| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle equation from centre and radius |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question requiring only routine techniques: midpoint formula, distance formula, and simultaneous equations. All parts are standard textbook exercises with no problem-solving insight needed, making it easier than average for A-level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | D = (5, 1) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | ( x−5 )2 +( y−1 )2 =20 oe | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iii) | ( x−1 )2 +( y−3 )2 =( 9−x )2 +( y+1 )2 soi | M1 |
| Answer | Marks |
|---|---|
| x2 −2x+1+ y2 −6y+9=x2 −18x+81+ y2 +2y+1 | A1 |
| y=2x−9 www AG | A1 |
| Answer | Marks |
|---|---|
| −½ | M1 |
| Equation of perp. bisector is y−1=2 ( x−5 ) | A1 |
| y=2x−9 www AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iv) | Eliminate y (or x) using equations in (ii) and (iii) | *M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(y‒1)2 = 80 | DM1 | Simplify to one of the forms shown on the right (allow arithmetic |
| Answer | Marks | Guidance |
|---|---|---|
| x = 3 and 7, or y = ‒3 and 5 | A1 | |
| (3, ‒3), (7, 5) | A1 | Both pairs of x & y correct implies A1A1. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(i) ---
7(i) | D = (5, 1) | B1
1
--- 7(ii) ---
7(ii) | ( x−5 )2 +( y−1 )2 =20 oe | B1 | FT on their D.
Apply ISW, oe but not to contain square roots
1
Question | Answer | Marks | Guidance
--- 7(iii) ---
7(iii) | ( x−1 )2 +( y−3 )2 =( 9−x )2 +( y+1 )2 soi | M1 | Allow 1 sign slip
For M1 allow with √ signs round both sides but sides must be
equated
x2 −2x+1+ y2 −6y+9=x2 −18x+81+ y2 +2y+1 | A1
y=2x−9 www AG | A1
Alternative method for question 7(iii)
−1
grad. of AB = ‒½ → grad of perp bisector =
−½ | M1
Equation of perp. bisector is y−1=2 ( x−5 ) | A1
y=2x−9 www AG | A1
3
--- 7(iv) ---
7(iv) | Eliminate y (or x) using equations in (ii) and (iii) | *M1 | To give an (unsimplified) quadratic equation
5x2 ‒50x + 105 (= 0) or 5(x‒5)2 = 20 or 5y2‒10y‒75 (= 0) or
5(y‒1)2 = 80 | DM1 | Simplify to one of the forms shown on the right (allow arithmetic
slips)
x = 3 and 7, or y = ‒3 and 5 | A1
(3, ‒3), (7, 5) | A1 | Both pairs of x & y correct implies A1A1.
SC B2 for no working
4
Question | Answer | Marks | GuidanceThe coordinates of two points $A$ and $B$ are $(1, 3)$ and $(9, -1)$ respectively and $D$ is the mid-point of $AB$. A point $C$ has coordinates $(x, y)$, where $x$ and $y$ are variables.
\begin{enumerate}[label=(\roman*)]
\item State the coordinates of $D$. [1]
\item It is given that $CD^2 = 20$. Write down an equation relating $x$ and $y$. [1]
\item It is given that $AC$ and $BC$ are equal in length. Find an equation relating $x$ and $y$ and show that it can be simplified to $y = 2x - 9$. [3]
\item Using the results from parts (ii) and (iii), and showing all necessary working, find the possible coordinates of $C$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q7 [9]}}