Mathematics · Oct / Nov 2025 · Paper 21 · Question 6

YMA01

As printed in the paper

The question as text

6. y C1 A(–2, 18) B(14, 6) x O Figure 2 The points A(–2, 18) and B(14, 6) lie on a circle C1 as shown in Figure 2. Given that AB is a diameter of the circle C1

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

6(a)[5]
−
2,18
6,12
(
)
(
)
Attempts radius2 or diameter2. E.g.
D =
2
−−
2
+
−
2
=
14
2
6 18
400
Radius2 = 100
(
)
2
(
)
2
x
−
6
+
y
−
12
=
100
6(b)[4]
Recognises equation of C2 is
2x
+
y
2
=
k
2
Attempts to find one value of k or
k Look for
2
2
2
''6''
+
''12''
±
''100''
(
)
(
)
2
2
x
2
+
y
2
=
6 5
+
10
or
x
2
+
y
2
=
6 5 10
−
o.e.
(
)
(
)
2
2
x
2
+
y
2
=
6 5
+
10
and
x
2
+
y
2
=
6 5
−
10
o.e.
(a) The demand here is that the method must be made clear
B1: Correctly finds the midpoint of the two given points to find the centre of the circle.
It can be implied within the circle formula(
)
2
(
)
2
2
x
−
6
+
y
−
12
=
r
M1: Attempts to find the radius, radius2 , diameter or diameter2.
It does not need to be correctly assigned to either the radius or the diameter
So (
)
2
(
)
2
2
2
2
2
14
−−
2
+
6
−
18
,
16
+
12
,
8
+
6
or radius = 10 are acceptable attempts
that score M1
A1: Correctly calculates the radius2 = 100 o.e. or the radius = 10
Alternatively states that the diameter = 20 or diameter2 = 400
It must be simplified and correctly assigned (radius or diameter) but may be
implied from within a circle formula(
)
(
)
x
±
'6'
2
+
y
±
'12'
2
=
100
Attempts correct form for circle using (
)
(
)
−
2
+
−
2
=
2
M1:
x
a
y
b
r
where (
)
,a b is their m.p. of AB and r is the value of what they think is the radius of
the circle.
If there is no statement i.e, radius(r) = … or diameter (d) = … , the value used must
be the equivalent of the distance between the midpoint of AB and either point
A or point B.
''400''
Look for (
)
2
(
)
2
or (
)
2
(
)
2
x
−
''6''
+
y
−
''12''
=
''100''
x
−
''6''
+
y
−
''12''
=
4
So if they clearly state something that is incorrect, e.g.
r
2
=
10
, thus losing the
previous A mark, they can score this method mark for (
)
2
(
)
2
x
−
6
+
y
−
12
(
A1: (
)
2
(
)
2
, (
)
2
(
)
2
2
but not (
)
2
(
)
2
x
−
6
+
y
−
12
=
100
x
−
6
+
y
−
12
=
10
x
−
6
+
y
−
12
=
100
ISW after a correct answer.
The method used must have been made clear for A1 to be scored. So, look for all
of the following
a correct calculation of the midpoint of AB or alternatively words such as
‘centre of circle is (
)
6,12 ’
a correct calculation for radius (radius2) or diameter (diameter2) or
alternatively words such as ‘radius of circle is 10’
a correct equation
(b)
B1: Recognises equation of C2 is
2x
+
y
2
=
k
2
. Allow for any
x
2
+
y
2
=
p
,
p
>
0
Accept (
)
(
)
Allow ‘letters’ here so long as the rhs is
x
±
0
2
+
y
±
0
2
=
p
,
p
>
0
positive, so
k is fine but p isn’t unless p is defined to be positive.
2
M1: Attempts to find at least one value for the radius of C2.
± for their (
)
(
)
For example, score for a radius of
+
−
2
+
−
2
=
a
2
b
2
r
x
a
y
b
r
(
)
(
)
2
2
A1:
2
+
2
=
+
2
+
2
=
−
or exact equivalents including versions
x
y
6 5
10
or
x
y
6 5 10
(
)
such as (
)
(
)
2
±
2
+
±
2
=
+
. ISW after sight of a correct answer.
x
0
y
0
6 5
10
Expanded equivalents are
2
2
2
2
but isw after
x
+
y
=
280
+
120 5 or
x
+
y
=
280
−
120 5
a correct answer
(
)
(
)
2
2
A1:
2
2
2
2
o.e. including versions such as
x
+
y
=
6 5
+
10
and
x
+
y
=
6 5
−
10
(
)
(
)
(
)
2
±
+
±
2
=
+
. ISW after sight of correct answers.
x
0
2
y
0
6 5
10
2
2
Expanded equivalents are
x
2
+
y
=
280
+
120 5 and
x
2
+
y
=
280
−
120 5
but isw
after a correct answer
(b) Longer alternative
into (
)
(
)
Substitutes
−
2
+
−
2
=
and solves
y
=
2
x
x
6
y
12
100
E.g. (
)
2
(
)
2
(
)
2
(
)
2
x
−
6
+
2
x
−
12
=
100
5
x
−
6
=
100
x
−
6
=
20
x
=
6
±
2 5
Hence points of intersection are(
)
and (
)
6
−
2 5,12
−
4 5
6
+
2 5,12
+
4 5
(
)
(
)
2
2
Radius for smaller circle is found via
r =
2
6
−
2 5
+
12
−
4 5
=
280 120 5
−
(
)
(
)
2
2
Radius for larger circle is found via
r =
2
6
+
2 5
+
12
+
4 5
=
280
+
120 5
The same scoring traits can be applied
B1: Recognises equation of C2 is
2x
2
2
. Allow for any (
)
(
)
+
y
=
k
x
±
0
2
+
y
±
0
2
=
p
,
M1: Attempts to find at least a value of r or a value of r2 for C2.
For example,
find at least one intersection point of the line through (0, 0) and their (6, 12) and
the circle
find the distance from the origin to one of the points of intersection
A1:
x
2
+
y
2
=
280
−
120 5 or
x
2
+
y
2
=
280
+
120 5
or exact equivalents
A1:
x
2
+
y
2
=
280
−
120 5 and
x
2
+
y
2
=
280
+
120 5
or exact equivalents
5
1
=
3
−
2
−
2
+
−
y
x
4
x
kx
28
x
44
5
1
3
1
−
d
y
k

Practise this paper →

Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.