Mathematics · May / June 2022 · Paper 21 · Question 10
YMA01
As printed in the paper
The question as text
10. The circle C has centre X (3, 5) and radius r The line l has equation y = 2x + k , where k is a constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(a)[3]
Equation of circle is
x
−
+
y
−
=
r
and line is
So intersect when (
)
2
(
)
2
x
−
3
+
2
x
+
k
−
5
=
r
2
2
6
9
4
2
4
(
5
)
(
5
)
2
2
x
−
x
+ +
x
+
k
−
x
+
k
−
=
r
(
)
5
x
2
+ −+
6
4
k
−
20
x
+ +
9
k
2
−
1
0
k
+
25
−
r
2
=
0
(
)
5
x
2
+
4
k
−
26
x
+
k
2
−
10
k
+
34
−
r
2
=
0*
(
)
(
)
10(b)[3]
Tangent to C
b
2
−
4
ac
=
0
4
k
−
26
2
−
4
××
5
k
2
−
10
k
+
34
−
r
2
=
0
1
6
k
2
−
208
k
+
676
−
20
k
2
+
200
k
−
6
80
+
20
r
2
=
0
5
r
2
=
. .
.
5
r
2
=
k
2
+
2
k
+ =
1
(
k
+
1)
2
1
1
(b)
(
)
Gradient of BX is
−
so equation of BX is
y
−= −
5
x
−
3
2
2
Way 2
1
13
−
2
k
26
+
k
5
(
3 ,
)
2
...,
...
,
y
−
= −
x
−
y
=
x
+
k
x
=
y
=
2
5
5
2
2
13
−
2
k
26
+
k
−
3
+
−
5
=
r
2
5
5
5
r
2
=
k
2
+
2
k
+ =
1
(
k
+
1)
2
(
)
2
(
)
2
10(c)[6]
Triangle AXB is right angled so
AB
2
+
r
2
=
XA
2
=
3
−
0
+
5
−
k
AB
2
=
4
r
2
so
AB
2
r
2
5
r
2
+
=
5
r
2
=
9
+
(
5
−
k
)
2
2
2
1
9
25 10
2
k
+
k
+ =
+
−
k
+
k
12
k
=
3
3
k
=
...
11
k =
4
Notes:
(a)
M1: Forms equation of circle and substitutes in equation of line.
(
3)
2
(
5)
2
2
The circle equation must be of the form
x
±
+
y
±
=
r
dM1: Expands both sets of brackets and collects terms in x2 and x.
A1*: Reaches the answer given with no errors seen.
Note that some candidates expand the brackets first before substitution e.g.:
(
)
2
(
(
x
−
3)
2
+
(
y
−
5)
2
=
r
2
x
2
−
6
x
+ +
9
y
2
−
10
y
+
25
=
r
2
x
2
−
6
x
+ +
9
2
x
+
k
−
10 2
x
+
k
This implies the first M and then the second M will score when terms in x2 and x are collected.
Note about poor squaring e.g.
(
x
−
3)
2
=
x
2
+
9
: The first M is available in both cases above but
the second M requires at least two x2 terms and at least two x terms from the expansions.
Note that it is acceptable to go from a completely correct full expansion to the printed answer e.g.
(
)
(
(
x
−
3)
2
+
(
y
−
5)
2
=
r
2
x
2
−
6
x
+ +
9
y
2
−
10
y
+
25
=
r
2
x
2
−
6
x
+ +
9
2
x
+
k
2
−
10 2
x
+
k
(
)
x
2
−
6
x
+ +
9
4
x
2
+
4
kx
+
k
2
−
20
x
−
10
k
+
25
=
r
2
5
x
2
+
4
k
−
26
x
+
k
2
−
10
k
+
34
−
Scores full marks in (a)
(b)
M1: Uses the discriminant is zero to form an equation in k and r
dM1: Expands and rearranges to make αr2 the subject
A1: Correct answer
Way 2
M1: Attempts the equation of BX and solves simultaneously with l to find the coordinates of B
b
13
−
2
k
Alternatively uses
x
= −
=
at B and uses this to find y at B
2
a
5
dM1: Correct use of Pythagoras for BX and sets = r2
A1: Correct answer
(c)
M1: Attempts XA2 correctly in terms of k (the k may appear as y but must be replaced by k later) and
uses it in Pythagoras theorem for triangle AXB.
May be implied.
2
2
M1: Applies AB = 2r to get
AB
+
r
in terms of r. Condone with AB2 = 2r2 used.
A1: Correct equation.
M1: Substitutes the result from (b) and expands brackets.
dM1: Solves a linear equation in k. Depends on the previous M.
A1: Correct value.
Alt (c)
b
13
−
2
k
13
−
2
k
At B
x
= −
=
,
y
=
2
+
k
2
a
5
5
13
2
2
26
4
2
−
k
−
k
AB
2
=
−
0
+
+
k
−
k
5
5
(
)
2
25 4
×
r
2
=
13
−
2
k
+
(26
−
4 )
k
2
=
...
(
)
2
20 (
×
k
+
1)
2
=
13
−
2
k
+
(26
−
4 )
k
2
=
20
k
2
−
260
k
+
845
169
k
2
+
2
k
+
1
=
k
2
−
13
k
+
k
=
..
.
4
11
k
=
4
Notes:
(c)
M1: Uses the result in (a) to find the x coordinate where the line and circle meet and then finds y
An alternative is to find the equation of BX as in (b) way 2 and solve with l to find x and y at B
(May have already found the coordinates of B in (a) but must be restated or used in (c) to score
this mark)
M1: Uses distance formula to find an expression in k for AB or AB2
A1: Correct expression for AB or AB2. Need not be simplified.
M1: Applies AB = 2r to the equation. Condone with AB2 = 2r2 used.
dM1: Substitutes the result from (b) and solves a linear equation in k. Depends on the previous M.
A1: Correct value.
Some more unusual methods for 3ii
Via perpendicular bisectors:
If AB is diameter centre must be midpoint of AB ie M 1 3 0 10
+
−
Way 5
,
2
2
=(
)
2, 5
−
Attempts 2 of:
−+
6 10
7
+
3
−−
6 10
=
(
)
and midpoint is
,
5, 8
−
m
=
=
1
BC
7
−
3
2
2
1
so perpendicular bisector is
(
)
y
+
"8"
= −
x
−
"5"
"1"
or
7 1
−
7
+
1
−
6
=
(
)
m
1
and midpoint is
,
4, 3
−
=
= −
AC
−−
6
0
2
2
1
so perpendicular bisector is
(
)
y
+
"3"
= −
x
−
"4"
" 1"
−
or
3 1
10
−
10
−
0
+
−
=
(
)
m
5
and midpoint is
,
2, 5
−
=
= −
AB
3 1
−
2
2
1
so perpendicular bisector is
(
)
y
+
"5"
= −
x
−
"2"
" 5"
−
1
(
)
oe or
y
+ =
3
x
− oe or
4
(
)
oe
y
+
8
= −
x
−
5
y
+
5
=
x
−
2
5
And solves simultaneously:
y
+ =
3
x
−
4,
y
+ = −−
8
5
x
5
x
−=
5
x
−
4
x
=
2
,
y
= −
5
E.g.
Hence centre of circle is (
)
2, 5
−
E.g. Midpoint of AB is the centre of the circle so AB is a diameter
(or equivalent)
M1: Attempts midpoint of AB. If no method is shown accept one correct coordinate as evidence.
A1: Correct midpoint
dM1: Attempts 2 perpendicular bisectors, and solves simultaneously
A1: Obtains (2, −5)
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded.
Via circle equation:
Uses(
)
2
(
)
2
Way 6
x
−
a
+
y
−
b
=
r
2
With (1, 0), (7, −6) and (3, −10)
To find (a, b) = … or r/r2 = …
Centre (2, −5) or radius
26
y
5
(
x
1
)
5 2 1
(
)
5
E.g. Equation of AB is
= −
−
and
−
−
= −
(
)
(
)
or
AB
=
3 1
−
2
+
−
1
0
−
0
2
=
10
4
=
2 2
6
or midpoint of AB is 1 3 0 10
+
−
= (
)
,
2, 5
−
2
2
So centre is on AB or AB is twice the radius or midpoint is the centre
hence AB is a diameter of the circle. (or equivalent)
M1: Uses all three points in circle equation to set up three equations in three unknowns to find
centre or radius.
A1: Correct centre or correct radius
dM1: Finds e.g. equation of AB, distance AB or midpoint of AB
A1: Correct equation of AB, distance AB or midpoint of AB
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded.
Via intersecting circles:
If AB is diameter centre must be midpoint of AB ie M 1 3 0 10
+
−
(ii)
,
2
2
Way 7
=(
)
2, 5
−
Circle centre C radius r is (
)
2
(
)
2
x
−
7
+
y
+
6
=
r
2
Circle centre B radius r is (
)
(
)
x
−
3
2
+
y
+
10
2
=
r
2
These intersect when (
)
2
(
)
2
(
)
2
(
)
2
x
−
7
+
y
+
6
=
x
−
3
+
y
+
10
x
+
y
= −
3
dM1A1
Circle centre A radius r is (
)
2
x
−
1
+
y
2
=
r
2
(
)
(
)
(
)
x
−
3
2
+
y
+
10
2
=
x
−
1
2
+
y
2
x
−
5
y
=
27
Solves simultaneously:
x
+
y
= −
3,
x
−
5
y
=
27
x
=
2,
y
= −
5
E.g. Midpoint of AB is the centre of the circle so AB is a diameter
(or equivalent)
M1: Attempts midpoint of AB. If no method is shown accept one correct coordinate as evidence.
A1: Correct midpoint
dM1: Attempts equations of 2 circles with A, B or C as centre with radius r, repeats the process for
2 different circles and finds the intersection of both straight lines and solves simultaneously
A1: Correct coordinates of centre
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded.
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