CAIE P1 2023 November — Question 11 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeCircle through three points using right angle in semicircle
DifficultyStandard +0.3 This is a straightforward multi-part question testing standard techniques: perpendicular gradients to find p, then finding a circle through three points (likely using perpendicular bisectors or the general circle equation), and finally finding a tangent using the radius perpendicular to tangent property. All are routine P1/AS-level procedures with no novel insight required, making it slightly easier than average.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents
The coordinates of points \(A\), \(B\) and \(C\) are \((6, 4)\), \((p, 7)\) and \((14, 18)\) respectively, where \(p\) is a constant. The line \(AB\) is perpendicular to the line \(BC\).
  1. Given that \(p < 10\), find the value of \(p\). [4]
A circle passes through the points \(A\), \(B\) and \(C\).
  1. Find the equation of the circle. [3]
  2. Find the equation of the tangent to the circle at \(C\), giving the answer in the form \(dx + ey + f = 0\), where \(d\), \(e\) and \(f\) are integers. [3]
Question 11:
AnswerMarks
11(a) 7−4  18−7
their their =−1
 p−6  14− p
AnswerMarks Guidance
OR Scalar product leading to (14− p)(6− p)−33=0*M1 Difference in the ys
Their gradients must both come from .
Difference in the xs
AnswerMarks Guidance
p2 −20p+84=33 leading to p2−20p+51 =0  or p2−20p=−51A1 Clearing of fractions and collecting terms to arrive at the
three-term quadratic. Allow integer multiples.
Alternative method for first 2 marks of Question 11(a)
(p−6)2+(7−4)2+(14− p)2+(18−7)2 =(14−6)2+(18−4)2
OR
AnswerMarks Guidance
E.g. (10− p)2 + 42 = 42+72*M1 For correct use of Pythagoras with A,B and C.
OR
For correct use of Pythagoras with the centre, B and one of
the other two points.
AnswerMarks Guidance
2p2−40p+102 =0 A1 OE
Collecting terms to arrive at the three-term quadratic.
20 202 −451
[2](p−3)(p−17) or
AnswerMarks Guidance
2DM1 OE
Solving their three-term quadratic.
AnswerMarks Guidance
p=3A1 If M1A1DM0 scored then SC B1 is available for final
answer.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
11(b)[Midpoint or Centre is] (10, 11) B1
1
(14−6)2 +(18−4)2 (18−their11)2 +(14−their10)2
or or
2
(their11−4)2 +(their10−6)2 r2 =65 or r= 65
AnswerMarks Guidance
 M1 Finding half of the length of AC or using their centre, which
cannot be A, B or C, to find r2 or r. Note: r=65 is M0.
AnswerMarks Guidance
(x−10)2+(y−11)2 =65 or x2 + y2 −20x−22y+156=0A1 (x−6)(x−14)+(y−4)(y−18)=0 scores 3/3.
3
AnswerMarks
11(c)18−their11 their11−4 18−4  7
or or =
 
AnswerMarks Guidance
14−their10 their10−6 14−6  4*M1 Gradient of their centre, which cannot be A, B or C, from
part (b), to A or C or the gradient of AC but working needed
if incorrect centre.
OR by clearly differentiating and substitution of (14,18).
1
y−18=− (x−14)
7
their
AnswerMarks Guidance
4DM1 OE
1
Using (14,18) and − to form the equation of a
7
their
4
straight line.
AnswerMarks Guidance
4x+7y−182=0A1 All terms on one side in any order. Allow multiples of this
format by an integer only.
3
Question 11:
--- 11(a) ---
11(a) |  7−4  18−7
their their =−1
 p−6  14− p
OR Scalar product leading to (14− p)(6− p)−33=0 | *M1 | Difference in the ys
Their gradients must both come from .
Difference in the xs
p2 −20p+84=33 leading to p2−20p+51 =0  or p2−20p=−51 | A1 | Clearing of fractions and collecting terms to arrive at the
three-term quadratic. Allow integer multiples.
Alternative method for first 2 marks of Question 11(a)
(p−6)2+(7−4)2+(14− p)2+(18−7)2 =(14−6)2+(18−4)2
OR
E.g. (10− p)2 + 42 = 42+72 | *M1 | For correct use of Pythagoras with A,B and C.
OR
For correct use of Pythagoras with the centre, B and one of
the other two points.
2p2−40p+102 =0  | A1 | OE
Collecting terms to arrive at the three-term quadratic.
20 202 −451
[2](p−3)(p−17) or
2 | DM1 | OE
Solving their three-term quadratic.
p=3 | A1 | If M1A1DM0 scored then SC B1 is available for final
answer.
4
Question | Answer | Marks | Guidance
--- 11(b) ---
11(b) | [Midpoint or Centre is] (10, 11) | B1 | SOI by final answer.
1
(14−6)2 +(18−4)2 (18−their11)2 +(14−their10)2
or or
2
(their11−4)2 +(their10−6)2 r2 =65 or r= 65
  | M1 | Finding half of the length of AC or using their centre, which
cannot be A, B or C, to find r2 or r. Note: r=65 is M0.
(x−10)2+(y−11)2 =65 or x2 + y2 −20x−22y+156=0 | A1 | (x−6)(x−14)+(y−4)(y−18)=0 scores 3/3.
3
--- 11(c) ---
11(c) | 18−their11 their11−4 18−4  7
or or =
 
14−their10 their10−6 14−6  4 | *M1 | Gradient of their centre, which cannot be A, B or C, from
part (b), to A or C or the gradient of AC but working needed
if incorrect centre.
OR by clearly differentiating and substitution of (14,18).
1
y−18=− (x−14)
7
their
4 | DM1 | OE
1
Using (14,18) and − to form the equation of a
7
their
4
straight line.
4x+7y−182=0 | A1 | All terms on one side in any order. Allow multiples of this
format by an integer only.
3
The coordinates of points $A$, $B$ and $C$ are $(6, 4)$, $(p, 7)$ and $(14, 18)$ respectively, where $p$ is a constant. The line $AB$ is perpendicular to the line $BC$.

\begin{enumerate}[label=(\alph*)]
\item Given that $p < 10$, find the value of $p$. [4]
\end{enumerate}

A circle passes through the points $A$, $B$ and $C$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the equation of the circle. [3]

\item Find the equation of the tangent to the circle at $C$, giving the answer in the form $dx + ey + f = 0$, where $d$, $e$ and $f$ are integers. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q11 [10]}}
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