| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| 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 structured multi-part question testing standard C1 circle and coordinate geometry techniques: finding line equations, perpendicular bisectors, verifying points on circles, and finding axis intersections. All parts are routine applications of formulas with clear guidance, requiring minimal problem-solving insight. Slightly easier than average due to the scaffolded structure and verification rather than discovery approach. |
| 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 |
|---|---|---|
| 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 subst coords of A or B |
| \(y = 2x + 3\) 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 in eqn |
| \(y - 5 = -\frac{1}{2}(x-1)\) o.e. or ft | M1 | ft their grad and/or midpt, but M0 if their midpt not used |
| at least one correct interim step towards \(2y + x = 11\) | M1 | no ft; correct eqn only |
| Alt method: \(y = \frac{11-x}{2}\) 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 \((1,5)\) is on \(2y+x=11\) |
| showing midpt of \(AB\) is \((1, 5)\) | M1 | [for both methods: for M4 must be fully correct] |
| Answer | Marks | Guidance |
|---|---|---|
| showing \((-1-5)^2 + (1-3)^2 = 40\) | M1 | at least one interim step needed for each mark; M0 for just \(6^2 + 2^2 = 40\) with no other evidence |
| showing \(B\) to centre \(= \sqrt{40}\) or verifying \((3,9)\) 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 \(\frac{10 \pm \sqrt{124}}{2}\) isw | A1 | or \(5 \pm \frac{\sqrt{124}}{2}\) |
# Question 12:
## Part (i)
| 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 subst coords of A or B |
| $y = 2x + 3$ 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 in eqn |
| $y - 5 = -\frac{1}{2}(x-1)$ o.e. or ft | M1 | ft their grad and/or midpt, but M0 if their midpt not used |
| at least one correct interim step towards $2y + x = 11$ | M1 | no ft; correct eqn only |
| **Alt method:** $y = \frac{11-x}{2}$ 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 $(1,5)$ is on $2y+x=11$ |
| showing midpt of $AB$ is $(1, 5)$ | M1 | [for both methods: for M4 must be fully correct] |
## Part (iii)
| showing $(-1-5)^2 + (1-3)^2 = 40$ | M1 | at least one interim step needed for each mark; M0 for just $6^2 + 2^2 = 40$ with no other evidence |
| showing $B$ to centre $= \sqrt{40}$ or verifying $(3,9)$ 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 $\frac{10 \pm \sqrt{124}}{2}$ isw | A1 | or $5 \pm \frac{\sqrt{124}}{2}$ |12 (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 2008 Q12 [12]}}