The Oakwood Academy
Circle Theorems
(Proof Questions/Linked with other
Topics)
Page 1
(G10)
The Oakwood Academy
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)
Page 2
The Oakwood Academy
(b)
The diagram shows a circle touching a square at A, B, C and D.
Not drawn accurately
Give reasons to show why y = 45°
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(3)
(Total 5 marks)
Page 3
The Oakwood Academy
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)
Page 4
The Oakwood Academy
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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(1)
(b)
Work out the value of x.
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Answer................................................................ degrees
(3)
(Total 4 marks)
Page 5
The Oakwood Academy
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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(Total 6 marks)
Page 6
The Oakwood Academy
Q5.
A, B and C are points on a circle.
•
•
BC bisects angle ABQ.
PBQ is a tangent to the circle.
Not drawn accurately
Angle CBQ = x
Prove that AC = BC
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(Total 3 marks)
Page 7
The Oakwood Academy
Q6.
(a)
A, B and C are points on a circle, centre O.
Not drawn accurately
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)
Page 8
The Oakwood Academy
(b)
L, M and N are points on a circle.
PLQ is a tangent.
Not drawn accurately
Work out angle MLN.
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Answer ........................................................ degrees
(3)
(Total 6 marks)
Page 9
The Oakwood Academy
Q7.
ABCD is a cyclic quadrilateral.
Work out the values of x and y.
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x = ............................. , y = ............................
(Total 5 marks)
Page 10
The Oakwood Academy
Q8.The diagram shows a circle centre O.
A and C are points on the circumference.
AB and CB are tangents.
Not drawn
accurately
(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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(1)
Page 11
The Oakwood Academy
(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.
Not drawn
accurately
Work out the length OB.
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Answer ................................................................ cm
(Total 4 marks)
Page 12
The Oakwood Academy
Q10.
In the diagram, AB = BC
Prove that ABCD is a cyclic quadrilateral.
Give reasons for any statements you make.
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(Total 3 marks)
Page 13
The Oakwood Academy
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]
Page 14
The Oakwood Academy
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]
Page 15
The Oakwood Academy
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
Page 16
The Oakwood Academy
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 BOD = 180 − 2x
Angle at centre = 2 × angle at circumference
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 2
Page 17
The Oakwood Academy
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 BOD = 180 − 2x
Angle at centre = 2 × angle at circumference
M1
Angle BOD = 360 − 90 − 90 − y
= 180 − y
hence 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 3
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
M1
y = 360 − 90 − (90 − 2x) − (90 − x) − x − 90
hence y = 2x
Angle sum of quadrilateral ABCD
y = 2x clearly shown from simplification
Page 18
The Oakwood Academy
A1
Must have at least two different reasons stated in the proof
B1ft
Alternative method 4
Angle BOD = 180 − y
Angle sum of quadrilateral ABOD
M1
Angle OCD = x
Isosceles Δ OCD
M1
Angle OBC = 90 − 2x
Tangent-radius property
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
Must have at least two different reasons stated in the proof
B1ft
[6]
M5.
angle ABC = x
M1
angle BAC = x and
alternate segment theorem
M1
Page 19
The Oakwood Academy
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
Page 20
The Oakwood Academy
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
Page 21
The Oakwood Academy
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]
Page 22
The Oakwood Academy
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
Page 23
The Oakwood Academy
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]
Page 24