept the same arc.

Case I.

The diameter lies along one ray of an inscribed angle (see figure below).
Step 1 Spot the isosceles triangle.  and  are both radii, so they have the same length. So, ∆𝐷𝐵𝐶 is an isosceles, which also means that its base
angles are congruent. Then, 𝑚(∠𝐵𝐶𝐷) = 𝑚(∠𝐵𝐷𝐶) = 𝑎.

Step 2 Spot the straight angle. So, 𝑚(∠𝐷𝐵𝐶) = 180° – 𝜃
Step 3 Write an equation and solve for 𝑎.

𝑎 + 𝑎 + (180° – 𝜃) = 180° which implies 2𝑎 – 𝜃 = 0

which again implies 2𝑎 = 𝜃. Proved.

Case II. The diameter is between the rays of the inscribed angle 𝑎 (see figure 5.20).
Step 1 Draw a diameter and using the diameter let’s break ‘𝑎’ into 𝑎1 and 𝑎2, and 𝜃 into 𝜃1 and 𝜃2 as shown in figure.

Step 2 Use what we learned from case I to establish two

Equations

𝑎₁ + 𝑎₁ + (180° – 𝜃₁) = 180°
2𝑎₁ – 𝜃₁ = 0 which implies 2𝑎₁ = 𝜃₁.

𝑎₂ + 𝑎₂ + (180° – 𝜃₂) = 180° which implies 𝜃₂ = 2𝑎₂

Step 3 Add the above two equations
                        𝜃₁ + 𝜃₂ = 2𝑎₁ + 2𝑎₂ which implies 𝜃 = 2(𝑎₁ + 𝑎₂) = 2𝑎.

Case III. The diameter is outside the rays of the inscribed angle.
Step 1 Draw a diameter and using the diameter let’s create two new angles
               𝜃₂ and 𝑎₂ as shown in figure below.
Step 2 Use what we learned from case I to establish two equations.

𝑎₂ + 𝑎₂ + (180° – 𝜃₂) = 180°

2𝑎₂ – 𝜃₂ = 0 which implies 2𝑎₂ = 𝜃₂ 𝑎 + 𝑎₂ + 𝑎 + 𝑎₂ + (180° – 𝜃₂ – 𝜃) = 180°
2𝑎 + 2𝑎₂ – 𝜃₂ – 𝜃 = 0 but 2𝑎₂ = 𝜃₂
∴ 𝜃 = 2𝑎. We prove that 𝜃 = 2𝑎.

Example 5:

Both drawings in Figure below illustrate this theorem.

Example 6:

In figure below, ∠𝐴𝐵𝐶 is an inscribed angle with an intercepted minor arc from  A to C. Find the value of 𝑥.

Solution:

∠A𝐵C is an inscribed angle.
So, m(∠ABC) = ½ m(arcAC)

= ½ (82° ) = 41°

Example 7:

In each of the following figures, O is the centre of the circle. Calculate the measure of the angles marked x.

a. x_o = ½ (80°) = 40°

b. . x_o = 2 × 40° = 80°

1.2.2. Measure of Central Angles and Inscribed Angles

Theorem:
Inscribed angles subtended by the same arc have the same measure.

Proof:

In Figure below, m(∠𝐴P𝐵) = ½ m(∠𝐴O𝐵)

m(∠𝐴Q𝐵) = ½ m(∠𝐴O𝐵)
Therefore, m(∠𝐴P𝐵) = m(∠𝐴Q𝐵).

Example 8:

1. Calculate the marked angles in each of the following figures:

2. In the figure below, what is the measure of angle 𝐶𝐵𝑋?

Solution:
1. By the above theorem x^o = 72^o

Given: m(∠𝐴D𝐵) = m(∠𝐴C𝐵) = 32° and ∠𝐴C𝐵 ≡ ∠𝑋C𝐵.
So, in 𝛥𝐵𝑋𝐶 we have 𝑚(∠𝐵X𝐶) = 85° and
𝑚(∠𝑋C𝐵) = 32°. By angle sum theorem,
𝑚(∠𝐶B𝑋) + 𝑚(∠𝐵X𝐶) + 𝑚(∠𝐵C𝑋) = 180°.
𝑚(∠𝐶𝐵𝑋) + 85° + 32° = 180°.
𝑚(∠𝐶𝐵𝑋) = 180° – 1170 = 63°.

Theorem: Angle in a semicircle (Thales’ Theorem)
An angle inscribed in a semicircle is a right angle.

Proof:

The given angle APB is subtended by a semicircle as shown in figure below. The corresponding central angle which is subtended by arc 𝐴𝐵 is a straight angle, that is, the central angle is 180°.

Hence, m(∠𝐴P𝐵) = ½ m(∠𝐴O𝐵)

      = ½ × 180°
      = 90°

Example 9:

In figure below, O is the centre of a circle. Find the measure of 𝑥^o.

Solution:

x^o = 180^o – (90^o + 30^o) = 60^o

1.2.3. Cyclic Quadrilateral

Definition:
A quadrilateral is said to be a cyclic quadrilateral if there is a circle passing through all its four vertices.

Theorem:

The sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

Proof:
Given: A cyclic quadrilateral WXYZ is inscribed in a circle with centre 𝑂 as shown in figure below.
Construction: Join the vertices 𝑊 and 𝑌 with centre 𝑂.

We want to show: m(∠WXY) + m(∠ WZY) = 180°. Consider arc WXY and arc WZY

1. ∠WOY ≡ 2∠WZY (The angle subtended by same arc is half of the angle subtended at the center)
2. Reflex angle WOY ≡ 2arcWXY (the angle subtended by same arc is half of the angle subtended at the center).
3. m(∠WOY) + Reflex m(∠WOY) ≡ 360° (Using steps 1 and 2).
4. 2m(∠WZY) + 2m(∠WXY) = 360° (Using steps 1 and 2).
5. 2(m(∠WZY) + m(∠WXY)) = 360°(Why?).
6. m(∠WZY) + m(∠WXY) = 180° (Why?)

Example 10:

If the measures of all four angles of a cyclic quadrilateral are given as (4𝑦 + 2), (𝑦 + 20), (5𝑦 – 2), and 7𝑦 respectively, find the value of 𝑦.

Solution:
The sum of all four angles of a cyclic quadrilateral is 360°. So, to find the value of 𝑦, we need to equate the sum of the given four angles to 360°.

(4𝑦 + 2) + (𝑦 + 20) + (5𝑦 – 2) + 7𝑦 = 360°
17𝑦 + 20 = 360°
17𝑦 = 340°.

Therefore, 𝑦 = 20°.