Mathematics · May / June 2021 · Paper 21 · Question 6
YMA01
As printed in the paper
The question as text
6. A circle has equation x2 – 6x + y2 + 8y + k = 0 where k is a positive constant. Given that the x-axis is a tangent to this circle,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
x
−
3
+
y
+
4
−
9 16
−
+
k
=
0
So radius must be 4
⇒
25
−
k
=
16
⇒
k
=
9
Way 2: y = 0
⇒
x −
2
6
x
+
k
=
0
has one solution
⇒
26
−×
4
1
×
k
=
0
⇒
k
=
9
6(b)[4]
C intersects x axis at (3,0)
Intersects y axis when
y
2
+
8
y +
9
=
0
⇒
y
=
.
..
or
y
= −
" 4"
±
"16" "9"
−
Or uses base of triangle is
2
"16" "9"
−
Area RST = 1
×
2 7
×
3
2
=
3
7
Notes:
(a) Way 1
M1: Attempts to complete square on both terms. May be implied by a centre of (
)
±
3, 4
±
M1: Uses a valid method to find k, e.g. y coordinate of centre and sets "9" "16" k
+
−
= |their “4”|2.
Can be implied by sight of k = (their x coordinate of centre)2
A1: Correct value for k
Way 2
M1: Substitutes equation for x axis (y = 0) into the circle equation.
M1: Sets the discriminant of the resulting quadratic to 0 (since as a tangent there is only one point of
intersection).
A1: Correct value for k
(b)
B1: Identifies or implies the coordinates where circle touches the x axis. Probably from having
completed the square on the equation. The 0 ordinate need not be seen and may be seen anywhere.
M1: Attempts the y coordinates of the intersections with the y axis, or the distance between these
two points. E.g. may set x = 0 in the circle equation, or may use centre is 3 units from y axis etc.
M1: Uses a correct method for the area of the triangle RST. Longer methods are possible, such as
use of cosine rule to find an angle in the triangle, followed by 1
2 ab
sin
C , or may use the shoelace
3
0
0
1
1 3
(
)
(
)
method, e.g.
4
7
0
0
0
0
3
4
7
=
−+
+
+
−
−
−
−−
2
0
−+
4
7
−−
4
7
2
A1: Correct answer.
oe seen or implied by working
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