| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question testing standard techniques: midpoint formula for the centre, distance formula for radius, perpendicular distance from centre to chord, and the angle-in-a-semicircle theorem. Part (c) requires Pythagoras on the right triangle formed, which is routine. All parts follow predictable patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03d 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/Working | Mark | Guidance |
| \(A\left(\frac{-9+15}{2}, \frac{8-10}{2}\right) = A(3,-1)\) | M1A1 | M1: Correct attempt to find midpoint between \(P\) and \(Q\). Can be implied by one of \(x\) or \(y\)-coordinates correctly evaluated. A1: \((3,-1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((-9-3)^2 + (8+1)^2\) or \(\sqrt{(-9-3)^2+(8+1)^2}\) or \((15-3)^2+(-10+1)^2\) or \(\sqrt{(15-3)^2+(-10+1)^2}\) | M1 | Uses Pythagoras correctly to find the radius. Must be identified as radius (may be implied by circle equation). Or uses Pythagoras for diameter: \((15+9)^2+(-10-8)^2\). Allow if 30 seen as diameter or 15 as radius |
| \((x-3)^2 + (y+1)^2 = 225\) \(\left(\text{or } (15)^2\right)\) | M1 | \((x \pm \alpha)^2 + (y \pm \beta)^2 = k^2\) where \(A(\alpha,\beta)\) and \(k\) is their radius |
| \((x-3)^2 + (y+1)^2 = 225\) | A1 | Allow \((x-3)^2+(y+1)^2 = 15^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(A(\pm\alpha, \pm\beta)\) and \(x^2+2ax+y^2+2by+c=0\), e.g. \(x^2+2(-3)x+y^2+2(1)y+c=0\) | M1 | |
| Uses \(P\) or \(Q\) and \(x^2+2ax+y^2+2by+c=0\), e.g. \((-9)^2+2(-3)(-9)+(8)^2+2(1)(8)+c=0 \Rightarrow c=-215\) | M1 | |
| \(x^2-6x+y^2+2y-215=0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(= \sqrt{15^2-10^2}\) | M1 | \(= \sqrt{(\text{their } r)^2 - 10^2}\) or correct method e.g. their \(r \times \cos\left[\sin^{-1}\left(\frac{10}{\text{their }r}\right)\right]\) |
| \(\{=\sqrt{125}\} = 5\sqrt{5}\) | A1 | \(5\sqrt{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin(A\hat{R}Q) = \frac{20}{30}\) or \(A\hat{R}Q = 90 - \cos^{-1}\left(\frac{10}{15}\right)\) | M1 | \(\sin(A\hat{R}Q) = \frac{20}{(2\times\text{their }r)}\) or \(\frac{10}{\text{their }r}\); or \(A\hat{R}Q = 90-\cos^{-1}\left(\frac{10}{\text{their }r}\right)\); or \(A\hat{R}Q = \cos^{-1}\left(\frac{\text{Part(c)}}{\text{their }r}\right)\); or \(20^2=15^2+15^2-2\times15\times15\cos(2ARQ)\); or \(15^2=15^2+(10\sqrt{5})^2-2\times15\times10\sqrt{5}\cos(ARQ)\). Must be correct statement involving angle \(ARQ\) where their \(r>10\) |
| \(A\hat{R}Q = 41.8103\ldots\) | A1 | awrt 41.8 |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A\left(\frac{-9+15}{2}, \frac{8-10}{2}\right) = A(3,-1)$ | M1A1 | M1: Correct attempt to find midpoint between $P$ and $Q$. Can be implied by one of $x$ or $y$-coordinates correctly evaluated. A1: $(3,-1)$ |
**[2 marks]**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(-9-3)^2 + (8+1)^2$ or $\sqrt{(-9-3)^2+(8+1)^2}$ or $(15-3)^2+(-10+1)^2$ or $\sqrt{(15-3)^2+(-10+1)^2}$ | M1 | Uses Pythagoras correctly to find the **radius**. Must be identified as radius (may be implied by circle equation). Or uses Pythagoras for **diameter**: $(15+9)^2+(-10-8)^2$. Allow if 30 seen as diameter or 15 as radius |
| $(x-3)^2 + (y+1)^2 = 225$ $\left(\text{or } (15)^2\right)$ | M1 | $(x \pm \alpha)^2 + (y \pm \beta)^2 = k^2$ where $A(\alpha,\beta)$ and $k$ is their radius |
| $(x-3)^2 + (y+1)^2 = 225$ | A1 | Allow $(x-3)^2+(y+1)^2 = 15^2$ |
**Alternative using $x^2+2ax+y^2+2by+c=0$:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $A(\pm\alpha, \pm\beta)$ and $x^2+2ax+y^2+2by+c=0$, e.g. $x^2+2(-3)x+y^2+2(1)y+c=0$ | M1 | |
| Uses $P$ or $Q$ and $x^2+2ax+y^2+2by+c=0$, e.g. $(-9)^2+2(-3)(-9)+(8)^2+2(1)(8)+c=0 \Rightarrow c=-215$ | M1 | |
| $x^2-6x+y^2+2y-215=0$ | A1 | |
**[3 marks]**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= \sqrt{15^2-10^2}$ | M1 | $= \sqrt{(\text{their } r)^2 - 10^2}$ or correct method e.g. their $r \times \cos\left[\sin^{-1}\left(\frac{10}{\text{their }r}\right)\right]$ |
| $\{=\sqrt{125}\} = 5\sqrt{5}$ | A1 | $5\sqrt{5}$ |
**[2 marks]**
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin(A\hat{R}Q) = \frac{20}{30}$ or $A\hat{R}Q = 90 - \cos^{-1}\left(\frac{10}{15}\right)$ | M1 | $\sin(A\hat{R}Q) = \frac{20}{(2\times\text{their }r)}$ or $\frac{10}{\text{their }r}$; or $A\hat{R}Q = 90-\cos^{-1}\left(\frac{10}{\text{their }r}\right)$; or $A\hat{R}Q = \cos^{-1}\left(\frac{\text{Part(c)}}{\text{their }r}\right)$; or $20^2=15^2+15^2-2\times15\times15\cos(2ARQ)$; or $15^2=15^2+(10\sqrt{5})^2-2\times15\times10\sqrt{5}\cos(ARQ)$. Must be correct statement involving angle $ARQ$ where their $r>10$ |
| $A\hat{R}Q = 41.8103\ldots$ | A1 | awrt 41.8 |
**[2 marks] — Total 9**\begin{enumerate}
\item The circle $C$, with centre $A$, passes through the point $P$ with coordinates ( $- 9,8$ ) and the point $Q$ with coordinates $( 15 , - 10 )$.
\end{enumerate}
Given that $P Q$ is a diameter of the circle $C$,\\
(a) find the coordinates of $A$,\\
(b) find an equation for $C$.
A point $R$ also lies on the circle $C$.\\
Given that the length of the chord $P R$ is 20 units,\\
(c) find the length of the shortest distance from $A$ to the chord $P R$.
Give your answer as a surd in its simplest form.\\
(d) Find the size of the angle $A R Q$, giving your answer to the nearest 0.1 of a degree.
\hfill \mbox{\textit{Edexcel C2 2014 Q10 [9]}}