CIRCLE GEOMETRY
Extended investigation
Part 1: Take-home Solutions
Question 1
Central Angle: An angle whose vertex is at the centre and whose sides are radii.
Chord: An interval joining two points on the circle.
Cyclic quadrilateral: A cyclic quadrilateral is a quadrilateral whose vertices all lie on
a circle.
Diameter: A chord that passes through the centre.
Radius: Any line segment joining a point on the circle to the centre.
Major/Minor Arc: A part of the circumference of the circle. A major arc represents
more than half of the circumference; a minor arc represents less than half of the
circumference.
Major/Minor Segment: The part of the circular region enclosed between a chord (not
a diameter) and the circle. A major segment encloses a region greater than that of the
semicircle; a minor segment encloses a region smaller than that of the semicircle.
Alternate Segment: The word ‘alternate’ means ‘other’. The chord AB divides the
circle into two segments and AU is tangent to the circle. Angle APB ‘lies in’ the
segment on the other side of chord AB from angle BAU. We say that it is in the
alternate segment.
Major/Minor Sector: The portion of a circular region bounded by two radii and an
arc. A major sector covers a region greater than that of the semicircle; a minor sector
covers a region smaller than that of the semicircle.
Secant: A line that cuts a circle at two distinct points.
Semicircle: Half a circular region, formed by the diameter and half the circle.
Tangent: A line that intersects a circle at one point.
Question 2
(a)
The angle formed by the diameter and two chords lies in a semicircle.
(b)
The size of the angle remains unchanged.
(c)
The angle in a semicircle is a right angle.
Question 3
Angle in a Semicircle Theorem
Given:
Circle centre A, diameter BD. E is any point on the circle, BED is
an angle in the semicircle DBE.
To Prove:
BED = 90°
Extension to
the diagram:
Draw AE.
Proof:
In ABE,


AE = AB
ABE is isosceles
ABE = AEB = 

DBE = 
In ADE,


AE = AD
ADE is isosceles
ADE = AED = 

BDE = 
In BDE,
BED = AEB + AED
=  +
BED + DBE +BDE = 180



         180
 + = 90
BED = 90
radii
angles opposite congruent sides
radii
angles opposite congruent sides
angle sum of 
Question 4
(a)
arc CD
(b)
CAD = 2 x CBD
(c)
The size of the angle at the centre subtended by an arc of the circle is twice the
size of the angle at the circumference subtended by the same arc.
Question 5
Central Angle Theorem
Given:
Circle centre A. CAD is the angle subtended by arc CD at the
centre and CBD is the angle subtended by arc CD at the
circumference
To Prove:
CAD  2  CBD
Extension to
the diagram:
Join BA and produce it to E.
Proof:
In ABC,



In ABD,




AB = AC
ABC is isosceles
ABC = ACB = 
EAC = ABC + ACB
EAC = 2
AB = AD
ABD is isosceles
ABD = ADB = 
EAD = ABD + ADB
 EAD = 2
CBD = ABC + ABD
CBD =   
CAD = EAC + EAD
CAD = 2  2 
 2(   )
 2  CBD
radii
angles opposite congruent sides
exterior angle
radii
angles opposite congruent sides
exterior angle
Question 6
Given:
Circle centre O. APB and AQB are angles at the circumference
subtended by arc AB.
To Prove:
APB = AQB
Proof:
and

APB = 2  AOB
AQB = 2  AOB
APB = AQB
Central angle theorem
Central angle theorem
Question 7
(a)
The two angles are the same size.
(b)
An angle between a chord and a tangent is equal to any angle in the alternate
segment.
Question 8
(a)
The tangents from C to the circle are equal in length.
(b)
Tangents drawn to a circle from an external point are equal in length.
Question 9
(a)
The products are the same.
(b)
When two chords of a circle intersect, the product of the lengths of the intervals
on one chord equals the product of the lengths of the intervals on the other
chord.
Question 10 (a)
AP = AB - BP
 15  9 cm
 6 cm
AQ = AP = 6 cm
BR = BP = 9 cm
CR = BC - BR
= 17 - 9 cm
= 8 cm
CQ = CR = 8 cm
AC = AQ + CQ
= 6 + 8 cm
= 14 cm
length of tangents theorem
length of tangents theorem
length of tangents theorem
Question 10 (b)
ABC = BDC = 74
angle in the alternate segment
ACB = BDC = 74
angle in the alternate segment
BAC = 180 - ABC - ACB
= 180 - 2  74
angle sum of 
= 32
Question 10 (c)
ACB = ADB = 55
In ABC,
ABC = 180 - ACB - BAC
= 180 - 55 - 42
= 83
angles subtended by same arc AB
CBD = ADB
= 55
AD parallel to PC
angle sum of 
ABD = ABC - CBD
= 83 - 55
= 28
EAD = ABD
= 28
angles in the alternate segment
PBC = 180
180 = ABP + ABC
180 = ABP + 83
ABP = 97
straight angle