CAIE P1 2024 June — Question 10 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicCircles
TypeTangent from external point - intersection or geometric properties
DifficultyStandard +0.8 This is a moderately challenging circle-tangent problem requiring substitution of the line into the circle equation, using the discriminant condition for tangency (b²-4ac=0), solving a quadratic in m, then finding two pairs of contact points. It involves multiple algebraic steps and the discriminant technique, placing it above average difficulty but within reach of competent P1 students.
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 \((x - 3)^2 + y^2 = 18\). The line with equation \(y = mx + c\) passes through the point \((0, -9)\) and is a tangent to the circle. Find the two possible values of \(m\) and, for each value of \(m\), find the coordinates of the point at which the tangent touches the circle. [8]
Question 10:
AnswerMarks Guidance
10x32 y2 18 ymx9 leading to x32 mx92 18 M1
equation.
Must be mx9.
x2 6x9m2x2 18mx8118 leading to
AnswerMarks Guidance
 m2 1  x2 618mx720M1 Brackets expanded and all terms collected on one side of
the equation. May be implied in the discriminant. m
cannot be numeric.
AnswerMarks Guidance
618m2 4  m2 1  720*M1 Use of b2  4ac . Not in quadratic formula. m cannot be
numeric, c must be numeric.
36m2216m2520 leading to m2 6m70
AnswerMarks Guidance
 DM1 Simplifies to 3 term quadratic.
m1 or m 7A1 Condone no method for solving quadratic shown.
m1 leading to 2x2 24x72  0 leading to x6DM1 Must be correct x for their quadratic.
6
m7 leading to 50x2 120x720 leading to x
AnswerMarks Guidance
5DM1 Must be correct x for their quadratic.
 6 3
6, 3,  ,  
AnswerMarks Guidance
 5 5A1
QuestionAnswer Marks
10Alternative Method 1 for first 4 marks of Question 10
3m109
AnswerMarks Guidance
m2 1(M1) Use of the formula for the length of a perpendicular from
a point to a line.
3m109
= 18
AnswerMarks Guidance
m2 1(M1) Equates length of a perpendicular from a point to a line to
the radius.
(3 – 9)2 = 18(m2 + 1)
AnswerMarks Guidance
m(M1) Squares and clears the fraction.
9m2- 54 m + 81 = 0   leading to m2 6m70 (M1)
Alternative Method 2 for first 3 marks of Question 10
(3 – )(9 + 6 - x2)-1/2 =
AnswerMarks Guidance
x x m(M1) OE
dy
Differentiates implicitly or otherwise and equates to
dx
m.
m2)x
AnswerMarks Guidance
(1m2) x2 – 6(1 + + 9(1 – m2)[ = 0](M1) Brackets expanded and all terms collected on one side of
the equation. May be implied in the discriminant.
AnswerMarks Guidance
36(1 m2)2 – 4(1 m2) × 9(1 – m2)[ = 0](M1) Use of b2 4ac.
8
AnswerMarks Guidance
QuestionAnswer Marks 
Question 10:
10 | x32 y2 18 ymx9 leading to x32 mx92 18 | M1 | Finding equation of tangent and substituting into circle
equation.
Must be mx9.
x2 6x9m2x2 18mx8118 leading to
 m2 1  x2 618mx720 | M1 | Brackets expanded and all terms collected on one side of
the equation. May be implied in the discriminant. m
cannot be numeric.
618m2 4  m2 1  720 | *M1 | Use of b2  4ac . Not in quadratic formula. m cannot be
numeric, c must be numeric.
36m2216m2520 leading to m2 6m70
  | DM1 | Simplifies to 3 term quadratic.
m1 or m 7 | A1 | Condone no method for solving quadratic shown.
m1 leading to 2x2 24x72  0 leading to x6 | DM1 | Must be correct x for their quadratic.
6
m7 leading to 50x2 120x720 leading to x
5 | DM1 | Must be correct x for their quadratic.
 6 3
6, 3,  ,  
 5 5 | A1
Question | Answer | Marks | Guidance
10 | Alternative Method 1 for first 4 marks of Question 10
3m109
m2 1 | (M1) | Use of the formula for the length of a perpendicular from
a point to a line.
3m109
= 18
m2 1 | (M1) | Equates length of a perpendicular from a point to a line to
the radius.
(3 – 9)2 = 18(m2 + 1)
m | (M1) | Squares and clears the fraction.
9m2- 54 m + 81 = 0   leading to m2 6m70  | (M1)
Alternative Method 2 for first 3 marks of Question 10
(3 – )(9 + 6 - x2)-1/2 =
x x m | (M1) | OE
dy
Differentiates implicitly or otherwise and equates to
dx
m.
m2)x
(1m2) x2 – 6(1 + + 9(1 – m2)[ = 0] | (M1) | Brackets expanded and all terms collected on one side of
the equation. May be implied in the discriminant.
36(1 m2)2 – 4(1 m2) × 9(1 – m2)[ = 0] | (M1) | Use of b2 4ac.
8
Question | Answer | Marks | Guidance
The equation of a circle is $(x - 3)^2 + y^2 = 18$. The line with equation $y = mx + c$ passes through the point $(0, -9)$ and is a tangent to the circle.

Find the two possible values of $m$ and, for each value of $m$, find the coordinates of the point at which the tangent touches the circle. [8]

\hfill \mbox{\textit{CAIE P1 2024 Q10 [8]}}
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