| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Geometric properties with circles |
| Difficulty | Standard +0.8 This question requires completing the square to find the circle's center and radius, finding the tangent equation using perpendicular gradients, calculating a triangle area from intercepts, and then using geometric properties of equilateral triangles inscribed in circles with trigonometry. Part (b) particularly requires spatial reasoning about the configuration and applying exact value calculations, making this notably harder than standard circle questions. |
| 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 | Mark | Guidance |
| Centre is \((3, -2)\) | B1 | |
| Gradient of radius \(= \frac{(\text{their}-2)-4}{(\text{their }3)-5} [=3]\) | *M1 | Finding gradient using their centre (not \((0,0)\)) and \(P(5,4)\). |
| Equation of tangent: \(y - 4 = -\frac{1}{3}(x-5)\) | DM1 | Using \(P\) and the negative reciprocal of their gradient to find the equation of \(AB\). |
| Sight of \([x=]17\) and \([y=]\frac{17}{3}\) | A1 | |
| \(\left[\text{Area} = \frac{1}{2} \times \frac{17}{3} \times 17 =\right] \frac{289}{6}\) | A1 | Or \(48\frac{1}{6}\) or AWRT 48.2. |
| Alternative method: | ||
| \(2x + 2y\frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0\) | B1 | |
| At \(P\): \(10 + 8\frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0 \left[\Rightarrow \frac{dy}{dx} = -\frac{1}{3}\right]\) | *M1 | Find the gradient using \(P(5,4)\) in their implicit differential (with at least one correctly differentiated \(y\) term). |
| Equation of tangent: \(y - 4 = -\frac{1}{3}(x-5)\) | DM1 | Using \(P\) and their value for the gradient to find the equation of \(AB\). |
| Sight of \([x=]17\) and \([y=]\frac{17}{3}\) | A1 | |
| \(\left[\text{Area} = \frac{1}{2} \times \frac{17}{3} \times 17 =\right] \frac{289}{6}\) | A1 | Or \(48\frac{1}{6}\) or AWRT 48.2. |
| Further alternative: | ||
| \(\left[y = -2 \pm \left(40-(x-3)^2\right)^{\frac{1}{2}}\right]\) OE leading to \(\frac{dy}{dx} = (3-x)\left(31+6x-x^2\right)^{-\frac{1}{2}}\) | B1 | OE. Correct differentiation of rearranged equation. |
| \(\frac{dy}{dx} = (3-5)\left(31+6(5)-(5)^2\right)^{-\frac{1}{2}} \left[\Rightarrow \frac{dy}{dx} = -\frac{1}{3}\right]\) | *M1 | Find the gradient using \(x=5\) in their differential (with clear use of chain rule). |
| Equation of tangent: \(y - 4 = -\frac{1}{3}(x-5)\) | DM1 | Using \(P\) and their value for the gradient to find the equation of \(AB\). |
| Sight of \([x=]17\) and \([y=]\frac{17}{3}\) | A1 | |
| \(\left[\text{Area} = \frac{1}{2} \times \frac{17}{3} \times 17 =\right] \frac{289}{6}\) | A1 | Or \(48\frac{1}{6}\) or AWRT 48.2. |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Radius of circle \(= \sqrt{40}\) | B1 | Or \(2\sqrt{10}\) or 6.32 AWRT or \(r^2 = 40\). |
| Area of \(\triangle CRQ = \frac{1}{2} \times (\text{their } r)^2 \sin 120 \left[= \frac{1}{2} \times 40 \times \frac{\sqrt{3}}{2}\right]\) OR Area of \(\triangle CQX = \frac{1}{2} \times \sqrt{40}\cos30 \times \sqrt{40}\cos60\) OE \(\left[= \frac{1}{2} \times \sqrt{30} \times \sqrt{10}\right]\) OR Area of circle \(- 3\times\) Area of segment \(= 40\pi - 3\times\left(40\frac{\pi}{3} - 10\sqrt{3}\right)\) OR \(QR = \sqrt{120}\) or \(2\sqrt{30}\) and area \(= \frac{1}{2}QR^2\sin60\) | M1 | Using \(\frac{1}{2}r^2\sin\theta\) with their \(r\) and 120 or 60 [\(\times 3\)]; Using \(\frac{1}{2}\times\text{base}\times\text{height}\) in a correct right-angled triangle [\(\times 6\)]; Use of cosine rule and area of large triangle. |
| \(30\sqrt{3}\) | A1 | AWRT 52[.0] implies B1M1A0. |
| Total: 3 marks |
## Question 12(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre is $(3, -2)$ | B1 | |
| Gradient of radius $= \frac{(\text{their}-2)-4}{(\text{their }3)-5} [=3]$ | *M1 | Finding gradient using their centre (not $(0,0)$) and $P(5,4)$. |
| Equation of tangent: $y - 4 = -\frac{1}{3}(x-5)$ | DM1 | Using $P$ and the negative reciprocal of their gradient to find the equation of $AB$. |
| Sight of $[x=]17$ and $[y=]\frac{17}{3}$ | A1 | |
| $\left[\text{Area} = \frac{1}{2} \times \frac{17}{3} \times 17 =\right] \frac{289}{6}$ | A1 | Or $48\frac{1}{6}$ or AWRT 48.2. |
| **Alternative method:** | | |
| $2x + 2y\frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0$ | B1 | |
| At $P$: $10 + 8\frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0 \left[\Rightarrow \frac{dy}{dx} = -\frac{1}{3}\right]$ | *M1 | Find the gradient using $P(5,4)$ in their implicit differential (with at least one correctly differentiated $y$ term). |
| Equation of tangent: $y - 4 = -\frac{1}{3}(x-5)$ | DM1 | Using $P$ and their value for the gradient to find the equation of $AB$. |
| Sight of $[x=]17$ and $[y=]\frac{17}{3}$ | A1 | |
| $\left[\text{Area} = \frac{1}{2} \times \frac{17}{3} \times 17 =\right] \frac{289}{6}$ | A1 | Or $48\frac{1}{6}$ or AWRT 48.2. |
| **Further alternative:** | | |
| $\left[y = -2 \pm \left(40-(x-3)^2\right)^{\frac{1}{2}}\right]$ OE leading to $\frac{dy}{dx} = (3-x)\left(31+6x-x^2\right)^{-\frac{1}{2}}$ | B1 | OE. Correct differentiation of rearranged equation. |
| $\frac{dy}{dx} = (3-5)\left(31+6(5)-(5)^2\right)^{-\frac{1}{2}} \left[\Rightarrow \frac{dy}{dx} = -\frac{1}{3}\right]$ | *M1 | Find the gradient using $x=5$ in their differential (with clear use of chain rule). |
| Equation of tangent: $y - 4 = -\frac{1}{3}(x-5)$ | DM1 | Using $P$ and their value for the gradient to find the equation of $AB$. |
| Sight of $[x=]17$ and $[y=]\frac{17}{3}$ | A1 | |
| $\left[\text{Area} = \frac{1}{2} \times \frac{17}{3} \times 17 =\right] \frac{289}{6}$ | A1 | Or $48\frac{1}{6}$ or AWRT 48.2. |
| **Total: 5 marks** | | |
---
## Question 12(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Radius of circle $= \sqrt{40}$ | B1 | Or $2\sqrt{10}$ or 6.32 AWRT or $r^2 = 40$. |
| Area of $\triangle CRQ = \frac{1}{2} \times (\text{their } r)^2 \sin 120 \left[= \frac{1}{2} \times 40 \times \frac{\sqrt{3}}{2}\right]$ **OR** Area of $\triangle CQX = \frac{1}{2} \times \sqrt{40}\cos30 \times \sqrt{40}\cos60$ OE $\left[= \frac{1}{2} \times \sqrt{30} \times \sqrt{10}\right]$ **OR** Area of circle $- 3\times$ Area of segment $= 40\pi - 3\times\left(40\frac{\pi}{3} - 10\sqrt{3}\right)$ **OR** $QR = \sqrt{120}$ or $2\sqrt{30}$ and area $= \frac{1}{2}QR^2\sin60$ | M1 | Using $\frac{1}{2}r^2\sin\theta$ with their $r$ and 120 or 60 [$\times 3$]; Using $\frac{1}{2}\times\text{base}\times\text{height}$ in a correct right-angled triangle [$\times 6$]; Use of cosine rule and area of large triangle. |
| $30\sqrt{3}$ | A1 | AWRT 52[.0] implies B1M1A0. |
| **Total: 3 marks** | | |12\\
\includegraphics[max width=\textwidth, alt={}, center]{10b2ec29-adca-4313-ae24-bab8b2d9f8a4-18_750_981_258_580}
The diagram shows the circle with equation $x ^ { 2 } + y ^ { 2 } - 6 x + 4 y - 27 = 0$ and the tangent to the circle at the point $P ( 5,4 )$.
\begin{enumerate}[label=(\alph*)]
\item The tangent to the circle at $P$ meets the $x$-axis at $A$ and the $y$-axis at $B$.
Find the area of triangle $O A B$, where $O$ is the origin.
\item Points $Q$ and $R$ also lie on the circle, such that $P Q R$ is an equilateral triangle.
Find the exact area of triangle $P Q R$.\\
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 2021 Q12 [8]}}