Circle Theorems - The Oakwood Academy HIGHER Circl… · O is the centre of the circle D is the point at the end of PA extended B2 Any 2 correct B1 Any 1 correct B3 Writes a correct

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Circle Theorems

(Proof Questions/Linked with other

Topics) (G10)

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Q1.(a) The diagram shows a circle, centre O, with diameter AB.

Not drawn accurately

Work out the size of angle x You must show your working, which may be on the diagram.

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Answer ........................................................ degrees (2)

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(b) The diagram shows a circle touching a square at A, B, C and D.


Give reasons to show why y = 45°

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(Total 5 marks)

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Q2.

A, B and C are points on the circumference of a circle.

• BC is a diameter

• BCP is a straight line

• AP is a tangent to the circle

• PC = CA

Work out the value of angle CPA, marked x on the diagram.

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x = ..................................................................... degrees (Total 5 marks)

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Page 5

Q3.

R, S and T are on the circumference of a circle, centre O.

(a) Give a reason why angle OTS = x

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(b) Work out the value of x.

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Answer................................................................ degrees (3)

(Total 4 marks)

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Page 6

Q4.

ABP and ADQ are tangents to the circle, centre O.

C lies on the circumference of the circle.

Prove that y = 2x

Give reasons for any statements you make.

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Q5. A, B and C are points on a circle.

• BC bisects angle ABQ. • PBQ is a tangent to the circle.


Angle CBQ = x

Prove that AC = BC

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Page 8

Q6. (a) A, B and C are points on a circle, centre O.


AB is a diameter.

The ratio of the size of angle x to the size of angle y is

x : y = 5 : 1

Work out the size of angle z.

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Answer ........................................................ degrees (3)

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(b) L, M and N are points on a circle. PLQ is a tangent.


Work out angle MLN.

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Answer ........................................................ degrees (3)

(Total 6 marks)

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Page 10

Q7.

ABCD is a cyclic quadrilateral.

Work out the values of x and y.

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x = ............................. , y = ............................ (Total 5 marks)

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Q8.The diagram shows a circle centre O.

A and C are points on the circumference. AB and CB are tangents.


(a) Work out the size of angle x.

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Answer ........................................................ degrees (2)

(b) Write down the length of BC. Give a reason for your answer.

Answer ........................................................................................... cm

Reason...........................................................................................................

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(c) Work out the radius of the circle.

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Answer ................................................................ cm (3)

(Total 6 marks)

Q9.The diagram shows a circle, centre O. AB is a tangent.


Work out the length OB.

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Answer ................................................................ cm (Total 4 marks)

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Page 13

Q10.

In the diagram, AB = BC

Prove that ABCD is a cyclic quadrilateral.

Give reasons for any statements you make.

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M1.(a) OCA = 36

or ACB = 90

or COA = 108

or COB = 72

or OBC = 54

or 90 − 36

or (180 − 72) ÷ 2

oe

May be on diagram M1

54 A1

(b) (Triangle) RDC is isosceles

or RC and RD are equal tangents

May be implied from 90 and 45 in triangle RDC B1

Angle RDC = y

or Angle RCD = y B1

Angle RDC or Angle RCD is 45

and

alternate segment (theorem) stated

Strand (ii)

Complete reasons with both B marks scored Q1

[5]

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M2. Correct expressions or value for any three of these angles

angle PAC = x

angle CAB = 90

angle PBA = x

angle PCA = 180 − 2x or 90 + x

angle ACB = 90 − x or 2x

angle COA = 2x or 90 − x

angle PAO = 90

angle CAO = 90 − x or 2x

angle BAD = 2x or 90 − x

angle AOB = 180 − 2x or 90 + x

angle OAB = x

O is the centre of the circle

D is the point at the end of PA extended

B2 Any 2 correct

B1 Any 1 correct B3

Writes a correct equation that has solution 30

e.g. 1 PAC + CAB + x + PBA = 180

e.g. 2 PCA + ACB = 180

e.g. 3 ACB + CAB + CBA = 180

e.g. 4 PAO + APC + POA = 180

oe M1

30 A1

[5]

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M3. (a) Valid reason

e.g.1 Triangle OTS is isosceles

e.g.2 OT = OS

e.g.3 OT and OS are radii B1

(b) Correct equation

e.g.1 5x = 2(x + 30)

e.g.2 2.5x = x + 30

e.g.3 (180 − 2x) + 120 + 5x = 360

e.g.4 x + 30 + x + 30 + 360 − 5x = 360

oe

Brackets not needed in e.g.3 M1

Collects terms for their initial equation

e.g.1 5x − 2x = 60

e.g.2 2.5x − x = 30

e.g.3 − 2x + 5x = 360 − 180 − 120

oe

their initial equation must have ≥ 2 terms in x

Any brackets must be expanded correctly M1

20 A1

[4]

M4. Join BD

Angle BDC = 2x

Alternate segment theorem M1

Angle BDO = x M1

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Angle DBO = x

Isosceles triangle BOD M1

Angle BOD = 180 − 2x

Angle sum of triangle BOD M1

y = 360 − 90 − 90 − (180 − 2x)

y = 2x

Angle sum of quadrilateral ABOD

y = 2x clearly shown from simplification A1

Must have at least two different reasons stated in the proof B1ft

Alternative method 1

Angle OBC = 90 − 2x

Tangent-radius property M1

Angle OCB = 90 − 2x

Isosceles Δ OBC M1

Angle OCD = x

Isosceles Δ OCD M1

Angle BCD = 90 – 2x + x

= 90 – x hence


Angle at centre = 2 × angle at circumference M1

y = 360 − 90 − 90 − (180 − 2x)

y = 2x





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Isosceles Δ OBC M1

Angle OCD = x

Isosceles Δ OCD M1

Angle BCD = 90 − 2x + x

= 90 − x hence


Angle at centre = 2 × angle at circumference M1

Angle BOD = 360 − 90 − 90 − y

= 180 − y

hence y = 2x








Isosceles Δ OBC M1

Angle OCD = x

Isosceles Δ OCD M1

Angle BCD = 90 − 2x + x

= 90 – x M1

y = 360 − 90 − (90 − 2x) − (90 − x) − x − 90

hence y = 2x

Angle sum of quadrilateral ABCD

y = 2x clearly shown from simplification

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Page 19

A1



Angle BOD = 180 − y

Angle sum of quadrilateral ABOD M1

Angle OCD = x

Isosceles Δ OCD M1



Angle BCO = 90 − 2x

hence

Angle BOD reflex = 360 − (90 − 2x) − (90 − 2x) − x − x = 180 + 2x

Isosceles Δ OBC

Angle sum of quadrilateral BODC

... this can also come from Angle BOC (4x) + Angle DOC

(180 − 2x) M1

180 − y + 180 + 2x = 360

hence y = 2x

Angles round a point

y = 2x clearly shown from rearranging A1


[6]

M5.

angle ABC = x M1

angle BAC = x and

alternate segment theorem M1

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angle ABC = x and

angle BAC = x and

alternate segment theorem and two equal angles so isosceles (AC = BC) A1

[3]

M6. (a) 90 seen or implied

B1

90 ÷ 6 or 15

or 90 ÷ 6 × 5 or 75

oe M1

30 A1

Additional Guidance 30 without working

B1M1A1

(b) Angle LMN = 80 or angle MLP = 58

May be on diagram M1

180 − 80 − 58

oe M1

42 A1

[6]

M7. Any one of these equations

2x + y + 20 = 180

or

x + 2y + y + 40 = 180

or

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Page 21

2x + y + 20 = x + 2y + y + 40

or

2x + y + 20 + x + 2y + y + 40 = 360

oe M1

Another of these equations

2x + y + 20 = 180

or

x + 2y + y + 40 = 180

or

2x + y + 20 = x + 2y + y + 40

or

2x + y + 20 + x + 2y + y + 40 = 360

oe

these simplify to ...

2x + y = 160 or

x + 3y = 140 or

x − 2y = 20 or

3x + 4y = 300 M1

equating coefficients and elimination of x or y for their equations

e.g.

x + 3y = 140 and 6x + 3y = 480

or

2x + 6y = 280 and 2x + y = 160

rearrangement and substitution for their equations

e.g.

y = 160 − 2x and x + 3(160 − 2x) = 140

or

x = 140 − 3y and 2(140 − 3y) + y = 160 M1dep

Allow one numerical error for the 3rd M1, but not an error in method (e.g. adding equations when they ought to be subtracted is an error in method, so M0)

5x = 340 or 5y = 120

ft their elimination or substitution M1dep

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Page 22

x = 68 and y = 24 A1

[5]

M8.(a) 180 − 90 − 74

or 90 − 74 M1

16 A1

(b) 8.7 and tangents from the same point (are equal) oe B1

(c) tan 74 =

=

or tan 16 = M1

or 8.7 tan 16 M1dep

2.49(…) or 2.5

ft from part (a) A1ft

[6]

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Page 23

M9.90 seen or implied

90 may be on diagram

or may implied by use of Pythagoras or trigonometry M1

8.32 + 5.22

sin 32.(067…) or cos 57.(9326…) =

or cos 32.(067…) or sin 57.(9326…) = M1

or M1dep

9.79 … or 9.8

Accept 10 if working seen A1

[4]

M10.

∠ACB = x and

(Triangle ABC is) isosceles

oe M1

∠ABC = 180 − 2x and Angle sum of triangle (is 180°)

oe

∠CAD + ∠ACD = 180 − 2x and Angle sum of triangle (is 180°)

M1

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Page 24

180 − 2x + 2x = 180

and Opposite angles of cyclic quadrilateral (add up to 180°)

Must have seen working for both M marks

oe e.g. ∠ABC + ∠ADC = 180 and

Opposite angles of cyclic quadrilateral

SC2 ‘Correct’ solution with one reason missing

SC1 ‘Correct’ solution with > 1 reason missing A1

[3]

Documents

Circle Theorems - The Oakwood Academy HIGHER Circl… · O is the centre of the circle D is the point at the end of PA extended B2 Any 2 correct B1 Any 1 correct B3 Writes a correct