| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using perpendicular bisectors |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring standard techniques: finding perpendicular bisector (given gradient), line intersection, and circle equation from center and radius. All steps are routine C1 content with clear scaffolding. The final 'show' part requires verification that C and D lie on the circle, which is straightforward substitution. Slightly above average due to length and the need to connect multiple concepts, but no novel insight required. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Midpoint of \(AB = \left(\frac{1}{2}, \frac{5}{2}\right)\) | B2 | Allow unsimplified; B1 for one coordinate correct; if working shown should come from \(\left(\frac{3+-2}{2}, \frac{4+1}{2}\right)\); NB B0 for \(x\) coord \(= \frac{5}{2}\) obtained from subtraction |
| \(\text{grad } AB = \frac{4-1}{3-(-2)}\) | M1 | Must be obtained independently of given line; accept 3 and 5 correctly shown e.g. in sketch, followed by \(\frac{3}{5}\); M1 for rise/run \(= \frac{3}{5}\); M0 for just \(\frac{3}{5}\) with no evidence |
| Using gradient of \(AB\) to obtain grad perp bisector | M1 | For use of \(m_1 m_2 = -1\) soi or ft their gradient \(AB\); M0 for just \(\frac{-5}{3}\) without \(AB\) grad found |
| \(y - 2.5 = \frac{-5}{3}(x - 0.5)\) | M1 | e.g. M1 for \(y = \frac{-5}{3}x + c\) and subst of midpt; ft their gradient of perp bisector and midpt; M0 for just rearranging given equation |
| Completion to \(3y + 5x = 10\) showing at least one interim step | M1 | Condone slight slip if they recover quickly; M0 if clearly 'fudging' |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3y + 5(4y - 21) = 10\) | M1 | Or other valid strategy for eliminating one variable attempted e.g. \(\frac{-5}{3}x + \frac{10}{3} = \frac{x}{4} + \frac{21}{4}\); condone one error |
| \((-1, 5)\) or \(y=5, x=-1\) | A2 | A1 for each value; if AO allow SC1 for both values correct but unsimplified fractions e.g. \(\left(\frac{-23}{23}, \frac{115}{23}\right)\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-a)^2 + (y-b)^2 = r^2\) seen or used | M1 | Or for \((x+1)^2 + (y-5)^2 = k\), or ft their \(E\), where \(k > 0\) |
| \(1^2 + 4^2\) (may be unsimplified), from clear use of \(A\) or \(B\) | M1 | For calculating \(AE\) or \(BE\) or their squares, or for substituting coords of \(A\) or \(B\) into circle equation to find \(r\) or \(r^2\); ft their \(E\); this M not earned for use of \(CE\) or \(DE\) or \(\frac{1}{2}CD\) |
| \((x+1)^2 + (y-5)^2 = 17\) | A1 | For equation of circle centre \(E\), through \(A\) and \(B\) |
| Showing midpoint of \(CD = (-1, 5)\) | M1 | |
| Showing \(CE\) or \(DE = \sqrt{17}\) or showing one of \(C\) and \(D\) on circle | M1 | Alt M1 for showing \(CD^2 = 68\); allow to be earned earlier as invalid attempt to find \(r\) |
| [5] |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoint of $AB = \left(\frac{1}{2}, \frac{5}{2}\right)$ | B2 | Allow unsimplified; B1 for one coordinate correct; if working shown should come from $\left(\frac{3+-2}{2}, \frac{4+1}{2}\right)$; NB B0 for $x$ coord $= \frac{5}{2}$ obtained from subtraction |
| $\text{grad } AB = \frac{4-1}{3-(-2)}$ | M1 | Must be obtained independently of given line; accept 3 and 5 correctly shown e.g. in sketch, followed by $\frac{3}{5}$; M1 for rise/run $= \frac{3}{5}$; M0 for just $\frac{3}{5}$ with no evidence |
| Using gradient of $AB$ to obtain grad perp bisector | M1 | For use of $m_1 m_2 = -1$ soi or ft their gradient $AB$; M0 for just $\frac{-5}{3}$ without $AB$ grad found |
| $y - 2.5 = \frac{-5}{3}(x - 0.5)$ | M1 | e.g. M1 for $y = \frac{-5}{3}x + c$ and subst of midpt; ft their gradient of perp bisector and midpt; M0 for just rearranging given equation |
| Completion to $3y + 5x = 10$ showing at least one interim step | M1 | Condone slight slip if they recover quickly; M0 if clearly 'fudging' |
| **[6]** | | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3y + 5(4y - 21) = 10$ | M1 | Or other valid strategy for eliminating one variable attempted e.g. $\frac{-5}{3}x + \frac{10}{3} = \frac{x}{4} + \frac{21}{4}$; condone one error |
| $(-1, 5)$ or $y=5, x=-1$ | A2 | A1 for each value; if AO allow SC1 for both values correct but unsimplified fractions e.g. $\left(\frac{-23}{23}, \frac{115}{23}\right)$ |
| **[3]** | | |
---
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-a)^2 + (y-b)^2 = r^2$ seen or used | M1 | Or for $(x+1)^2 + (y-5)^2 = k$, or ft their $E$, where $k > 0$ |
| $1^2 + 4^2$ (may be unsimplified), from clear use of $A$ or $B$ | M1 | For calculating $AE$ or $BE$ or their squares, or for substituting coords of $A$ or $B$ into circle equation to find $r$ or $r^2$; ft their $E$; this M not earned for use of $CE$ or $DE$ or $\frac{1}{2}CD$ |
| $(x+1)^2 + (y-5)^2 = 17$ | A1 | For equation of circle centre $E$, through $A$ and $B$ |
| Showing midpoint of $CD = (-1, 5)$ | M1 | |
| Showing $CE$ or $DE = \sqrt{17}$ or showing one of $C$ and $D$ on circle | M1 | Alt M1 for showing $CD^2 = 68$; allow to be earned earlier as invalid attempt to find $r$ |
| **[5]** | | |
---10 (i) Points A and B have coordinates $( - 2,1 )$ and $( 3,4 )$ respectively. Find the equation of the perpendicular bisector of AB and show that it may be written as $5 x + 3 y = 10$.\\
(ii) Points C and D have coordinates $( - 5,4 )$ and $( 3,6 )$ respectively. The line through C and D has equation $4 y = x + 21$. The point E is the intersection of CD and the perpendicular bisector of AB . Find the coordinates of point E .\\
(iii) Find the equation of the circle with centre E which passes through A and B . Show also that CD is a diameter of this circle.
\hfill \mbox{\textit{OCR MEI C1 2013 Q10 [14]}}