SAINT JOSEPH SCHOOL
MATHEMATICS GRADE 9 NOTE 3
Unit 5 (PAGE 221-232)
5.4 Circles
Lesson 1
Symmetrical properties of circle
 A circle is symmetrical about any diameter. (It has infinite number of line of symmetry).
 The locus of a point equidistant from two fixed points is perpendicular dissector of the line joining
the two fixed point.
1. If 𝑃𝐴 = 𝑄𝐴, then A lies on the perpendicular bisector of ̅̅̅̅
𝑃𝑄 .
̅̅̅̅, then 𝑃𝐵
̅̅̅̅.
̅̅̅̅ ≡ 𝑄𝐵
2. If B lies on the perpendicular of 𝑃𝑄
̅̅̅̅ , radii. Therefore,
If P and Q are any two points on the circumference of circle, center O, then ̅̅̅̅
𝑂𝑃 ≡ 𝑂𝑄
̅̅̅̅ .
O lies on the perpendicular of 𝑃𝑄
Theorem
The line segment joining the center of a circle to the midpoint of a chord is perpendicular to the chord.
Proof:
We want to prove that < 𝑂𝑃𝐵 is a right angle.
𝐴𝑃 = 𝐵𝑃 and the circle is symmetrical about the line ̅̅̅̅
𝑂𝑃.
̅̅̅̅
So, 𝑂𝑃 is the perpendicular bisector of 𝐴𝐵.
Theorem
The line segment drawn from the center of the circle perpendicular to a chord bisects the chord.
Proof:
We want to prove that 𝐴𝑁 = 𝑁𝐵 .
Draw a line segment 𝑂𝑁 through AB.
Then the figure is symmetrical about 𝑂𝑁.
𝑂𝑁 ⊥ 𝐴𝐵 and A and B are on the circle.
Therefore, 𝑂𝑁 is symmetrical bisector of 𝐴𝐵.
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SAINT JOSEPH SCHOOL
MATHEMATICS GRADE 9 NOTE 3
Theorem
i)
If two chords of a circle are equal, then they are equidistant from the center.
Proof:
We want to prove that 𝑂𝑃 = 𝑂𝑄 .
𝐴𝐵 = 𝐶𝐷 , 𝑂𝑃 𝑎𝑛𝑑 𝑂𝑄 are symmetrical bisector of
𝐴𝐵 𝑎𝑛𝑑 𝐶𝐷 respectively.
Therefore, 𝑂𝑃 𝑎𝑛𝑑 𝑂𝑄 are equal from the center.
ii)
If two chords of a circle are equidistant from the center, then their lengths are equal.
Proof:
We want to prove that 𝐴𝐵 = 𝐶𝐷 .
𝑂𝑃 𝑎𝑛𝑑 𝑂𝑄 are equidistant from the center and perpendicular bisector of
𝐴𝐵 𝑎𝑛𝑑 𝐶𝐷respectively.
𝑆𝑜, 𝐴𝐵 𝑎𝑛𝑑 𝐶𝐷 are equal in length.
Theorem
If two tangent segments are drawn to a circle from the external point. Then
i.
ii.
The tangents are equal in length
The line segment joining the center to the external point bisects the angle between the
tangents.
Proof:
∆𝑂𝐶𝐴 𝑎𝑛𝑑 ∆𝑂𝐶𝐵 are right angled triangles with right
angle at A and B. 𝑂𝐴 = 𝑂𝐵 .
∆𝑂𝐶𝐴 ≡ ∆𝑂𝐶𝐵 (by RHS)
Therefore, 𝐶𝐴 = 𝐶𝐵 𝑎𝑛𝑑 𝑚(< 𝑂𝐶𝐴) = 𝑚(< 𝑂𝐶𝐵)
Supplementary Exercise
1. A chord of length 8cm, is at a distance of 3 cm from the center of the circle. Find the radius.
2. A chord of circle radius 10 cm is 16 cm. Find the distance of the chord from the center.
3. In a circle of radius 13 cm, there are two parallel chords of lengths 10 cm and 24 cm. Find the distance
between the chords.
4. In the adjacent figure, ̅̅̅̅
𝑃𝑇 𝑎𝑛𝑑 ̅̅̅̅
𝑄𝑇 are tangent segments to circle O. If 𝑂𝑃 = 5 𝑐𝑚 𝑎𝑛𝑑 𝑄𝑇 = 12 𝑐𝑚,
then find 𝑂𝑇.
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SAINT JOSEPH SCHOOL
MATHEMATICS GRADE 9 NOTE 3
Lesson 2
Angle properties of circles
v
 Arc: A part of a circle (part of its circumference) between any two points on the circle.
̂ (Or 𝐴𝐶𝐵
̂ ) is an arc.
Example: In fig a, 𝐴𝑋𝐵
 Semicircle: Half of the circumference of the circle.
̅̅̅̅ is a diameter of a circle. 𝐴𝑋𝐵
̂ (𝑂𝑟 𝐴𝐶𝐵
̂ ) is semicircle.
Example: In fig b, 𝐴𝐵
 Major arc: An arc greater than a semicircle.
̂ is major arc.
Example: In fig c, 𝐴𝑋𝐵
 Minor arc: An arc less than a semicircle.
̂ is minor arc.
Example: In fig d, 𝐴𝑋𝐵
 Central angle: an angle whose vertex is at the center of the circle and whose sides are radii of the
circle.
Example: The angle < 𝐴𝑂𝐵 is Central angle.
̂
< 𝐴𝑂𝐵 intercepts 𝐴𝑋𝐵
̂
< 𝐴𝑂𝐵 subtended by 𝐴𝑋𝐵
̂ subtends < 𝐴𝑂𝐵
𝐴𝑋𝐵
̂)
𝑚(< 𝐴𝑂𝐵) = 𝑚(𝐴𝑋𝐵
 Inscribed angle: an angle whose vertex is on the circle and whose sides are chords of the circle.
Example: The angle < 𝐴𝐵𝐶 is Inscribed angle.
̂
< 𝐴𝐵𝐶 intercepts 𝐴𝑋𝐶
̂
< 𝐴𝐵𝐶 subtended by 𝐴𝑋𝐶
̂
< 𝐴𝐵𝐶 inscribed on 𝐴𝐵𝐶
𝑚(< 𝐴𝐵𝐶) =
1
̂)
𝑚(𝐴𝑋𝐶
2
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SAINT JOSEPH SCHOOL
MATHEMATICS GRADE 9 NOTE 3
Theorem
The measure of a central angle subtended by an arc is twice the measure of an inscribed angle in the circle
subtended by the same arc.
 Dear Student: Please take time to practice the following Exercise.
Page 226
Exercise 5.10
# a, b, c, d, e, f, g
Lesson 3
Corollary
An angle inscribed in the same arc of a circle (i.e. subtended by the same arc) are equal.
̂.
Both < 𝐴𝐷𝐵 and < 𝐴𝐶𝐵 are inscribed in 𝐴𝐷𝐵
So, 𝑚(< 𝐴𝐷𝐵) = 𝑚(< 𝐴𝐶𝐵)
Corollary
 The angle inscribed in a semicircle is a right angle.
 An angle inscribed in a minor arc (subtended by major arc) obtuse angle.
 An angle inscribed in a major arc (subtended by minor arc) acute angle.
Corollary
Points 𝑃, 𝑄, 𝑅 𝑎𝑛𝑑 𝑆 all lie on a circle. They are called concyclic points.
𝑃𝑄𝑅𝑆 is a cyclic quadrilateral.
The opposite angle of a cyclic quadrilateral are supplementary
i.e. 𝑚(< 𝑃) + 𝑚(< 𝑅) = 1800 and
𝑚(< 𝑆) + 𝑚(< 𝑄) = 1800
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SAINT JOSEPH SCHOOL
MATHEMATICS GRADE 9 NOTE 3
 Dear Student: Please take time to practice the following Exercise.
Page 227-228
Exercise 5.11
# 1, 2
Activity 5.15
# a, b, c, d
Lesson 4
Arc lengths, Perimeters and
Areas of Segments and Sectors
 Circumference of a circle 𝐶 = 2𝜋𝑟 𝑜𝑟 𝐶 = 𝜋𝑑.
 Area of a circle 𝐴 = 𝜋𝑟 2 .
 Arc is part of the circumference of a circle.
 A segment of a circle is a region bounded by a chord and an arc.
 A sector of a circle is bounded by two radii and an arc.
 Arc length
The length 𝑙 of an arc of a circle of radius 𝑟 that subtends an angle of 𝜃 at the center is given by:
𝑙 = 2𝜋𝑟
𝜃
𝜋𝑟𝜃
=
360° 180°
 The Area and Perimeter of Sector
𝜋𝑟 2 𝜃
𝐴=
360°
𝜋𝑟𝜃
𝑃=
+ 2𝑟
180°
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SAINT JOSEPH SCHOOL
MATHEMATICS GRADE 9 NOTE 3
 The Area and Perimeter of Segment
𝜋𝑟 2 𝜃
𝐴=
1
− 𝑟 2 sin 𝜃
360°
𝜋𝑟𝜃
𝑃=
2
(Sector area - triangle area)
𝜃
+ 2𝑟 sin ( )
180°
2
(Arc length + chord length)
Example
A segment of a circle of radius 12𝑐𝑚 is cut off by a chord subtending an angle 60° at the center of the
circle. Find
a) The area and perimeter of the sector
b) The area and perimeter of the segment
a) Area of sector
𝐴=
𝜋𝑟 2 𝜃
360°
60°
= 𝜋 × 122 × 360° = 24𝜋 𝑐𝑚2
𝑃=
Perimeter of sector
=
𝜋𝑟𝜃
180°
+ 2𝑟
𝜋×12×60°
180°
+ 2 × 12
= 4𝜋 + 24 = 4(𝜋 + 6) 𝑐𝑚
b) Area of segment
𝐴=
=
𝜋𝑟 2 𝜃
360°
1
− 𝑟 2 sin 𝜃
2
𝜋(122 )(60°)
360°
1
− (122 ) sin 60°
2
1
= 24𝜋 − × 122 ×
2
√3
2
= 24𝜋 − 36√3 = 12(2𝜋 − 3√3) 𝑐𝑚2
Perimeter of segment
𝑃=
=
𝜋𝑟𝜃
𝜃
+ 2𝑟 sin ( )
180°
2
𝜋(12)(60°)
180°
+ 2(12) sin 60°
= 4𝜋 + 12 = 4(𝜋 + 3) 𝑐𝑚
 Dear Student: Please take time to practice the following Exercise.
Page 231-232
Exercise 5.12
# 1, 2, 3, 4, 5
Tr. Abel Kassu
6