Geometry Of The Circle Tangents & Secants GOT O T CC TGTS ...

SOLUTIONS

TANGENTS & SECANTSGEOMETRY OF THE CIRCLE

Geom

etry

Of

The

Circ

leTa

ngen

ts &

Sec

ants

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13 100% Geometry of the Circle - Tangents and Secants Solutions

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Tangents and Secants

SERIES TOPIC

K 116

Solutions Basics

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



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.




SERIES TOPIC

K 316

Solutions Knowing More

Page 7 questions




(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+ += .




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+ =

= =

=

= +

= -

= =


(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= .




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.




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

+ +

+

=

=






(Equal angles of isosceles OFDT )


(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




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

+ + + += + +

= + +

=





(RS bisects QSC+ ; QSC 70c+ = )

( PQST ; is iscosceles)

(Given)

(Angles opposite equal sides)

(Interior angles of PQST )



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




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




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)




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

=

=

=




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

` # #

`

`

=

=

=

=


(Products of intercepts on intersecting chords)




SERIES TOPIC

K 1116

Solutions Thinking Even More

Page 21 questions

(Alternate Segment Angle)

(Alternate segment angle)

(Angles in same segment on same arc)




(Right angle)

(Angles in T sum to 180c )


(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 .




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



(Rhombus has parallel sides)

(Alternate angles are equal)

(Given)

(Rhombus)

(Common side)

(SSS)




(Supplementary cointerior angles, ||EB DC )


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




SERIES TOPIC

K 1316


(Given)

(Given)

(RHS)


(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.




SERIES TOPIC

K 1614


Page 23 questions

J

KL

.13 6

M

37c.7 65

-

-

P

.10 2

O

N

(Equal radii)

(Given)

(Given)

(Given)

(RHS)

(RHS)



(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)




SERIES TOPIC

K 1516


Page 23 questions

c

d

Find the length of LK.

Find JOL+ .


(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` =


(Common side and radii)

(Given)

(RHS)

(RHS)

(SSS)



(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)




SERIES TOPIC

K 1616


Page 24 questions

(Alternate 's+ , ||JK LM )



(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+ + ++ + =




SERIES TOPIC

K 1716


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)



(Angles on a straight line)

(Vertically opposite)


(Substitution)


(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+ + += = =




SERIES TOPIC

K 1618

Notes




SERIES TOPIC

K 1916

Notes




SERIES TOPIC

K 1620

Notes

Geometry Of The Circle Tangents & Secants GOT O T CC TGTS ...

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Transcript of Geometry Of The Circle Tangents & Secants GOT O T CC TGTS ...