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Circles III

SPM Practice (Short Questions)

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8.4 Circles, SPM Practice (Short Questions)
Question 9:

In figure above, ABC is a tangent to the circle centre O at B. AED is a straight line.
Find the value of y.

Solution:
∠ABO = 90o
∠BOE = 2 × 40o = 80o
In quadrilateral AEOB,
∠AEO = 360o  – ∠ABO  – ∠BOE – 35o
= 360o  – 90o – 80o – 35o = 155o
yo+ ∠AEO = 180o 
yo+ 155o = 180o
yo = 180o  – 155o
y o = 25



Question 10:
 

In figure above, ABC is a tangent to the circle centre O, at point B.
The value of x is

Solution:
∠OBC = 90o 
∠BOD = 2 × 50o = 100o
In quadrilateral BODC,
xo = 360o  – ∠BOD – ∠OBC – 120o
= 360o  – 100o – 90o – 120o
= 50

8.3 Common Tangents (Part 3)

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3. Do not intersect
(a) Circles of the same size



Number of common tangents
Properties of common tangents
Four common tangents:
AB, CD, PS and RQ
AB = CD = OV
PS = RQ
AB // OV // CD


(b) Circles of different sizes

 

Number of common tangents
Properties of common tangents
Four common tangents:
AB, CD, PS and RQ
AB = CD
BT = DT
PS = RQ
OA // VB
OC // VD

8.3 Common Tangents (Part 2)

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2. Intersect at one point
(a) Circles of the same size


 
Number of common tangents
Properties of common tangents
Three common tangents:
AB, CD and PQ
AC = PQ = BD
AB = OR = CD
AB // OR // CD
AC // PQ // BD
PQ perpendicular to OR
 

(b) Circles of different sizes
(i) External contact
 


Number of common tangents
Properties of common tangents
Three common tangents:
ABE, CDE and PQ
AB = CD
BE = DE
OA // RB
OC // RD
PQ perpendicular to ORE


(ii) Internal contact
 

Number of common tangents
Properties of common tangents
One common tangent: PQR
ONQ perpendicular to PR

8.2 Angle between Tangent and Chord (Example 3 & 4)

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Example 3:
In the diagram, PQR is a tangent to the circle QSTU at Q.
Find the values of
(a) x   (b) y
 
Solution:
(a)
∠UTS + ∠UQS = 180o ←(opposite angle in cyclic quadrilateral QSTU)
105o + ∠ UQS = 180o
∠ UQS = 75o
x+ 75o + 20o = 180o←(the sum of angles on a straight line PQR = 180o)
x+ 95o = 180o
x = 85o
 
(b) 
∠ PQU = ∠ QSU  ← (angle in alternate segment)
85o = 35o + y
y = 50o



Example 4:

In the diagram, ABC is a tangent to the circle BDE with centre O, at B.
Find the value of x.
 
Solution:


∠ B E D = ∠ C B D = 54 ∘ ∠ B D E = 180 ∘ − 54 ∘ 2 = 63 ∘ ← Isosceles triangle ∠ E B D = ∠ E D B ∠ A B E = ∠ B D E = 63 ∘ In △ A B E , x ∘ + 45 ∘ + 63 ∘ = 180 ∘ x ∘ + 108 ∘ = 180 ∘ x = 72

8.2 Angle between Tangent and Chord (Example 1 & 2)

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Example 1:

In the diagram, ABC is a tangent to the circle BDE at B.
The length of arc BD is equal to the length of arc DE.
Find the value of p.
 
Solution:
Angle BED = 82o  ← (angle in alternate segment)
Angle DBE = 82o  ← (Arc BD = Arc DE, BDE is an isosceles triangle)
Therefore p= 180o – 82o – 82o = 16o



Example 2:
In the diagram, PQR is a tangent to the circle QSTU at Q.

Find the value of y.
 
Solution:
Angle QUT 
= 180o– 98o ← (opposite angle in cyclic quadrilateral QSTU )
= 82o
Angle QTU = 75o  ← (angle in alternate segment)
Therefore y= 180o – (82o + 75o)  ← (Sum of interior angles in ∆ QTU)
= 23o

8.1 Tangents to a Circle (Examples)

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Example 1:

In the diagram, O is the centre of a circle. ABC and CDE are two tangents to the circle at points B and D respectively. Find the length of OC.
 
Solution:
OC2= OB2 + BC2 ← (Pythagoras’ Theorem)
= 62 + 82
= 100
OC = √100 = 10 cm


Example 2:

In the diagram, AB and BC are two tangents to a circle with centre O. Calculate the values of
(a) x   (b) y
 
Solution:
(a)
AB = BC
7 + x = 12
x = 5

(b)
∠ OBA = ∠ OBC = 21o
∠ OAB = 90o ← (OA is perpendicular to AB)
yo= 180o – 21o – 90o
y = 69


Example 3:

In the diagram, ABC is a tangent to the circle with centre Oat point B. CDE  is a straight line. Find the value of x.
 
Solution:
∠ CBO = 90o ← (OB is perpendicular to BC)
In ∆ BCE,
x =  180o – 30o – 50o– 90o
x =  10o