| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Standard +0.3 This is a structured multi-part question on circle equations and line-circle intersections using standard techniques. Part (a) is direct formula application, (b) uses perpendicular gradient for tangent, (c) involves simultaneous equations with verification provided, and (d) requires substitution and solving a quadratic. All methods are routine for P1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2+(y-2)^2=100\) | B1 | OE e.g. \((x-0)^2+(y-2)^2=10^2\) ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient of radius \(= \left[\frac{10-2}{6-0}=\right]\frac{4}{3}\) or gradient of tangent \(= -\frac{3}{4}\) | M1 | OE SOI Use coordinates to find gradient of radius or differentiate to find \(m_T\). e.g. \(2x+2(y-2)\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{-3}{4}\) at \((6,10)\); \(y=2+\sqrt{100-x^2}\Rightarrow\frac{dy}{dx}=\frac{1}{2}(100-x^2)^{-\frac{1}{2}}(-2x)=-\frac{3}{4}\) |
| Equation of tangent is \(y-10=-\frac{3}{4}(x-6)\) \(\left[\Rightarrow y=-\frac{3}{4}x+\frac{29}{2}\right]\) | A1 | OE ISW Allow e.g. \(\frac{58}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Coordinates of centre of circle \(Q\) are \(\left(0, \textit{their}\ \frac{29}{2}\right)\) | M1 | SOI From a linear equation in (b) |
| Equation of circle \(Q\) is \(x^2+\left(y-\textit{their}\ \frac{29}{2}\right)^2=\left(\frac{5\sqrt{5}}{2}\right)^2\left[=\frac{125}{4}\right]\) | A1 FT | OE e.g. \((x-0)^2+(y-14.5)^2=31.25\) ISW |
| \(x^2+(11-2)^2=100 \Rightarrow x^2=19\) and \(x^2+\left(11-\frac{29}{2}\right)^2=\frac{125}{4} \Rightarrow x^2=19\) OR e.g. \(\frac{125}{4}-\left(y-\frac{29}{2}\right)^2+(y-2)^2=100 \Rightarrow 25y=275 \Rightarrow y=11\) | B1 | OE e.g. \(x=[\pm]\sqrt{19}\), \(x^2-19=x^2-19\). Correct argument to verify both \(y\)-coords are 11 ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2+\left(-\frac{3}{4}x+\frac{29}{2}-\frac{29}{2}\right)^2=\frac{125}{4}\left[\Rightarrow\frac{25}{16}x^2=\frac{125}{4}\Rightarrow x^2=20\right]\) or \(y^2-29y+199[=0]\) | M1 | Substitute equation of *their* tangent into equation of *their* circle. May see \(y=\sqrt{31.25-x^2}+14.5\) |
| \(x=\pm2\sqrt{5}\) or \(y=\frac{29\mp3\sqrt{5}}{2}\) | A1 | OE e.g. \(x=\pm\sqrt{20}\). For 2 \(x\)-values or 2 \(y\)-values or correct \((x,y)\) pair |
| \(y\left[=\left(-\frac{3}{4}\times\pm\sqrt{20}\right)+\frac{29}{2}\right]=\frac{29\mp3\sqrt{5}}{2}\) | A1 | OE e.g. \(\frac{58}{4}+\frac{3\sqrt{20}}{4}\), \(\frac{58}{4}-\frac{3\sqrt{20}}{4}\). Correct \((x,y)\) pairs |
## Question 12(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2+(y-2)^2=100$ | B1 | OE e.g. $(x-0)^2+(y-2)^2=10^2$ ISW |
**Total: 1 mark**
---
## Question 12(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of radius $= \left[\frac{10-2}{6-0}=\right]\frac{4}{3}$ or gradient of tangent $= -\frac{3}{4}$ | M1 | OE SOI Use coordinates to find gradient of radius or differentiate to find $m_T$. e.g. $2x+2(y-2)\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{-3}{4}$ at $(6,10)$; $y=2+\sqrt{100-x^2}\Rightarrow\frac{dy}{dx}=\frac{1}{2}(100-x^2)^{-\frac{1}{2}}(-2x)=-\frac{3}{4}$ |
| Equation of tangent is $y-10=-\frac{3}{4}(x-6)$ $\left[\Rightarrow y=-\frac{3}{4}x+\frac{29}{2}\right]$ | A1 | OE ISW Allow e.g. $\frac{58}{4}$ |
**Total: 2 marks**
---
## Question 12(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Coordinates of centre of circle $Q$ are $\left(0, \textit{their}\ \frac{29}{2}\right)$ | M1 | SOI From a linear equation in **(b)** |
| Equation of circle $Q$ is $x^2+\left(y-\textit{their}\ \frac{29}{2}\right)^2=\left(\frac{5\sqrt{5}}{2}\right)^2\left[=\frac{125}{4}\right]$ | A1 FT | OE e.g. $(x-0)^2+(y-14.5)^2=31.25$ ISW |
| $x^2+(11-2)^2=100 \Rightarrow x^2=19$ and $x^2+\left(11-\frac{29}{2}\right)^2=\frac{125}{4} \Rightarrow x^2=19$ OR e.g. $\frac{125}{4}-\left(y-\frac{29}{2}\right)^2+(y-2)^2=100 \Rightarrow 25y=275 \Rightarrow y=11$ | B1 | OE e.g. $x=[\pm]\sqrt{19}$, $x^2-19=x^2-19$. Correct argument to verify both $y$-coords are 11 ISW |
**Total: 3 marks**
---
## Question 12(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2+\left(-\frac{3}{4}x+\frac{29}{2}-\frac{29}{2}\right)^2=\frac{125}{4}\left[\Rightarrow\frac{25}{16}x^2=\frac{125}{4}\Rightarrow x^2=20\right]$ or $y^2-29y+199[=0]$ | M1 | Substitute equation of *their* tangent into equation of *their* circle. May see $y=\sqrt{31.25-x^2}+14.5$ |
| $x=\pm2\sqrt{5}$ or $y=\frac{29\mp3\sqrt{5}}{2}$ | A1 | OE e.g. $x=\pm\sqrt{20}$. For 2 $x$-values or 2 $y$-values or correct $(x,y)$ pair |
| $y\left[=\left(-\frac{3}{4}\times\pm\sqrt{20}\right)+\frac{29}{2}\right]=\frac{29\mp3\sqrt{5}}{2}$ | A1 | OE e.g. $\frac{58}{4}+\frac{3\sqrt{20}}{4}$, $\frac{58}{4}-\frac{3\sqrt{20}}{4}$. Correct $(x,y)$ pairs |
**Total: 3 marks**12\\
\includegraphics[max width=\textwidth, alt={}, center]{77f27b11-b931-481f-b4ef-5e549eff8086-18_1006_938_269_591}
The diagram shows a circle $P$ with centre $( 0,2 )$ and radius 10 and the tangent to the circle at the point $A$ with coordinates $( 6,10 )$. It also shows a second circle $Q$ with centre at the point where this tangent meets the $y$-axis and with radius $\frac { 5 } { 2 } \sqrt { 5 }$.
\begin{enumerate}[label=(\alph*)]
\item Write down the equation of circle $P$.
\item Find the equation of the tangent to the circle $P$ at $A$.
\item Find the equation of circle $Q$ and hence verify that the $y$-coordinates of both of the points of intersection of the two circles are 11.
\item Find the coordinates of the points of intersection of the tangent and circle $Q$, giving the answers in surd form.\\
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}
\hfill \mbox{\textit{CAIE P1 2023 Q12 [9]}}