Mathematics · Oct / Nov 2024 · Paper 21 · Question 10
YMA01
As printed in the paper
The question as text
10. The circle C has equation 2 + y 2 + 4x – 30y + 209 = 0 x
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(a)[3]
x
2
+
y
2
+
4
x
−
30
y
+
209
=
0
(
)
(
)
±
2
2
+
±
15
2
...
=
0
x
y
(i)
Centre (−2, 15)
(ii)
Radius
20
10(b)[3]
(
)
2
(
)
2
y
=
mx
+
1
x
+
2
+
mx
+ −
1 15
=
20
or
1
2
(
1
)
2
4
30
(
1
)
209
0
y
=
mx
+
x
+
mx
+
+
x
−
mx
+
+
=
2
+
2
2
+
4
−
28
+
180
=
0
x
m x
x
mx
(
)
(
)
2
−
4
=
0
4
−
28
2
−
4 1
+
2
×
180
=
0
b
ac
m
m
(
)
(
)
2
4
−
28
−
4 1
+
2
180
=
0
16
−
224
+
784
2
−
720 720
−
2
=
0
m
m
m
m
m
2
2
−
7
−
22
=
0*
m
m
10(c)[4]
11
2
2
−
7
−
22
=
0
=
,
−
2
m
m
m
2
11
125
12
71
m
=
x
2
−
150
x
+
180
=
0
x
=
y
=
2
4
5
5
or
m
= −
2
5
x
2
+
60
x
+
180
=
0
x
= −
6
y
=
13
12
71
(
)
,
or
−
6, 13
oe
5
5
12
71
,
and
(
6, 13
)
−
oe
5
5
(a) Look out for answers to part (a) written in the body of the question.
M1:
Attempts to complete the square for both variables or states a centre of (±2, ±15)
For completing the square allow (
)
2
(
)
2
x
±
2
...
y
±
15
...
=
...
A1:
Centre (−2, 15) or e.g.
= −
2,
=
15.
Condone e.g. −2, 15 or [−2, 15]
x
y
20 or 2 5 which may be scored following (
)
2
(
)
2
A1:
For
x
±
2
+
y
±
15
=
20
Do not allow
±
20
but apply isw if a correct radius is seen which is then e.g.
converted to a decimal or simplified incorrectly.
(b) Mark (b) and (c) together.
M1:
Attempts to substitute y = mx + 1 into the given equation or their rearranged equation
to obtain an equation in m and x only. Condone slips as long as the intention is clear.
dM1: Attempts
b
2
−
4
ac
...0
where … is “=” or “>” or “<” etc. or equivalent e.g.
b
where
a
=
A
+
Bm
2,
b
=
C
+
Dm c
,
=
E
where A, B, C, D and E are nonzero
constants. Condone copying slips if the intention is clear if the above conditions are
met. Depends on the first M mark.
A1*: Obtains the printed answer with no errors e.g. must have had
2
−
4
=
0
or
b
ac
equivalent throughout and before the final line e.g. not
2
4
0
and brackets
b
−
ac
>
expanded. Note this is a given answer so must follow correct work.
(b) Alternative: Perpendicular distance from a point to a line:
ax
+
by
+
c
(
2, 15 ,
)
1
0
with
=
0
0
−
mx
−
y
+ =
d
2
+
2
a
b
2
15 1
−
m
−
+
(
)
20
=
20
2
+
1
=
(
2
+
14
)
2
m
m
2
+
1
m
16
m
2
−
56
m
−
176
=
0
2
m
2
−
7
m
−
22
=
0*
M1:
Substitutes into a correct distance formula with their centre and the given L and sets
equal to their radius.
dM1: Squares, multiplies up to obtain (
)
(
)
2
α
2
+
=
β
+
oe,
where A, B, C and D
Am
B
Cm
D
are nonzero.
A1*: Obtains the printed answer with no errors
(c)
M1:
Solves the given quadratic equation by any valid means, including calculator, to
obtain at least one value for m. May be implied by their value(s).
One correct value only with no incorrect work is sufficient for M1
Condone x = … for this mark.
M1:
Uses at least one of their values of m to attempt one position for P.
This must be a complete and correct method to find a position for P i.e. finds at
least one value for x and then the corresponding value for y correctly.
E.g. substitutes at least one value for x into their (
)
2
(
)
2
x
+
2
+
mx
+ −
1 15
=
20,
the resulting 3TQ by any method including a calculator and then uses the correct
y = mx + 1 to find the y value. Condone slips when e.g. simplifying their 3TQ
providing the complete method is correct.
A1:
At least one correct point. Allow as a coordinate pair or as x = …, y = … and allow
equivalent values e.g. (2.4, 14.2) etc.
A1:
Both points correct and no others. Allow as coordinate pairs or as x = …, y = …
(c) Alternative 1:
Finds the intersections of the possible tangents with the perpendiculars passing through the
centre of the circle:
11,
M1:
m
=
−
2
As above
2
11
2
2
161
(
)
For
m =
perpendicular is
−
15
= −
+
2
= −
+
y
x
y
x
2
11
11
11
2
161
11
12
71
−
+
=
+
1
=
,
=
x
x
x
y
11
11
2
5
5
or
1
1
(
)
For
m = − perpendicular is
2
−
15
=
+
2
=
+
16
y
x
y
x
2
2
1
+
16
= −
2
+
1
= −
6,
=
13
2 x
x
x
y
M1: For a complete method by
forming the equation of at least one of the perpendiculars, with the negative reciprocalgradient and the coordinates of their centre correctly placed
solving simultaneously with the corresponding tangent to find x or yfinding the corresponding x or y coordinateA1:
At least one correct point. Allow as a coordinate pair or as x = …, y = … and allow
equivalent values e.g. (2.4, 14.2) etc.
A1:
Both points correct and no others. Allow as coordinate pairs or as x = …, y = …
(c) Alternative 2:
11
=
+
1
y
x
2
(
)
−
2, 15
11
,
+
1
b
b
2
(
,
−
2
+
1
)
a
a
(
)
0, 1
= −
2
+
1
y
x
(
)
2
(
)
2
(
)
2
15 1
−
+
2
2
=
a
2
+
2
a
+
20
or
11
2
(
)
2
(
)
2
15 1
−
+
2
2
=
2
+
+
20
b
b
2
Then
a
2
=
36
a
= −−
6
2
a
+ =
1
13
or
144
12
11
71
b
2
=
b
=
b
+ =
1
25
5
2
5
Score as:
M1: As above
M1: For a complete method by
using Pythagoras correctly to find the distance or distance2 between (0, 1) and theircentre
applying Pythagoras correctly with a general point on either tangent and their radiusto obtain an equation in one variable and solves
finding the corresponding x or y coordinateA1:
At least one correct point. Allow as a coordinate pair or as x = …, y = … and allow
equivalent values e.g. (2.4, 14.2) etc.
A1:
Both points correct and no others. Allow as coordinate pairs or as x = …, y = …
General diagram for reference:
11
y
=
x
+
1
2
(
)
−
2, 15
y
= −
12 71
,
(
)
5
5
−
6, 13
= −
2
+
1
y
x
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