| 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 -0.5 This is a structured multi-part question covering standard C1 circle and coordinate geometry techniques: finding line equations, perpendicular bisectors, verifying points on circles, and finding axis intersections. While it requires multiple steps, each part uses routine methods with clear guidance, making it slightly easier than average for A-level. |
| 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^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{grad } AB = \dfrac{9-1}{3--1}\) or \(2\) | M1 | |
| \(y - 9 = 2(x-3)\) or \(y - 1 = 2(x+1)\) | M1 | ft their \(m\), or subst coords of A or B in \(y = mx + c\) |
| \(y = 2x + 3\) o.e. | A1 | Or B3 |
| Answer | Marks | Guidance |
|---|---|---|
| Mid pt of \(AB = (1, 5)\) | M1 | Condone not stated explicitly, but used in eqn |
| Grad perp \(= -1/\text{grad } AB\) | M1 | soi by use e.g. in eqn |
| \(y - 5 = -\frac{1}{2}(x-1)\) o.e. | M1 | ft their grad and/or midpt |
| At least one correct interim step towards \(2y + x = 11\), and correct completion | M1 | No ft; correct eqn only |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \dfrac{11-x}{2}\) o.e. | M1 | |
| Grad perp \(= -1/\text{grad } AB\) and showing 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 midpt of \(AB\) is \((1,5)\) | M1 | For both methods: M4 must be fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Showing \((-1-5)^2 + (1-3)^2 = 40\) | M1 | At least one interim step needed |
| Showing \(B\) to centre \(= \sqrt{40}\) or verifying that \((3,9)\) fits given circle | M1 | Condone marks earned in reverse order |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-5)^2 + 3^2 = 40\) | M1 | For subst \(y=0\) in circle eqn |
| \((x-5)^2 = 31\) | M1 | Condone slip on rhs; or for rearrangement to zero and attempt at quad. formula |
| \(x = 5 \pm \sqrt{31}\) or \(\dfrac{10 \pm \sqrt{124}}{2}\) isw | A1 | Or \(5 \pm \dfrac{\sqrt{124}}{2}\) |
## Question 5:
### Part (i)
| $\text{grad } AB = \dfrac{9-1}{3--1}$ or $2$ | M1 | |
|---|---|---|
| $y - 9 = 2(x-3)$ or $y - 1 = 2(x+1)$ | M1 | ft their $m$, or subst coords of A or B in $y = mx + c$ |
| $y = 2x + 3$ o.e. | A1 | Or B3 |
### Part (ii)
| Mid pt of $AB = (1, 5)$ | M1 | Condone not stated explicitly, but used in eqn |
|---|---|---|
| Grad perp $= -1/\text{grad } AB$ | M1 | soi by use e.g. in eqn |
| $y - 5 = -\frac{1}{2}(x-1)$ o.e. | M1 | ft their grad and/or midpt |
| At least one correct interim step towards $2y + x = 11$, and correct completion | M1 | No ft; correct eqn only |
**Alt method (working back from answer):**
| $y = \dfrac{11-x}{2}$ o.e. | M1 | |
|---|---|---|
| Grad perp $= -1/\text{grad } AB$ and showing 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 midpt of $AB$ is $(1,5)$ | M1 | For both methods: M4 must be fully correct |
### Part (iii)
| Showing $(-1-5)^2 + (1-3)^2 = 40$ | M1 | At least one interim step needed |
|---|---|---|
| Showing $B$ to centre $= \sqrt{40}$ or verifying that $(3,9)$ fits given circle | M1 | Condone marks earned in reverse order |
### Part (iv)
| $(x-5)^2 + 3^2 = 40$ | M1 | For subst $y=0$ in circle eqn |
|---|---|---|
| $(x-5)^2 = 31$ | M1 | Condone slip on rhs; or for rearrangement to zero and attempt at quad. formula |
| $x = 5 \pm \sqrt{31}$ or $\dfrac{10 \pm \sqrt{124}}{2}$ isw | A1 | Or $5 \pm \dfrac{\sqrt{124}}{2}$ |5 (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 Q5 [12]}}