CAIE P1 2024 November — Question 10 9 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypePerpendicular bisector of chord
DifficultyStandard +0.3 This is a standard circle geometry problem requiring perpendicular bisector concepts and simultaneous equations. Part (a) involves finding the perpendicular bisector of AB (routine calculation), while part (b) requires solving a quadratic to find two circle centers. The techniques are straightforward applications of GCSE/AS coordinate geometry with no novel insight needed, making it slightly easier than average for A-level.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle
Points \(A\) and \(B\) have coordinates \((4, 3)\) and \((8, -5)\) respectively. A circle with radius 10 passes through the points \(A\) and \(B\).
  1. Show that the centre of the circle lies on the line \(y = \frac{1}{2}x - 4\). [4]
  2. Find the two possible equations of the circle. [5]
Question 10:
AnswerMarks
10(a)−5−3
Gradient of AB = =−2 
AnswerMarks
8−4M1*
8+4 −5+3
Midpoint AB =  ,    (6,−1) 
AnswerMarks
 2 2 M1
1  1
Gradient of normal =− = and an attempt to find the required
 
−2  2
AnswerMarks Guidance
equationDM1 Must be used to find equation of perpendicular through their
(6,−1).
1 1
Equation of perpendicular bisector is y+1= (x−6), so y= x−4
AnswerMarks Guidance
2 2A1 WWW
AG – working involving the perpendicular bisector must be
seen.
Alternative Method for Question 10(a)
AC2=(a−4)2 +(b−3)2 BC2=(a−8)2+(b+5)2
AnswerMarks Guidance
, both expandedM1*
Solving AC = BC [= 10]DM1 Only allow a single sign error.
Eliminating a2 and b2DM1 May be awarded before the previous DM1.
x
a=2b+8, concluding y= −4
AnswerMarks Guidance
2A1 WWW
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
10(b) 1 
Using the centre as a, a−4
AnswerMarks Guidance
 2 M1 May see centre as (2y + 8, y) OE.
May be seen in an incorrect equation.
AnswerMarks Guidance
(4−a)2+ (3−0.5a+4)2 =100M1 Sub in (4, 3) or (8, −5).
Could use circle with(6,−1) and r= 80.
AnswerMarks Guidance
1.25a2 −15a−35 =0   a2 −12a−28 =0  ( or b2 +2b−15 =0 )DM1 Obtain a 3-term quadratic in their x or y.
 (a−14)(a+2)=0  a=14, a=−2
AnswerMarks Guidance
 A1 Or   (b−3)(b+5)= 0    b=3, b=−5.
⇒ (x−14)2+ (y−3)2 =100 and (x+2)2+ (y+5)2 =100A1
Alternative Method 1 for the first 3 marks:
AnswerMarks Guidance
Make a or b the subject from a circle centre(a,b)using A or BM1 E.g.b= 100−(y−3)2 +4from circle through A.
These equations may have been found in part (a).
AnswerMarks Guidance
Form an equation in a or b onlyM1 Substitute their a or b into their second circle equation.
Simplify to a quadratic in a or bDM1 Expect a2 −12a−28=0 or b2 +2b−15=0, OE.
Alternative Method 2 for the first 3 marks:
AnswerMarks Guidance
Obtaining CM (C, centre; M, mid-point of AB)M1 Expect 80.Must be clear this is CM, not AB.
Using the triangle CMT, where CT is parallel to the x-axis, to find the
AnswerMarks Guidance
vertical distance of C from M, MTDM1 Expect MT =4.
Using the triangle CMT, where MT is parallel to the y-axis, to find the
AnswerMarks Guidance
horizontal distance of C from M, CTDM1 ExpectCT =8.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 10:
--- 10(a) ---
10(a) | −5−3
Gradient of AB = =−2 
8−4 | M1*
8+4 −5+3
Midpoint AB =  ,    (6,−1) 
 2 2  | M1
1  1
Gradient of normal =− = and an attempt to find the required
 
−2  2
equation | DM1 | Must be used to find equation of perpendicular through their
(6,−1).
1 1
Equation of perpendicular bisector is y+1= (x−6), so y= x−4
2 2 | A1 | WWW
AG – working involving the perpendicular bisector must be
seen.
Alternative Method for Question 10(a)
AC2=(a−4)2 +(b−3)2 BC2=(a−8)2+(b+5)2
, both expanded | M1*
Solving AC = BC [= 10] | DM1 | Only allow a single sign error.
Eliminating a2 and b2 | DM1 | May be awarded before the previous DM1.
x
a=2b+8, concluding y= −4
2 | A1 | WWW
4
Question | Answer | Marks | Guidance
--- 10(b) ---
10(b) |  1 
Using the centre as a, a−4
 2  | M1 | May see centre as (2y + 8, y) OE.
May be seen in an incorrect equation.
(4−a)2+ (3−0.5a+4)2 =100 | M1 | Sub in (4, 3) or (8, −5).
Could use circle with(6,−1) and r= 80.
1.25a2 −15a−35 =0   a2 −12a−28 =0  ( or b2 +2b−15 =0 ) | DM1 | Obtain a 3-term quadratic in their x or y.
 (a−14)(a+2)=0  a=14, a=−2
  | A1 | Or   (b−3)(b+5)= 0    b=3, b=−5.
⇒ (x−14)2+ (y−3)2 =100 and (x+2)2+ (y+5)2 =100 | A1
Alternative Method 1 for the first 3 marks:
Make a or b the subject from a circle centre(a,b)using A or B | M1 | E.g.b= 100−(y−3)2 +4from circle through A.
These equations may have been found in part (a).
Form an equation in a or b only | M1 | Substitute their a or b into their second circle equation.
Simplify to a quadratic in a or b | DM1 | Expect a2 −12a−28=0 or b2 +2b−15=0, OE.
Alternative Method 2 for the first 3 marks:
Obtaining CM (C, centre; M, mid-point of AB) | M1 | Expect 80.Must be clear this is CM, not AB.
Using the triangle CMT, where CT is parallel to the x-axis, to find the
vertical distance of C from M, MT | DM1 | Expect MT =4.
Using the triangle CMT, where MT is parallel to the y-axis, to find the
horizontal distance of C from M, CT | DM1 | ExpectCT =8.
5
Question | Answer | Marks | Guidance
Points $A$ and $B$ have coordinates $(4, 3)$ and $(8, -5)$ respectively. A circle with radius 10 passes through the points $A$ and $B$.

\begin{enumerate}[label=(\alph*)]
\item Show that the centre of the circle lies on the line $y = \frac{1}{2}x - 4$. [4]

\item Find the two possible equations of the circle. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q10 [9]}}
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