How many marks are G10 questions worth?

In verified past-paper appearances, G10 questions on Edexcel GCSE Maths 1MA1 have been worth 3–5 marks.

Which past exam papers has G10 appeared in?

G10 has appeared in 11 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G10

Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results [Higher only]

Higher only

Circle theorems

Worked answers, methods and verified real exam appearances for G10 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only.
Standard circle theorems connect angles, radii, tangents and chords.
The angle at the centre is twice the angle at the circumference standing on the same arc; angles in the same segment are equal; an angle in a semicircle is $90^\circ$ ; and opposite angles in a cyclic quadrilateral total $180^\circ$ .
A radius is perpendicular to a tangent at the contact point, tangents from one external point are equal, and the alternate segment theorem equates the tangent-chord angle with the angle in the opposite segment.
A proof must identify the relevant common arc or chord and name each theorem used.

The angle at the centre is twice the angle at the circumference on the same arc.

Worked example

Points $A,B,C,D$ lie on a circle. $\angle ABC=73^\circ$ and $\angle BCD=41^\circ$ . Find $\angle ADC$ and $\angle BAD$ .

1.Opposite angles in cyclic quadrilateral $ABCD$ total $180^\circ$ .
2. $\angle ADC=180^\circ-73^\circ=107^\circ$ .
3. $\angle BAD=180^\circ-41^\circ=139^\circ$ .

Answer: $\angle ADC=107^\circ$ and $\angle BAD=139^\circ$ .

Common mistakes

Don't halve a central angle without checking that both angles stand on the same arc.
Don't use 'angles in the same segment' for angles on opposite sides of a chord.

Exam tip

A circle-theorem 'give a reason' mark needs the theorem's name or an unambiguous full statement.

Worked practice

Tier 1 · Easy

AB

is a diameter of a circle and

C

is another point on the circumference. Find

\angle ACB

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$90^\circ$	1	The angle in a semicircle is $90^\circ$ , so $\angle ACB=90^\circ$ .

Tier 2 · Standard

In a circle with centre

O

, the angle

AOB

124^\circ

. Point

C

lies on the major arc

AB

. Work out

\angle ACB

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$62^\circ$	2	Both angles stand on the minor arc $AB$ . The angle at the centre is twice the angle at the circumference, so $\angle ACB=124^\circ\div2=62^\circ$ .

Tier 3 · Hard

From a point

X

outside a circle with centre

W

, tangents touch the circle at

Y

and

Z

. Prove that

XY=XZ

. Given that

\angle YXW=34^\circ

, work out

\angle YWZ

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$XY=XZ$ $\angle YWZ=112^\circ$	5	Radii meet tangents at right angles, so triangles $WYX$ and $WZX$ are right-angled. They have equal hypotenuse $WX$ and equal radii $WY=WZ$ , so they are congruent by RHS. Corresponding parts give $XY=XZ$ and $\angle YXW=\angle WXZ=34^\circ$ . In triangle $WYX$ , $\angle YWX=180^\circ-90^\circ-34^\circ=56^\circ$ . Congruence gives $\angle XWZ=56^\circ$ , so $\angle YWZ=56^\circ+56^\circ=112^\circ$ .

Tier 1 · Easy

A tangent touches a circle at

T

, radius

OT

is drawn, and

L

is another point on the tangent. Write down

\angle OTL

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$90^\circ$	1	A radius is perpendicular to a tangent at the point of contact. Therefore $OT\perp TL$ and $\angle OTL=90^\circ$ .

Tier 2 · Standard

Opposite angles of a cyclic quadrilateral are

(3x+12)^\circ

and

(5x-8)^\circ

. Work out

x

and the sizes of these two angles.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x=22$ ; the angles are $78^\circ$ and $102^\circ$	3	Opposite angles in a cyclic quadrilateral total $180^\circ$ , so $3x+12+5x-8=180$ . Hence $8x+4=180$ , giving $x=22$ . The angles are $3(22)+12=78^\circ$ and $5(22)-8=102^\circ$ .

Tier 3 · Hard

A chord

AB

of a circle is not a diameter. Its midpoint is

M

and the centre is

O

. Prove that

OM

is perpendicular to

AB

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$OM\perp AB$	4	Draw the perpendicular from $O$ to $AB$ , meeting it at $N$ . Right-angled triangles $OAN$ and $OBN$ have equal hypotenuses $OA=OB$ because they are radii, and the common shorter side $ON$ . RHS congruence gives $AN=NB$ , so $N$ is the midpoint of $AB$ . The midpoint is unique, hence $N=M$ and the perpendicular $ON$ is $OM$ . Therefore $OM\perp AB$ .

Tier 2 · Standard

A tangent

AT

touches a circle at

A

. Chord

AB

is drawn, and point

C

lies on the circle in the opposite segment. Points

T

and

C

are on opposite sides of

AB

. Given

\angle TAB=43^\circ

and

\angle ABC=72^\circ

, work out

\angle ACB

and

\angle BAC

. Give a reason for each answer.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\angle ACB=43^\circ$ (alternate segment theorem); $\angle BAC=65^\circ$ (angles in a triangle total $180^\circ$ )	3	By the alternate segment theorem, the angle between tangent $AT$ and chord $AB$ equals the angle in the opposite segment, so $\angle ACB=43^\circ$ . Then $\angle BAC=180^\circ-43^\circ-72^\circ=65^\circ$ .

Tier 3 · Hard

The two tangents from a point

P

meet a circle, centre

O

, at the points

A

and

B

. The point

C

lies on the major arc

AB

. Given

\angle APB=74^\circ

, work out

\angle AOB

and

\angle ACB

. Give a reason for each stage.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$\angle AOB=106^\circ$ ; $\angle ACB=53^\circ$	4	Radii are perpendicular to tangents, so $\angle OAP=\angle OBP=90^\circ$ . The angles in quadrilateral $OAPB$ total $360^\circ$ , giving $\angle AOB=360^\circ-90^\circ-90^\circ-74^\circ=106^\circ$ . Since $C$ is on the major arc, $\angle ACB$ and $\angle AOB$ stand on the same minor arc $AB$ . The angle at the circumference is half the angle at the centre, so $\angle ACB=53^\circ$ .

Tier 3 · Hard

A circle contains diameter

AB

and another point

C

on its circumference. The tangent at

C

meets the extension of

AB

beyond

B

P

. Given

\angle CAB=28^\circ

, work out

\angle CPB

. Give a reason for each stage.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\angle CPB=34^\circ$	4	The angle in a semicircle gives $\angle ACB=90^\circ$ , so $\angle ABC=180^\circ-90^\circ-28^\circ=62^\circ$ . Since $BP$ is the extension of $BA$ , $\angle CBP=180^\circ-62^\circ=118^\circ$ . By the alternate segment theorem, the angle between tangent $CP$ and chord $CB$ is $\angle BCP=\angle CAB=28^\circ$ . Hence $\angle CPB=180^\circ-118^\circ-28^\circ=34^\circ$ . On the unit circle, $A=(-1,0)$ , $B=(1,0)$ , $C=(\cos56^\circ,\sin56^\circ)$ and $P=(\sec56^\circ,0)$ verify existence with $P$ beyond $B$ ; the stated arc side and extension fix the requested angle uniquely.

Q10

Tier 3 · Hard

A circle has centre

O

and radius

10

cm.

AB

is a horizontal diameter with

A

to the left of

O

. The tangent at

A

is drawn. Point

E

lies on

OB

with

OE=6

cm. The line through

E

parallel to the tangent meets the circle at

C

above

AB

and

D

below

AB

. Prove that

E

is the midpoint of

CD

, then work out

CD

and prove that triangle

ACD

is isosceles.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$EC=ED=8$ cm, so $E$ is the midpoint of $CD$ ; $CD=16$ cm; $AC=AD$ , so triangle $ACD$ is isosceles	5	A radius is perpendicular to a tangent, so $AB$ is perpendicular to the tangent at $A$ . The parallel through $E$ is therefore perpendicular to $AB$ , and $OE$ is the perpendicular from the centre to chord $CD$ ; it bisects the chord, so $EC=ED$ . In right triangle $OEC$ , $EC=\sqrt{10^2-6^2}=8$ cm, hence $CD=16$ cm. Since $AB$ is the perpendicular bisector of $CD$ and $A$ lies on $AB$ , $AC=AD$ . Coordinates $O=(0,0)$ , $A=(-10,0)$ , $E=(6,0)$ , $C=(6,8)$ and $D=(6,-8)$ verify the unique labelled configuration.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-11	2H	Q17	3	Allowed	Higher	QP MS
2024-11	1H	Q17	4	Non-calculator	Higher	QP MS
2022-11	1H	Q18	3	Non-calculator	Higher	QP MS
2019-11	1H	Q22	4	Non-calculator	Higher	QP MS
2023-11	3H	Q21	4	Allowed	Higher	QP MS
2022-06	2H	Q20	4	Allowed	Higher	QP MS
2019-06	2H	Q18	5	Allowed	Higher	QP MS
2024-06	2H	Q22	4	Allowed	Higher	QP MS
2021-11	2H	Q14	4	Allowed	Higher	QP MS
2022-06	3H	Q15	3	Allowed	Higher	QP MS
2022-11	3H	Q16	4	Allowed	Higher	QP MS

Other points in G Geometry and measures

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