Question 8:
--- 8(a) ---
8(a) | 2
  1   (y−(−1))2
 x− − p   + OE
  2  | B1* | 1  1 
Allow a=− p and b=−1,or centre is − p,−1.
2  2 
  1   2 (y−(−1))2  1  2
 x− − p   + =−q+1+ − P  OE
  2   2  | DB1
2
--- 8(b)(i) ---
8(b)(i) | 1
[Gradient of tangent =] −
2 | B1 | OE
SOI
[Gradient of normal =] 2 | M1 | Use of m m = −1with their numeric tangent gradient.
1 2
y−3
=2 y=2x−5 
x−4 | A1 | OE
ISW
Allowy=2x+c, 3=24+c  c=−5.
3
Question | Answer | Marks | Guidance
--- 8(b)(ii) ---
8(b)(ii) | Method 1 for the first two marks:
 1 
−1−3=2− p−4 or −1=−p−5
 2  | M1* |  p 
Using their stated centre or  , 1 in their equation of the
 2 
normal.
p=−4 | A1
Method 2 for the first two marks:
1
−1=2x−5  x=2  − p=2
2 | M1* | Using their normal equation and their stated centre or
 p 
 , 1.
 2 
p=−4 | A1
Method 3 for the first two marks:
dy dy  dy
2x+2y + p+2 =0  p=−8−8
 
dx dx  dx | M1*
dy 1 
=−  p=−4
 
dx 2  | A1
Question | Answer | Marks | Guidance
8(b)(ii) | Method 1 for the last 3 marks:
r2 =(4−2)2 + ( 3−(−1))2 =20  | M1* | their p 
Using (4, 3) and their centre or  , 1 to find r2 or r.
 2 
1
−q+1+ p2 =20
4 | DM1 | OE
Using their expression for r2 (from (a)) equated to their 20.
q=−15 | A1
Method 2 for the last 3 marks:
2−2−10  10
r= =
 
5  5 | M1* | Using (2,−1) and x+2y−10=0 (distance from a point to a
line).
2
1 10 
−q+1+ p2 =
 
4  5 | DM1 | OE
2
10 
Using their expression for r2 equated to their   .
 5
q=−15 | A1
Method 3 for the last 3 marks:
42 +32 +4p+6+q=0  4p+q+31=0 
OR
 4−   − 1 p     2 + ( 3−(−1))2 =−q+1+   − 1 p   2
  2   2  | M1* | (4,3)
Substituting into their circle equation.
4(−4)+q+31=0 | DM1 | Substituting their p=−4.
q=−15 | A1
Question | Answer | Marks | Guidance
8(b)(ii) | Alternative Method for Question 8(b)(ii)
42 +32 +4p+6+q=0
x2 +(2x−5)2+ px+2(2x−5)+q=0 with x=4
10−x 2 10−x
x2 +  + px+2 +q=0 with x=4
 2   2 
 y+5 2  y+5
  + y2 + p  +2y+q=0 with y=3
 2   2 
(10−2y)2 +y2 + p(10−2y)+2y+q=0 with y=3
Each of these  4p+q+31=0 | M1* | Substituting (4,3) into their circle equation, or
replacing y with 2x−5 from the normal equation, or
10−x
replacing y with from the tangent equation, or
2
y+5
replacing x with from the normal equation, or
2
replacing x with10−2y from the tangent equation, and using
either x=4 or y=3to form an equation in p and q.
5 5
x2 +(p−6)x+35+q=0  (p−6)2 −4 (35+q) = 0
4 4
OR
5y2−y(38+2p)+100+10p+q=0  (38+2p)2−45(100+10p+q)=0
 Each of these p2 −12p−139−5q=0  | M1* | Solving the tangent and circle equations simultaneously to form
a quadratic equation in either x or y.
Then usingb2 −4ac=0 on their quadratic to form an equation
in p and q.
Solving the equations simultaneously to find p or q | DM1
p=−4 | A1
q=−15 | A1
5
Question | Answer | Marks | Guidance