| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle equation from centre and radius |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic circle and line concepts at C1 level. Parts (i)-(iii) involve routine calculations (gradient, perpendicular bisector, substitution into circle equation), while part (iv) requires solving a quadratic to find x-intercepts. All techniques are standard with clear signposting and no problem-solving 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 | Mark | Guidance |
| \(\text{grad } AB = \frac{9-1}{3--1}\) or \(2\) | M1 | |
| \(y - 9 = 2(x-3)\) or \(y - 1 = 2(x+1)\) | M1 | ft their \(m\), or substitute coords of A or B in \(y = \text{their } mx + c\) |
| \(y = 2x + 3\) oe | A1 | Or B3 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Midpoint of \(AB = (1, 5)\) | M1 | Condone not stated explicitly, but used in equation |
| Grad perp \(= -1/\text{grad }AB\) | M1 | Soi by use e.g. in equation |
| \(y - 5 = -\frac{1}{2}(x-1)\) oe or ft [no ft for just grad AB used] | M1 | ft their grad and/or midpoint, but M0 if their midpoint not used; allow M1 for \(y = -\frac{1}{2}x + c\) and then their midpoint substituted |
| At least one correct interim step towards given answer \(2y + x = 11\), and correct completion. NB answer \(2y + x = 11\) given | M1 | No ft; correct equation only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| showing \((-1-5)^2 + (1-3)^2 = 40\) | M1 | at least one interim step needed for each mark; M0 for just \(6^2 + 2^2 = 40\) |
| showing B to centre \(= \sqrt{40}\) or verifying that \((3, 9)\) fits given circle | M1 | with no other evidence such as a first line of working or a diagram; condone marks earned in reverse order |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \((x-5)^2 + 3^2 = 40\) | M1 | for subst \(y = 0\) in circle eqn |
| \((x-5)^2 = 31\) | M1 | condone slip on rhs; or for rearrangement to zero (condone one error) and attempt at quad. formula [allow M1 M0 for \((x-5)^2 = 40\) or for \((x-5)^2 + 3^2 = 0\)] |
| \(x = 5 \pm \sqrt{31}\) or \(\dfrac{10 \pm \sqrt{124}}{2}\) isw | A1 | or \(5 \pm \dfrac{\sqrt{124}}{2}\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(y = \dfrac{11-x}{2}\) o.e. | M1 | alt method working back from ans |
| grad perp \(= -1/\text{grad } AB\) and showing/stating same as given line | M1 | e.g. stating \(-\frac{1}{2} \times 2 = -1\) |
| finding intersection of \(y = 2x + 3\) and \(2y + x = 11\) \([= (1, 5)]\) | M1 | or showing that \((1, 5)\) is on \(2y + x = 11\), having found \((1, 5)\) first |
| showing midpoint of \(AB\) is \((1, 5)\) | M1 | [for both methods: for M4 must be fully correct] |
| Total | 4 | mark one method or the other, to benefit of candidate, not a mixture |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{grad } AB = \frac{9-1}{3--1}$ or $2$ | M1 | |
| $y - 9 = 2(x-3)$ or $y - 1 = 2(x+1)$ | M1 | ft their $m$, or substitute coords of A or B in $y = \text{their } mx + c$ |
| $y = 2x + 3$ oe | A1 | Or B3 |
| **[3]** | | |
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Midpoint of $AB = (1, 5)$ | M1 | Condone not stated explicitly, but used in equation |
| Grad perp $= -1/\text{grad }AB$ | M1 | Soi by use e.g. in equation |
| $y - 5 = -\frac{1}{2}(x-1)$ oe or ft [no ft for just grad AB used] | M1 | ft their grad and/or midpoint, but M0 if their midpoint not used; allow M1 for $y = -\frac{1}{2}x + c$ and then their midpoint substituted |
| At least one correct interim step towards given answer $2y + x = 11$, and correct completion. NB answer $2y + x = 11$ given | M1 | No ft; correct equation only |
## Part iii:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| showing $(-1-5)^2 + (1-3)^2 = 40$ | M1 | at least one interim step needed for each mark; M0 for just $6^2 + 2^2 = 40$ |
| showing B to centre $= \sqrt{40}$ or verifying that $(3, 9)$ fits given circle | M1 | with no other evidence such as a first line of working or a diagram; condone marks earned in reverse order |
| **Total** | **2** | |
---
## Part iv:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $(x-5)^2 + 3^2 = 40$ | M1 | for subst $y = 0$ in circle eqn |
| $(x-5)^2 = 31$ | M1 | condone slip on rhs; or for rearrangement to zero (condone one error) and attempt at quad. formula [allow M1 M0 for $(x-5)^2 = 40$ or for $(x-5)^2 + 3^2 = 0$] |
| $x = 5 \pm \sqrt{31}$ or $\dfrac{10 \pm \sqrt{124}}{2}$ isw | A1 | or $5 \pm \dfrac{\sqrt{124}}{2}$ |
| **Total** | **3** | |
---
## Part ii (perpendicular bisector continuation):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $y = \dfrac{11-x}{2}$ o.e. | M1 | alt method working back from ans |
| grad perp $= -1/\text{grad } AB$ and showing/stating same as given line | M1 | e.g. stating $-\frac{1}{2} \times 2 = -1$ |
| finding intersection of $y = 2x + 3$ and $2y + x = 11$ $[= (1, 5)]$ | M1 | or showing that $(1, 5)$ is on $2y + x = 11$, having found $(1, 5)$ first |
| showing midpoint of $AB$ is $(1, 5)$ | M1 | [for both methods: for M4 must be fully correct] |
| **Total** | **4** | mark one method or the other, to benefit of candidate, not a mixture |7 (i) Find the equation of the line passing through $\mathrm { A } ( - 1,1 )$ and $\mathrm { B } ( 3,9 )$.\\
(ii) Show that the equation of the perpendicular bisector of AB is $2 y + x = 11$.\\
(iii) A circle has centre $( 5,3 )$, so that its equation is $( x - 5 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = k$. Given that the circle passes through A , show that $k = 40$. Show that the circle also passes through B .\\
(iv) Find the $x$-coordinates of the points where this circle crosses the $x$-axis. Give your answers in surd form.
\hfill \mbox{\textit{OCR MEI C1 Q7 [12]}}