| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using right angle in semicircle |
| Difficulty | Moderate -0.3 This is a straightforward multi-part coordinate geometry question requiring standard techniques: distance formula for equal lengths, perpendicular gradient condition, and finding a circle equation through three points. All methods are routine P1 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((5-2p)^2 + (p+2)^2 = (10-2p)^2 + (3-p)^2\) | M1 A1 | Allow one sign error for M mark only |
| \(25 - 20p + 4p^2 + p^2 + 4p + 4 = 100 - 40p + 4p^2 + 9 - 6p + p^2\) | A1 | Allow 2.67 AWRT |
| \(30p = 80 \rightarrow p = \frac{8}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m_{AC} = \frac{p+2}{2p-5}\), \(m_{BC} = \frac{p-3}{2p-10}\) | M1 | Allow a sign error |
| \(\frac{p+2}{2p-5} \times \frac{p-3}{2p-10} = -1\) | M1 | Use of \(m_1 m_2 = -1\) with their \(m_{AC}\) and \(m_{BC}\) |
| \(p^2 - p - 6 = -(4p^2 - 30p + 50) \rightarrow 5p^2 - 31p + 44 (=0)\) | A1 | |
| \(p = 4\) (Ignore \(p = \frac{11}{5}\)) | A1 | Factors \((p-4)(5p-11)\), or formula or completing square must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mid-point of \(AB = (7\frac{1}{2},\ \frac{1}{2})\) | B1 | SOI |
| \(r^2 = 2\frac{1}{2}^2 + 2\frac{1}{2}^2 \left[=\frac{50}{4}\right]\) or \(r = \sqrt{(2\frac{1}{2}^2 + 2\frac{1}{2}^2)}\left[=\frac{5\sqrt{2}}{2}\right]\) | \*M1 | Or \(r^2 = \frac{1}{4}(5^2+5^2)\left[=\frac{50}{4}\right]\) etc. |
| Equation of circle is \(\left(x - their\ 7\frac{1}{2}\right)^2 + \left(y - their\ \frac{1}{2}\right)^2 = their\ \frac{50}{4}\) | DM1 | Must use \(r^2\) not \(r\) or \(d\) or \(d^2\) |
| \(x^2 + y^2 - 15x - y + 44 = 0\) | A1 | CAO |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(5-2p)^2 + (p+2)^2 = (10-2p)^2 + (3-p)^2$ | M1 A1 | Allow one sign error for M mark only |
| $25 - 20p + 4p^2 + p^2 + 4p + 4 = 100 - 40p + 4p^2 + 9 - 6p + p^2$ | A1 | Allow 2.67 AWRT |
| $30p = 80 \rightarrow p = \frac{8}{3}$ | | |
---
## Question 11(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m_{AC} = \frac{p+2}{2p-5}$, $m_{BC} = \frac{p-3}{2p-10}$ | M1 | Allow a sign error |
| $\frac{p+2}{2p-5} \times \frac{p-3}{2p-10} = -1$ | M1 | Use of $m_1 m_2 = -1$ with their $m_{AC}$ and $m_{BC}$ |
| $p^2 - p - 6 = -(4p^2 - 30p + 50) \rightarrow 5p^2 - 31p + 44 (=0)$ | A1 | |
| $p = 4$ (Ignore $p = \frac{11}{5}$) | A1 | Factors $(p-4)(5p-11)$, or formula or completing square must be seen |
---
## Question 11(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mid-point of $AB = (7\frac{1}{2},\ \frac{1}{2})$ | B1 | SOI |
| $r^2 = 2\frac{1}{2}^2 + 2\frac{1}{2}^2 \left[=\frac{50}{4}\right]$ or $r = \sqrt{(2\frac{1}{2}^2 + 2\frac{1}{2}^2)}\left[=\frac{5\sqrt{2}}{2}\right]$ | \*M1 | Or $r^2 = \frac{1}{4}(5^2+5^2)\left[=\frac{50}{4}\right]$ etc. |
| Equation of circle is $\left(x - their\ 7\frac{1}{2}\right)^2 + \left(y - their\ \frac{1}{2}\right)^2 = their\ \frac{50}{4}$ | DM1 | Must use $r^2$ not $r$ or $d$ or $d^2$ |
| $x^2 + y^2 - 15x - y + 44 = 0$ | A1 | CAO |11 The coordinates of points $A , B$ and $C$ are $A ( 5 , - 2 ) , B ( 10,3 )$ and $C ( 2 p , p )$, where $p$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Given that $A C$ and $B C$ are equal in length, find the value of the fraction $p$.
\item It is now given instead that $A C$ is perpendicular to $B C$ and that $p$ is an integer.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $p$.
\item Find the equation of the circle which passes through $A , B$ and $C$, giving your answer in the form $x ^ { 2 } + y ^ { 2 } + a x + b y + c = 0$, where $a , b$ and $c$ are constants.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q11 [11]}}