Mathematics · Oct / Nov 2020 · Paper 21 · Question 4
YMA01
As printed in the paper
The question as text
4. The points P and Q have coordinates (–11, 6) and (–3, 12) respectively. Given that PQ is a diameter of the circle C,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[4]
x = −or
y =
9
7
(−7, 9) or e.g. x = −7, y = 9
x = −and
y =
9
7
Award the marks in (a) once correct answers are seen.
Special case: If all you see is (9, −7) award B1B0
(a)(ii)
Examples:
(
(
)
)
2
(
)
2
r =
−−
3
" 7"
−
+
12 "9"
−
Correct strategy for the radius. Must be
a correct method for their centre (if
or
used) but allow 1 sign slip within one
(
(
)
)
2
(
)
r =
−
11
−
" 7"
−
+
6 "9"
−
2
of the brackets. A correct answer scores
both marks. Must see the ½ if finding
or
the length of the diameter.
1
(
)
(
)
r =
−+
3 11
2
+
12
−
6
2
2
r = 5
Correct radius
(
)
2
(
)
2
x
+
7
+
y
−
9
=
5
2
or e.g.
x
2
+
y
2
+ ×
2 7
x
−×
2 9
y
+
7
2
+
9
2
−
5
2
=
0
M1: Correct attempt at circle equation using their values.
Allow for
(
(
)
)
2
(
(
)
)
2
(
)
2
x
±
their
−
7
+
y
±
their
9
=
their numerical r
or e.g.
(
)
(
)
(
)
(
)
(
)
x
2
+
y
2
± ×
2
their
−
7
x
± ×
2
their
9
y
+
their
−
7
2
+
their
9
2
−
their numerical r
2
=
0
A1: Correct equation in any form
12
−
9
3
m
=
=
or
radius
−+
3
7
4
This mark is for an attempt to find the
12
−
9
4
radius gradient or the tangent gradient.
m
= −÷
1
= −
or
tangent
−+
3
7
3
If the method is not clear allow one sign
error in the numerator or denominator.
−+
3
7
4
m
= −
= −
tangent
12
−
9
3
Alternative for the first M:
d
y
(
x
7
)
2
(
y
9
)
2
5
2
2
(
x
7
)
2
(
y
9
)
0
+
+
−
=
⇒
+
+
−
=
d
x
d
y
x
+
7
−+
3
7
4
=
=
= −
d
x
9
y
9 12
3
−
−
d
y
d
y
Allow for (
)
(
)
(
)
(
)
x
+
7
2
+
y
−
9
2
=
5
2
⇒
α
x
+
7
+
β
y
−
9
=
0
⇒
=
...
d
d
x
x
4
(
)
y
−
12
= −
x
+
3
3
Uses a correct straight line method for the tangent using the point Q. Must be fully correct
work here so must be a clear attempt at the tangent not the radius. So if the radius gradient
is found previously, must apply negative reciprocal rule to their radius gradient.
If using y = mx + c must reach as far as c = …
4
x
+
3
y
−
24
=
0
Allow any integer multiple
4(b)[2]
(
)
2
(
)
2
x
+
7
+
y
−
9
=
5
2
or e.g.
x
2
+
y
2
+ ×
2 7
x
−×
2 9
y
+
7
2
+
9
2
−
5
2
=
0
M1: Correct attempt at circle equation using their values.
Allow for
(
(
)
)
2
(
(
)
)
2
(
)
2
x
±
their
−
7
+
y
±
their
9
=
their numerical r
or e.g.
(
)
(
)
(
)
(
)
(
)
x
2
+
y
2
± ×
2
their
−
7
x
± ×
2
their
9
y
+
their
−
7
2
+
their
9
2
−
their numerical r
2
=
0
A1: Correct equation in any form
4(c)[3]
12
−
9
3
m
=
=
or
radius
−+
3
7
4
This mark is for an attempt to find the
12
−
9
4
radius gradient or the tangent gradient.
m
= −÷
1
= −
or
tangent
−+
3
7
3
If the method is not clear allow one sign
error in the numerator or denominator.
−+
3
7
4
m
= −
= −
tangent
12
−
9
3
Alternative for the first M:
d
y
(
x
7
)
2
(
y
9
)
2
5
2
2
(
x
7
)
2
(
y
9
)
0
+
+
−
=
⇒
+
+
−
=
d
x
d
y
x
+
7
−+
3
7
4
=
=
= −
d
x
9
y
9 12
3
−
−
d
y
d
y
Allow for (
)
(
)
(
)
(
)
x
+
7
2
+
y
−
9
2
=
5
2
⇒
α
x
+
7
+
β
y
−
9
=
0
⇒
=
...
d
d
x
x
4
(
)
y
−
12
= −
x
+
3
3
Uses a correct straight line method for the tangent using the point Q. Must be fully correct
work here so must be a clear attempt at the tangent not the radius. So if the radius gradient
is found previously, must apply negative reciprocal rule to their radius gradient.
If using y = mx + c must reach as far as c = …
4
x
+
3
y
−
24
=
0
Allow any integer multiple
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