CAIE P1 2024 June — Question 7 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicCircles
TypeFind parameter values for tangency using discriminant
DifficultyStandard +0.3 This is a standard circle-tangent problem requiring the perpendicular distance formula and solving a quadratic equation. Part (a) uses the condition that distance from center to tangent equals radius (routine technique), while part (b) applies perpendicular gradient and point-slope form. Slightly above average due to the algebraic manipulation with parameter 'a', but follows well-established methods with no novel insight required.
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 circle
The equation of a circle is \((x-6)^2 + (y+a)^2 = 18\). The line with equation \(y = 2a - x\) is a tangent to the circle.
  1. Find the two possible values of the constant \(a\). [5]
  2. For the greater value of \(a\), find the equation of the diameter which is perpendicular to the given tangent. [3]
Question 7:
AnswerMarks Guidance
7(a)x62 2axa2
 18M1* Replacing y with 2a – x in the circle equation,
condone incorrect expansion before substitution.
AnswerMarks Guidance
2x2 12x6ax9a2 3618 0A1 All terms collected on one side of the equation. May
be implied by the discriminant.
AnswerMarks Guidance
126a2 42  9a2 18  0DM1 Correct use of “b2 — 4ac” from their 3 term
quadratic equation in x, with an x term of the form
mnax with both m and n  0.
AnswerMarks
36a2 144a 00A1
a0, a4A1
5
AnswerMarks Guidance
7(b)[Centre is] (6, −4) or [Point of intersection is] (9, −1) B1
[Gradient of diameter] 1B1
y4 x6 or y1 x9 leading to y x10B1FT FT on their point of intersection or their centre with
an x co-ordinate of ±6 and gradient = 1.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a) ---
7(a) | x62 2axa2
 18 | M1* | Replacing y with 2a – x in the circle equation,
condone incorrect expansion before substitution.
2x2 12x6ax9a2 3618 0 | A1 | All terms collected on one side of the equation. May
be implied by the discriminant.
126a2 42  9a2 18  0 | DM1 | Correct use of “b2 — 4ac” from their 3 term
quadratic equation in x, with an x term of the form
mnax with both m and n  0.
36a2 144a 00 | A1
a0, a4 | A1
5
--- 7(b) ---
7(b) | [Centre is] (6, −4) or [Point of intersection is] (9, −1) | B1
[Gradient of diameter] 1 | B1
y4 x6 or y1 x9 leading to y x10 | B1FT | FT on their point of intersection or their centre with
an x co-ordinate of ±6 and gradient = 1.
3
Question | Answer | Marks | Guidance
The equation of a circle is $(x-6)^2 + (y+a)^2 = 18$. The line with equation $y = 2a - x$ is a tangent to the circle.

\begin{enumerate}[label=(\alph*)]
\item Find the two possible values of the constant $a$. [5]

\item For the greater value of $a$, find the equation of the diameter which is perpendicular to the given tangent. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q7 [8]}}
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