Sec 3 G2 Maths Circle properties worksheet with answers

A free Sec 3 G2 Maths worksheet on circle properties: 8 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Circle properties notes. Or practise it online with step-by-step help when you get one wrong.

Try a question: Circle properties

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 3 G2 Maths worksheet: Circle properties

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

In the diagram,

A

B

and

C

lie on the circle with centre

O

. Find

x

Answer°[1]

In the diagram,

A

B

C

and

D

lie on the circle with centre

O

. Find

x

Answer°[2]

In the diagram,

PA

and

PB

are tangents to the circle with centre

O

. Find

x

Answer°[2]

In the diagram,

A

B

C

and

D

lie on the circle and

BC

is produced to

E

. Find

x

Answer°[2]

In the diagram,

AB

is a chord of the circle with centre

O

, and

OM

is perpendicular to

AB

AB = 16

cm. Find

x

Answercm[3]

In the diagram,

AT

is a tangent to the circle with centre

O

A

, and

AB

is a chord. Find

x

Answer°[3]

Section B

In the diagram,

AB

and

CD

are parallel chords of the circle with centre

O

, on opposite sides of

O

AB = 8

cm and

CD = 6

cm. The line through

O

perpendicular to the chords meets

AB

M

and

CD

N

MN = 7

cm and

OM = x

cm.

(a)Write down the length of

AM

, giving a reason.

Answer (a)cm[1]

(b)Show that

x = 3

[3]

(c)Find the radius of the circle.

Answer (c)cm[2]

(d)Find angle

AOB

, correct to 1 decimal place.

Answer (d)°[2]

In the diagram,

AB

is a diameter of the circle with centre

O

and radius

5

cm.

C

is a point on the circle. The tangent to the circle at

C

meets

AB

produced at

T

. Angle

BAC = 24^\circ

(a)Find angle

ABC

Answer (a)°[2]

(b)Find angle

BOC

Answer (b)°[1]

(c)Find angle

BTC

Answer (c)°[2]

(d)Find angle

BCT

Answer (d)°[2]

(e)Find the length of

CT

, correct to 3 significant figures.

Answer (e)cm[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

x = 71

°[1]

\mathstrut \angle ACB = \dfrac{142^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

x = 71

Question 2

Answer:

x = 108

°[2]

\mathstrut \angle ADC = \dfrac{144^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

\angle ADC = 72^\circ

\mathstrut x = 180^\circ - 72^\circ

(∠s in opp. segments)

x = 108

Question 3

Answer:

x = 59

°[2]

PA = PB

(tangents from ext. pt.)So triangle PAB is isosceles.

\mathstrut x = \dfrac{180^\circ - 62^\circ}{2}

(base ∠s of isos. △)

x = 59

Question 4

Answer:

x = 79

°[2]

\mathstrut \angle BCD = 180^\circ - 79^\circ

(∠s in opp. segments)

\angle BCD = 101^\circ

\mathstrut x = 180^\circ - 101^\circ

(adj. ∠s on a str. line)

x = 79

So the exterior angle equals the interior opposite angle (ext. ∠ of cyclic quad.)

Question 5

Answer:

x = 6

cm[3]

AM = MB

(⊥ from centre bisects chord)

AM = 8

OM^2 = OA^2 - AM^2

\mathstrut x^2 = 10^2 - 8^2

x^2 = 36

x = \sqrt{36}

x = 6

Question 6

Answer:

x = 86

°[3]

\angle OAT = 90^\circ

(tan. ⊥ rad.)

\mathstrut \angle OAB = 90^\circ - 43^\circ

\angle OAB = 47^\circ

\angle OBA = 47^\circ

(base ∠s of isos. △)

\mathstrut x = 180^\circ - 2 \times 47^\circ

(∠ sum of △)

x = 86

Question 7

(a)Answer:

4

cm[1]

AM = \dfrac{8}{2}

\phantom{AM} = 4

cm (⊥ from centre bisects chord)

(b)[3]

CN = 3

(⊥ from centre bisects chord)

OA^2 = AM^2 + OM^2

\phantom{OA^2} = 4^2 + x^2

(Pythagoras' theorem)

ON = 7 - x

OC^2 = CN^2 + ON^2

\phantom{OC^2} = 3^2 + (7 - x)^2

OA = OC

(radii)

4^2 + x^2 = 3^2 + (7 - x)^2

16 + x^2 = 9 + 49 - 14x + x^2

14x = 42

x = 3

(shown)

(c)Answer:

5

cm[2]

OA^2 = 4^2 + 3^2

\phantom{OA^2} = 25

OA = \sqrt{25}

\phantom{OA} = 5

(d)Answer:

106.3^\circ

°[2]

\tan \angle AOM = \dfrac{AM}{OM}

\phantom{\tan \angle AOM} = \dfrac{4}{3}

\angle AOM = 53.1301\ldots^\circ

\angle AOB = 2 \times 53.1301\ldots^\circ

\phantom{\angle AOB} = 106.2602\ldots^\circ

\angle AOB = 106.3^\circ

Question 8

(a)Answer:

66^\circ

°[2]

\angle ACB = 90^\circ

(rt. ∠ in semicircle)

\angle ABC = 180^\circ - 90^\circ - 24^\circ

\phantom{\angle ABC} = 66^\circ

(∠ sum of △)

(b)Answer:

48^\circ

°[1]

\angle BOC = 2 \times 24^\circ

\phantom{\angle BOC} = 48^\circ

(∠ at centre = 2 ∠ at circumference)

(c)Answer:

42^\circ

°[2]

\angle OCT = 90^\circ

(tangent ⊥ radius)

\angle BTC = 180^\circ - 90^\circ - 48^\circ

\phantom{\angle BTC} = 42^\circ

(∠ sum of △)

(d)Answer:

24^\circ

°[2]

OB = OC

(radii)

\angle OCB = \angle OBC

\phantom{\angle OCB} = 66^\circ

(base ∠s of isos. △)

\angle BCT = 90^\circ - 66^\circ

\phantom{\angle BCT} = 24^\circ

(e)Answer:

5.55

cm[2]

\triangle OCT

\angle OCT = 90^\circ

and

\angle COT = 48^\circ

\tan 48^\circ = \dfrac{CT}{5}

CT = 5 \tan 48^\circ

\phantom{CT} = 5.5530\ldots

CT = 5.55

Sec 3 G2 Maths Circle properties worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Circle properties notes. Or practise it online with step-by-step help when you get one wrong.

Section B

In the diagram,

AB

and

CD

are parallel chords of the circle with centre

O

, on opposite sides of

O

AB = 8

cm and

CD = 6

cm. The line through

O

perpendicular to the chords meets

AB

M

and

CD

N

MN = 7

cm and

OM = x

cm.

(a)Write down the length of

AM

, giving a reason.

Answer (a)cm[1]

(b)Show that

x = 3

[3]

(c)Find the radius of the circle.

Answer (c)cm[2]

(d)Find angle

AOB

, correct to 1 decimal place.

Answer (d)°[2]

Answers and worked solutions

Question 1

Answer:

x = 71

°[1]

\mathstrut \angle ACB = \dfrac{142^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

x = 71

Question 2

Answer:

x = 108

°[2]

\mathstrut \angle ADC = \dfrac{144^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

\angle ADC = 72^\circ

\mathstrut x = 180^\circ - 72^\circ

(∠s in opp. segments)

x = 108

Question 3

Answer:

x = 59

°[2]

PA = PB

(tangents from ext. pt.)So triangle PAB is isosceles.

\mathstrut x = \dfrac{180^\circ - 62^\circ}{2}

(base ∠s of isos. △)

x = 59

Question 4

Answer:

x = 79

°[2]

\mathstrut \angle BCD = 180^\circ - 79^\circ

(∠s in opp. segments)

\angle BCD = 101^\circ

\mathstrut x = 180^\circ - 101^\circ

(adj. ∠s on a str. line)

x = 79

So the exterior angle equals the interior opposite angle (ext. ∠ of cyclic quad.)

Question 5

Answer:

x = 6

cm[3]

AM = MB

(⊥ from centre bisects chord)

AM = 8

OM^2 = OA^2 - AM^2

\mathstrut x^2 = 10^2 - 8^2

x^2 = 36

x = \sqrt{36}

x = 6

Question 6

Answer:

x = 86

°[3]

\angle OAT = 90^\circ

(tan. ⊥ rad.)

\mathstrut \angle OAB = 90^\circ - 43^\circ

\angle OAB = 47^\circ

\angle OBA = 47^\circ

(base ∠s of isos. △)

\mathstrut x = 180^\circ - 2 \times 47^\circ

(∠ sum of △)

x = 86

Question 7

(a)Answer:

4

cm[1]

AM = \dfrac{8}{2}

\phantom{AM} = 4

cm (⊥ from centre bisects chord)

(b)[3]

CN = 3

(⊥ from centre bisects chord)

OA^2 = AM^2 + OM^2

\phantom{OA^2} = 4^2 + x^2

(Pythagoras' theorem)

ON = 7 - x

OC^2 = CN^2 + ON^2

\phantom{OC^2} = 3^2 + (7 - x)^2

OA = OC

(radii)

4^2 + x^2 = 3^2 + (7 - x)^2

16 + x^2 = 9 + 49 - 14x + x^2

14x = 42

x = 3

(shown)

(c)Answer:

5

cm[2]

OA^2 = 4^2 + 3^2

\phantom{OA^2} = 25

OA = \sqrt{25}

\phantom{OA} = 5

(d)Answer:

106.3^\circ

°[2]

\tan \angle AOM = \dfrac{AM}{OM}

\phantom{\tan \angle AOM} = \dfrac{4}{3}

\angle AOM = 53.1301\ldots^\circ

\angle AOB = 2 \times 53.1301\ldots^\circ

\phantom{\angle AOB} = 106.2602\ldots^\circ

\angle AOB = 106.3^\circ

Question 8

(a)Answer:

66^\circ

°[2]

\angle ACB = 90^\circ

(rt. ∠ in semicircle)

\angle ABC = 180^\circ - 90^\circ - 24^\circ

\phantom{\angle ABC} = 66^\circ

(∠ sum of △)

(b)Answer:

48^\circ

°[1]

\angle BOC = 2 \times 24^\circ

\phantom{\angle BOC} = 48^\circ

(∠ at centre = 2 ∠ at circumference)

(c)Answer:

42^\circ

°[2]

\angle OCT = 90^\circ

(tangent ⊥ radius)

\angle BTC = 180^\circ - 90^\circ - 48^\circ

\phantom{\angle BTC} = 42^\circ

(∠ sum of △)

(d)Answer:

24^\circ

°[2]

OB = OC

(radii)

\angle OCB = \angle OBC

\phantom{\angle OCB} = 66^\circ

(base ∠s of isos. △)

\angle BCT = 90^\circ - 66^\circ

\phantom{\angle BCT} = 24^\circ

(e)Answer:

5.55

cm[2]

\triangle OCT

\angle OCT = 90^\circ

and

\angle COT = 48^\circ

\tan 48^\circ = \dfrac{CT}{5}

CT = 5 \tan 48^\circ

\phantom{CT} = 5.5530\ldots

CT = 5.55