SOLUTIONS
TANGENTS & SECANTSGEOMETRY OF THE CIRCLE
Geom
etry
Of
The
Circ
leTa
ngen
ts &
Sec
ants
www.mathletics.com.au
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 116
Solutions Basics
Page 3 questions
A secant passes through a circle. The two points it passes through are called intercepts. While a tangent is a straight line that touches the circle once. This point is called the point of contact.
1. What is the difference between a secant and a tangent?
2. Draw a secant and a tangent to the circle below. Label the "point of contact".
3. What’s the difference between an intercept and a "point of contact"?
F
D
EB
D
SPRQ
A
C
DF is a tangent, E is point of contact AD is a secant, PS is also a secant
A point of contact touches the circle once. An intercept goes through the the circle.
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 162
Knowing MoreSolutions
Show that OT bisects UTV+ .a bShow that TUO TVOT T/ (SSS).
OT
V
U
Page 6 questions
(Tangent perpendicular to radius at point of contact)
(Pythagoras)
(Given)
(Tangents from common source are equal)
(Equal Radii)
(SSS)
(Corresponding sides of congruent triangles)
(Corresponding angles of congruent triangles)
In TUOT and OVTT
is common
TU TV
OU OV
OT
=
=
TUO TVO
UT VT
`
`
T T/
=
bisects
OTU OTV
OT UTV`
+ +
+
=
O
Y
cm24 cm.62 4
Z
is right angled
cm
.
. .
OY ZY
ZYO
OYZ
OZ OY YZ
YZ
YZ
90
62 4 24
3317 76 57 6
2 2 2
2 2 2
`
`
`
`
`
=
c+
T
=
= +
= -
= =
a
b
1. O is the centre of the circle below.
O is the centre
2. Find the length of tangent YZ if O is the centre of the circle.
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 316
Solutions Knowing More
Page 7 questions
(Tangent perpendicular to radius at point of contact)
(Tangent perpendicular to radius at point of contact)
(Tangent perpendicular to radius at point of contact)
(Alternate angles are equal)
(Alternate angles on parallel lines are equal)
reflex
PR OQ
OQR OQP
OQA
OQB
AQB
AOB
90
90 15 75
90 30 60
75 60 135
2 135 270
`
`
`
`
` #
=
c
c c c
c c c
c c c
c c
+ +
+
+
+
+
= =
= - =
= - =
= + =
= =
;;
90
AB ST
BST
CD ST
CTS
AB CD
BSD TDS
90`
`
`
=
c
=
c
+
+
+ +
=
=
=
OA
B
D
C
S
T
O
QP R
BA
(Angle at centre is twice the angle at circumference on same arc)
3. PR is a tangent to the circle with point of contact Q. O is the centre of the circle. 15PQA c+ = and 30RQB c+ = . Find reflex AOB+ .
4. O is the centre of the circle below. AB and CD are tangents with points of contact S and T respectively.Show that BSD TDS+ += .
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 164
Knowing MoreSolutions
O
M
BC
A
cm
cm
19.5
. .
. .
AM OA
OAM
AM MB MC
AM
OM OA AM
OA
OA
90
22 1 19 5
108 16 10 4
2 2 2
2 2 2
`
`
`
`
`
=
c+ =
= =
=
= +
= -
= =
(Tangent perpendicular to radius at point of contact)
(Tangents from common source are equal)
(Pythagoras)
Page 8 questions
5. MA is tangent to the larger circle and MC is tangent to the smaller circle. MB is a common tangent. O is the centre of the larger circle. Find the radius of the larger circle if cm.MC 19 5= and cm.MO 22 1= .
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 516
Solutions Using Our Knowledge
65
LKJ
LKJ
LKJ LJI
LJI
180 71 44
65`
`
c c c
c
c
+
+
+ +
+
= - -
=
=
=
(Sum of angles in a triangle)
(Alternate segment angles)
a
Page 11 questions
c
b
2. Find the size of LJI+ if IJ is tangent to the circle below.
L
J
I44c
71cK
1. Identify the angles equal to the labelled angles.
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 166
Using Our KnowledgeSolutions
3. CE and AC are both tangents to the circle below. Find DBC+ and FDE+ .
CD
E
F 68c
B87c
A
Page 12 questions
DBC BFD
DBC
FDB ABF
FDB
DBF
DBF
68
87
180 68 87
25
`
`
`
c
c
c c c
c
+ +
+
+ +
+
+
+
=
=
=
=
= - -
=
FED PFD
FED
OED
OED
50
50 30
20
`
`
`
c
c c
c
+ +
+
+
+
=
=
= -
=
20 40
EDF EDO ODF
60
c c
c
+ + += +
= +
=
EDF EFQ
EFQ 60` c
+ +
+
=
=
OFD DFP90
90 50
40
` c
c c
c
+ += -
= -
=
ODF 40` c+ =
is isoscelesOFDT
90OFP c+ =
FDE DBF
FDE 25` c
+ +
+
=
=
(Alternate segment angles)
(Alternate segment angles)
(Sum of angles in a triangle)
(Alternate segment angles)
(Alternate segment angles)
(Equal angles of isosceles OFDT )
(Alternate segment angles)
(Tangent = to radius at point of contact)
(OD = OF = equal radii)
.OED+ .ODF+ .EFQ+a b c
D
E
Q
P
F
O
50c
30c
a
b
c
4. The circle below has centre O and tangent PQ with point of contact F.Find
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 716
Solutions Using Our Knowledge
Page 13 questions
PQS BPS
PQS
PQS SPQ
70` c
+ +
+
+ +
=
=
=
180 140 40QSP
DSP SQP
DSP
QRS DSQ
QRS
70
70 40 110
`
c c c
c
c c c
+
+ +
+
+ +
+
= - =
=
=
=
= + =
QSC SPQ
QSC 70` c
+ +
+
=
=
bisectsRS QSC
RSQ
RSQ21 70
35
`
`
c
c
+
+
+
=
=
^ h
70
180 70 70
PS QS
PQS QPS
QSP
QSP 40
`
`
`
c
c c c
c
+ +
+
+
=
= =
= - -
=
PSD PQS
PSD 70` c
+ +
+
=
=
RSD RSQ QSP PSD
35 40 70
145
c c c
c
+ + + += + +
= + +
=
(Alternate segment angles)
(Alternate segment angles)
(Alternate segment angles)
(Alternate segment angles)
(RS bisects QSC+ ; QSC 70c+ = )
( PQST ; is iscosceles)
(Given)
(Angles opposite equal sides)
(Interior angles of PQST )
(Alternate segment angles)
(Sum of angles in a triangle)
a
b
c
5. AB and CD are tangents to the circle with points of contact P and S respectively. RS bisects QSC+ .
Show that PQST is an Isosceles triangle.
Find the size of QRS+ .
Find the size of RSD+ .
a
b
c
PQS` T is an Isosceles triangle
A
P B
70c70c
Q
R
S
D
C
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 168
Thinking MoreSolutions
Page 17 questions
1. Find x in each of the following (all measurements in cm).
A
C
B
E
40 60
20
x
D
a
bV
U
x3
9
5
T
cm
x
x
40 20 60
401200
30
` # #
`
=
=
=
AE x CD ED# #=
3x TU VT# #=
x PS PR2 #=
x x
x x
x x
5 3 12
5 36 0
9 4 0
2
` # #
`
`
+ =
+ - =
+ - =
^
^ ^
h
h h
cm
25 9 225
15
x
x
x
225
2`
`
`
#= =
=
=
or4 9x x` = =-
(Product of intercepts on intersecting chords)
(Products of intercepts of intersecting secants from external point)
(Square of tangent equal product of secant from common point)
Since length is always positive: cm4x =
c
Q
S
x
9
R
16
P
Given: PQ is a tangent
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 916
Solutions Thinking More
Page 18 questions
CE x CB CA# #=
MN x MK2 #=
x x
x x
x x
8 7 12
8 84 0
14 6 0
2
` #
`
`
+ =
+ - =
+ - =
^
^ ^
h
h h
12x x
x x
x x
8
12 64 0
16 4 0
2
2
`
`
`
= +
+ - =
+ - =
^
^ ^
h
h h
orx x6 14` = =-
orx x4 16` = =-
(Products of intercepts of intersecting secants from external point)
(Square of tangent equal product of secant from common point)
Since length is always positive: cm6x =
Since length is always positive: cm4x =
2. Find the missing lengths in each of the following (all measurements in cm).
E
D
C
B
A
a
Find
8
7
5
ED
BC
AB
CD x
=
=
=
=
K
L
M
N
b
Find
KL
MN
LM x
12
8
=
=
=
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1610
Thinking MoreSolutions
3. Find x and y in the diagram below:
C
B
4
8
5
yx
6
EA
D
Page 19 questions
4 24
CB x
x
x
x
4 10
8 4 40
6
2
2
` #
`
`
`
= +
= +
=
=
^ h
x y
y
y
6 5
36 5
536
751
` # #
`
`
=
=
=
=
(Square of tangent equal product of secant from common point)
(Products of intercepts on intersecting chords)
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1116
Solutions Thinking Even More
Page 21 questions
(Alternate Segment Angle)
(Alternate segment angle)
(Angles in same segment on same arc)
(Tangent perpendicular to radius at point of contact)
(Angles in same segment on same arc)
(Angles in same segment on same arc)
(Right angle)
(Angles in T sum to 180c )
(Angles in same segment on same arc)
(Angle in a semi circle)
(Angles in T sum to 180c )
Here is a mix of more difficult problems combining all the theorems for Circle Geometry.
A
O
D B
CP Q
= =
30c
CDB QCB
CDB
CAB
30
30
`
`
c
c
+ +
+
+
=
=
=
PCO QCO
BCA
BDA
90
90 30 60
60
`
`
`
c
c c c
c
+ +
+
+
= =
= - =
=
CDA CBA 90c+ += =
DCO
DBA
CBD
CAD
PCD
60
60
30
30
30
`
`
`
`
`
c
c
c
c
c
+
+
+
+
+
=
=
=
=
=
QP OC=
DB OC= (Line from centre to midpoint is perpendicular to chord)
1. O is the centre of the circle below. PQ is a tangent with point of contact C. BCQ 30c+ = .Find 5 other angles which equal 30c .
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1612
Thinking Even MoreSolutions
2. BEDC is a Rhombus and GD is a tangent to the circle at E.
a
b
c
Show GEA BED+ += .
Show CE bisects BED+ .
Show 180EAB EDC c+ ++ = .
=
=
=
=A
GE
D
C
B
a
b
c
(Alternate Segment Angle)
(Alternate Segment Angle)
(Rhombus has parallel sides)
(Alternate angles are equal)
(Given)
(Rhombus)
(Common side)
(SSS)
(Corresponding angles of congruent triangles)
(Rhombus has parallel sides)
(Rhombus has parallel sides)
(Supplementary cointerior angles, ||EB DC )
(Alternate Segment Angle)
GEA EBA
BED BAE
+ +
+ +
=
=
EB BC ED DC
EBC EDC
EC EC
+ +
= = =
=
=
;;AC GD
GEA BAE
GEA BED
`
`
+ +
+ +
=
=
;;
;;
AC GD
EB DC
BED EAB
BED EDC
EAB EDC
180
180`
c
c
+ +
+ +
+ +
=
+ =
+ =
In andEBC EDCT T
bisects
EBC EDC
CED BEC
CE BED
`
`
`
T T
+ +
+
/
=
Page 22 questions
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1316
Solutions Thinking Even More
(Given)
(Given)
(RHS)
(Corresponding sides of congruent triangles)
(Equal radii)
is commonOP
ON OJ
OPN OPJ 90c+ +
=
= =
In andOPN OPJT T
cm10.2
OPN OPJ
PN PJ
`
`
T T/
= =
cm10.2
NP JP
NP
`
`
=
=
bisectsOP NJ`
a
Given cm
cm
cm
.
.
.
JK
JP
OP OM
PNO
13 6
10 2
7 65
37c+
=
=
= =
=
Find the length of PN.
J
KL
.13 6
M
37c.7 65
-
-
P
.10 2
O
N
Page 23 questions
OP = NJ and OM = NL
OR
OP NJ=
(Line from centre to a chord, = to the chord, bisects it)
3. In the diagram below, O is the centre of the circle and J is the point of contact of tangent KJ.
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1614
Thinking Even MoreSolutions
Page 23 questions
J
KL
.13 6
M
37c.7 65
-
-
P
.10 2
O
N
(Equal radii)
(Given)
(Given)
(Given)
(RHS)
(RHS)
(Corresponding sides of congruent triangles)
(Corresponding sides of congruent triangles)
(Given)
is commonON
OM OP
OMN OPN 90c+ +
=
= =
is common
ON OL
OM
OMN OML 90c+ +
=
= =
OM NL=
bisectsOM NL`
MN ML` =
cm
2
20.4
NL MN` #=
=
In andOPN OMNT T
In andOML OMNT T
cm10.2
OPN OMN
MN PN
`
`
T T/
= =
.
OML OMN
LM MN 10 2
`
`
T T/
= =
cm10.2 10.2 20.4LN` = + =
b Find the length of LN.
OR
(Line from centre perpendicular to chord theorem)
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1516
Solutions Thinking Even More
Page 23 questions
c
d
Find the length of LK.
Find JOL+ .
(Square of tangent equal product of secant from common point)
(Re-arranging)
(Re-arranging)
(Only positive values)
.
. .
JK LK KN
LK KN
LK LK
13 6
13 6 20 4
2
2
2
#
#
`
=
=
= +^ h
. .
. .
LK LK
LK LK
20 4 184 96 0
6 8 27 2 0
2`
`
+ - =
- + =^ ^h h
cm6.8LK` =
(Corresponding angles of congruent triangles)
(Common side and radii)
(Given)
(RHS)
(RHS)
(SSS)
(Corresponding sides of congruent triangles)
(Sum of angles in a triangle)
(Previous result)
(Given)
PJO ONP
JON
37
180 37 37 106
`
`
c
c c c c
+ +
+
= =
= - - =
JOL 360 106 106 148` c c c c+ = - - =
ON ON
OM OP
OMN OPN 90c+ +
=
=
= =
In OPN OPJT T/
In andOPN OMNT T
OPN OMN
OML OMN
OJN OLN
LON JON 106
`
`
`
` c
T T
T T
T T
+ +
/
/
/
= =
OPN OMNT T/
PNM MNO
PNM
37
2 37
74
`
` #
c
c
c
+ +
+
= =
=
=
2
2 74
JOL JNL
148
c
c
+ +#
#
=
=
=
OR
(Angle at centre is equal to twice the angle at the circumference)
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1616
Thinking Even MoreSolutions
Page 24 questions
(Alternate 's+ , ||JK LM )
(Sum of angles in a triangle)
(Sum of angles in a triangle)
(Both equal x180 2c- )
(Alternate angles, ;;LM KJ )
( JLM LJM+ += , proved above in a )
(Angles opposite equal sides of isosceles JMLT )LJM JLM
JLM LJK
LJK LJM`
+ +
+ +
+ +
=
=
=
180 2
LJM x
JML x
`
` c
+
+
=
= -
180LKJ x x
LKJ x180 2
`
`
c
c
+
+
+ + =
= -
JML LKJ`+ +=
4. JM and LM are tangents with points of contact J and L respectively. | |JK ML and KPM is a straight line.
JN
M
L
K
P
a
b
Show LJK LJM+ += .
Show JML LKJ+ += .
Let JLM x+ =
Also, KJL MLJ x+ += =
and LKJ KJL JLM 180c+ + ++ + =
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1716
Solutions Thinking Even More
5. In the diagram below QR and QP are tangents with points of contact R and P respectively.
a Show PTQ PRQ+ += .
b Show ;;TQ UR .
c Show VPW URT 180c+ += = .
d Show URS WQR+ += .
(Isosceles triangle)
(Alternate segment angle)
(Alternate segment angle)
(Angles on a straight line)
(Vertically opposite)
(Sum of angles in a triangle)
(Substitution)
(Alternate segment angle)
(Previous result)
(Theorem 10)
but
QP QR
PRQ x
PTQ x
PTQ PRQ
`
`
+
+
+ +
=
=
=
=
;;
but
PUR x
PUR PTQ
TQ UR
`
`
+
+ +
=
=
VPW x y
VTP y
UTR VTP y
PUR x
UTR PUR URT
y x UTR
y x UTR VPW x y
UTR VPW
180
180
180
`
`
`
`
`
`
c
c
c
+
+
+ +
+
+ + +
+
+ +
+ +
+ + =
=
= =
=
+ + =
+ + =
+ + = + +
=
URS UPR+ +=
Page 25 questions
PQ O
S
U
R
W
z
T
V
yx
Let and; .QPW x OPV y WQR z+ + += = =
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1618
Notes
13 100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1916
Notes
100% Geometry of the Circle - Tangents and Secants Solutions
Mathletics 100% © 3P Learning
Tangents and Secants
SERIES TOPIC
K 1620
Notes