| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Moderate -0.5 This is a straightforward multi-part question testing standard circle equation manipulation and perpendicular bisector concepts. Part (i) requires reading the center and radius directly from the equation. Part (ii) involves substituting y=0 or x=0 and solving simple quadratics. Part (iii) applies the standard perpendicular bisector method (midpoint and negative reciprocal gradient) with verification. All techniques are routine C1 procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{radius} = \sqrt{125}\) or \(5\sqrt{5}\) | B1 | |
| \(C = (10, 2)\) | B1 | condone \(x=10\), \(y=2\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Verifying/deriving that \((21, 0)\) is one of the intersections with the axes | B1 | using circle equation or Pythagoras; putting \(y=0\) in circle equation and solving to get 21 and \(-1\); condone omission of brackets |
| \((-1, 0)\) | B1 | condone not written as coordinates |
| \((0, -3)\) and \((0, 7)\) | B2 | B1 each; condone not written as coordinates; condone not identified as D and E; condone \(D=(0,7)\), \(E=(0,-3)\) – will penalise themselves in (iii) |
| [4] | if B0 for D and E, then M1 for substitution of \(x=0\) into circle equation or use of Pythagoras showing \(125-10^2\) or \(h^2+10^2=125\) ft their centre and/or radius |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| midpoint \(BE = \left(\frac{21}{2}, \frac{7}{2}\right)\) oe | B1 | ft their E; or stating that the perpendicular bisector of a chord always passes through the centre of the circle (must be explicit generalised statement) |
| \(\text{grad } BE = \dfrac{7-0}{0-21}\) oe isw | M1 | ft their E; M0 for using grad \(BC\) \((= -\frac{2}{11})\) |
| grad perp bisector \(= 3\) oe | M1 | for use of \(m_1 m_2 = -1\) oe soi; ft their grad BE; no ft from grad BC used; condone \(3x\) oe; allow M1 for e.g. \(-\frac{1}{3} \times 3 = -1\) |
| \(y - \frac{7}{2} = 3\left(x - \frac{21}{2}\right)\) oe | M1 | ft; M0 for using grad BE or perp to BC; allow this M1 for C used instead of midpoint |
| \(y = 3x - 28\) oe | A1 | must be a simplified equation; no ft |
| verifying that \((10, 2)\) is on this line | A1 | no ft; A0 if C used to find equation of line, unless B1 earned for correct argument |
| [6] |
## Question 11:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{radius} = \sqrt{125}$ or $5\sqrt{5}$ | B1 | |
| $C = (10, 2)$ | B1 | condone $x=10$, $y=2$ |
| | **[2]** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Verifying/deriving that $(21, 0)$ is one of the intersections with the axes | B1 | using circle equation or Pythagoras; putting $y=0$ in circle equation and solving to get 21 and $-1$; condone omission of brackets |
| $(-1, 0)$ | B1 | condone not written as coordinates |
| $(0, -3)$ and $(0, 7)$ | B2 | **B1** each; condone not written as coordinates; condone not identified as D and E; condone $D=(0,7)$, $E=(0,-3)$ – will penalise themselves in (iii) |
| | **[4]** | if B0 for D and E, then **M1** for substitution of $x=0$ into circle equation or use of Pythagoras showing $125-10^2$ or $h^2+10^2=125$ ft their centre and/or radius |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| midpoint $BE = \left(\frac{21}{2}, \frac{7}{2}\right)$ oe | B1 | ft their E; or stating that the perpendicular bisector of a chord always passes through the centre of the circle (must be explicit generalised statement) |
| $\text{grad } BE = \dfrac{7-0}{0-21}$ oe isw | M1 | ft their E; M0 for using grad $BC$ $(= -\frac{2}{11})$ |
| grad perp bisector $= 3$ oe | M1 | for use of $m_1 m_2 = -1$ oe soi; ft their grad BE; no ft from grad BC used; condone $3x$ oe; allow M1 for e.g. $-\frac{1}{3} \times 3 = -1$ |
| $y - \frac{7}{2} = 3\left(x - \frac{21}{2}\right)$ oe | M1 | ft; M0 for using grad BE or perp to BC; allow this M1 for C used instead of midpoint |
| $y = 3x - 28$ oe | A1 | must be a simplified equation; no ft |
| verifying that $(10, 2)$ is on this line | A1 | no ft; A0 if C used to find equation of line, unless B1 earned for correct argument |
| | **[6]** | |
---11
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c55e1f96-670a-4bc3-9e77-92d28769b7f5-3_700_751_906_641}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows a sketch of the circle with equation $( x - 10 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 125$ and centre C . The points $\mathrm { A } , \mathrm { B }$, D and E are the intersections of the circle with the axes.\\
(i) Write down the radius of the circle and the coordinates of C .\\
(ii) Verify that B is the point $( 21,0 )$ and find the coordinates of $\mathrm { A } , \mathrm { D }$ and E .\\
(iii) Find the equation of the perpendicular bisector of BE and verify that this line passes through C .
\hfill \mbox{\textit{OCR MEI C1 2015 Q11 [12]}}