Mathematics · Oct / Nov 2023 · Paper 21 · Question 7
YMA01
As printed in the paper
The question as text
7. y l C O x X (4, –3) N Figure 2 Figure 2 shows a sketch of • the circle C with centre X (4, –3) 5 55 • the line l with equation y x 2 2 Given that l is the tangent to C at the point N,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[3]
States or implies that gradient XN =
−
5
2
Uses (
)
4,
−
3
with a gradient of
±
to form equation of XN
5
2
(
)
y
+
3
= −
x
−
4
5
5
+
15
= −
2
+
8
2
+
5
+
7
=
0
*
y
x
x
y
5
55
(i) Solves 2
5
7
0
and
...,
...
7(b)[5]
x
+
y
+
=
y
=
x
−
x
=
y
=
2
2
(
)
N =
9,
−
5
(ii) Attempts (
)
(
)
''9'' 4
−
2
+
'' 5'' 3
−
+
2
Correct form of equation E.g.
(
)
2
(
)
2
(
)
2
(
)
2
−
4
+
+
3
=
''9'' 4
−
+
'' 5'' 3
−
+
x
y
(
)
(
)
−
4
2
+
+
3
2
=
29
x
y
(a) This is a given answer so be careful of students who work backwards.
2
B1: States or implies that gradient XN =
−
o.e such as 0.4
−
5
This cannot be scored from candidates who show you they are just working back from the given answer.
2
M1: Uses (
)
4,
−
3
with a gradient of
±
o.e. to form equation of XN
5
(
)
2
−−
3
2
y
E.g
±
=
. If the form y
=
mx
+ is used look for
c
−= ±
3
×+
4
leading to c = ....
c
5
−
4
5
x
A1*: Shows sufficient working and proceeds to 2
+
5
+
7
=
0
. There should be one correct line
x
y
2
(
)
showing working between
+
3
= −
−
4
and 2
+
5
+
7
=
0
, for example 5
+
15
y
x
x
y
y
5
7
2
7
If y
=
+ is used expect to see
= −
and
= −
−
before 2
+
5
+
7
=
0
.
mx
c
c
y
x
x
y
5
5
5
Note that this is a given answer so there cannot be incorrect work used leading to this answer.
(b)(i)
5
55
M1: Attempts to solve 2
5
7
0
and
leading to
x
+
y
+
=
y
=
x
−
x
=
...,
y
=
...
2
2
Allow the values to be just written down, for example via use of a calculator
(
)
A1:
N =
9,
−
5
which may be written
=
9,
= −
5
x
y
(b)(ii)
M1: Attempts radius or radius 2. Condone slips on lhs, for example
(
''9'' 4
)
2
(
'' 5'' 3
r or d =
−
+
−
You must see an attempt to take the difference for both components before each are squared.
M1: Correct form of equation E.g. (
)
(
)
2
−
4
2
+
+
3
2
=
'' ''
x
y
r
Look for (
)
(
)
(
)
(
)
−
4
2
+
+
3
2
=
''9'' 4
−
2
+
'' 5'' 3
−
+
2
x
y
Condone for this mark a slip on one of the components when finding their r 2
A1: (
)
2
(
)
2
or exact equivalent. The constant terms must be collected and not 25
x
−
4
+
y
+
3
=
29
ISW after a correct answer.
Alt (b)(ii)
Note that an alternative version is
2
+
2
−
8
+
6
−
4
=
0
. It is possible to proceed using this alternative
x
y
x
y
version of the formula for a circle
2
2
M1: States equation of circle is
x
+
y
−
8
x
+
6
y
+
c
=
0
(
9,
5
)
2
2
M1: Substitutes their
N =
−
into
x
+
y
−
8
x
+
6
y
+
c
=
0
o.e. and finds ''c''
A1:
2
+
2
−
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