| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Moderate -0.8 This is a straightforward C2 question requiring completing the square to find centre and radius (standard technique), then substituting x=4 to find intersection points. All steps are routine applications of well-practiced methods with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks |
|---|---|
| (a) Uses any appropriate method to find coordinates of centre, e.g. achieves \((x\pm 5)^{2}+(y\pm 3)^{2}=\ldots\) Accept \((\pm 5,\pm 3)\) as indication of this | M1 |
| Centre is \((5,-3)\) | A1 |
| (b) Way 1: Uses \((x\pm"5")^{2}-"5^{2}"+(y\pm"3")^{2}-3^{2}+30=0\) to give | M1 |
| \(r=\sqrt{"25"+"9"-30}\) or \(r^{2}="25"+"9"-30\) (not \(30-25-9\)) | M1 |
| \(r=2\) | A1cao |
| Way 2: Using \(\sqrt{g^{2}+f^{2}-c}\) from \(x^{2}+y^{2}+2gx+2fy+c=0\) (Needs formula stated or correct working) | M1 |
| \(r=2\) | A1 |
| (c) Way 1: Use \(x=4\) in an equation of circle and obtain equation in \(y\) only | M1 |
| e.g. \((4-5)^{2}+(y+3)^{2}=4\) or \(4^{2}+y^{2}-10\times 4+6y+30=0\) | |
| Solve their quadratic in \(y\) and obtain two solutions for \(y\) | dM1 |
| e.g. \((y+3)^{2}=3\) or \(y^{2}+6y+6=0\) so \(y=-3\pm\sqrt{3}\) | A1 |
| Way 2: Divide triangle \(PTQ\) and use Pythagoras with "\(r^{2}-("5"-4)^{2}=h^{2}\)" | M1 |
| Find \(h\) and evaluate "\(3"±h\). May recognise \((1,\sqrt{3}, 2)\) triangle | dM1 |
| So \(y=-3\pm\sqrt{3}\) | A1 |
| (a) Uses any appropriate method to find coordinates of centre, e.g. achieves $(x\pm 5)^{2}+(y\pm 3)^{2}=\ldots$ Accept $(\pm 5,\pm 3)$ as indication of this | M1 |
|---|---|
| Centre is $(5,-3)$ | A1 |
| (b) **Way 1:** Uses $(x\pm"5")^{2}-"5^{2}"+(y\pm"3")^{2}-3^{2}+30=0$ to give | M1 |
| $r=\sqrt{"25"+"9"-30}$ or $r^{2}="25"+"9"-30$ (not $30-25-9$) | M1 |
| $r=2$ | A1cao |
| **Way 2:** Using $\sqrt{g^{2}+f^{2}-c}$ from $x^{2}+y^{2}+2gx+2fy+c=0$ (Needs formula stated or correct working) | M1 |
| $r=2$ | A1 |
| (c) **Way 1:** Use $x=4$ in an equation of circle and obtain equation in $y$ only | M1 |
| e.g. $(4-5)^{2}+(y+3)^{2}=4$ or $4^{2}+y^{2}-10\times 4+6y+30=0$ | |
| Solve their quadratic in $y$ and obtain two solutions for $y$ | dM1 |
| e.g. $(y+3)^{2}=3$ or $y^{2}+6y+6=0$ so $y=-3\pm\sqrt{3}$ | A1 |
| **Way 2:** Divide triangle $PTQ$ and use Pythagoras with "$r^{2}-("5"-4)^{2}=h^{2}$" | M1 |
| Find $h$ and evaluate "$3"±h$. May recognise $(1,\sqrt{3}, 2)$ triangle | dM1 |
| So $y=-3\pm\sqrt{3}$ | A1 |
**Notes:**
- **(a) and (b) can be marked together:** M1 as in scheme and can be implied by $(\pm 5,\pm 3)$. May be awarded for writing LHS as $(x\pm 5)^{2}+(y\pm 3)^{2}=\ldots$ or by comparing with $x^{2}+y^{2}+2gx+2fy+c=0$ to write down centre $(-g,-f)$ directly; A1: $(5,-3)$. This correct answer implies M1A1
- **(b):** M1: for a full correct method leading to $r=\ldots$ or $r^{2}=$ with their 5, their $-3$, their 25 and their 9 and their "$-30"$. Completion of square method errors result in M0 here. Usually $r=4$ or $r=16$ imply M0A0; A1 2 cao. Do not accept $r=\pm 2$ unless it is followed by $(r=)2$. The correct answer with no wrong work seen implies M1A1
- **Special case:** if centre is given as $(-5,-3)$ or $(5,3)$ or $(-5,3)$ allow M1A1 for $r=2$ worked correctly. i.e. $r^{2}="25"+"9"-30$
- **(c):** M1: **Way 1:** Use $x=4$ in a circle equation (may have wrong centre and/or radius) to obtain an equation in $y$ only or **Way 2:** Uses geometry to find equation in $h$ (ft on their radius and centre); dM1: (needs first method mark) Solve their quadratic in $y$ or **Way 2:** Uses their $h$ and their $y$ coordinate correctly; A1: cao5. The circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } - 10 x + 6 y + 30 = 0$$
Find
\begin{enumerate}[label=(\alph*)]
\item the coordinates of the centre of $C$,
\item the radius of $C$,
\item the $y$ coordinates of the points where the circle $C$ crosses the line with equation $x = 4$, giving your answers as simplified surds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2017 Q5 [7]}}