| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using right angle in semicircle |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: calculating gradients to show perpendicularity (product = -1), using the diameter form of a circle equation, and finding the other end of a diameter using midpoint. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-step nature and 12 marks total. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| grad \(AB = \frac{8}{4}\) or \(2\) or \(y = 2x - 10\) | 1 mark | or M1 for \(AB^2 = 4^2 + 8^2\) or \(80\) and \(BC^2 = 2^2 + 1^2\) or \(5\) and \(AC^2 = 6^2 + 7^2\) or \(85\); M1 for \(AC^2 = AB^2 + BC^2\) and 1 for [Pythag.] true so \(AB\) perp to \(BC\); if \(0\), allow G1 for graph of A, B, C |
| grad \(BC = \frac{1}{-2}\) or \(-\frac{1}{2}\) or \(y = -\frac{1}{2}x + 2.5\) | 1 mark | |
| product of grads \(= -1\) [so perp] | 1 mark | |
| 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| midpt E of \(AC = (6, 4.5)\) | 1 mark | allow seen in (i) only if used in (ii); or \(AE^2 = (9 - \text{their } 6)^2 + (8 - \text{ their } 4.5)^2\) or |
| \(AC^2 = (9 - 3)^2 + (8 - 1)^2\) or \(85\) | M1 mark | |
| rad \(= \frac{1}{2}\sqrt{85}\) o.e. \((x - 6)^2 + (y - 4.5)^2 = \frac{85}{4}\) o.e. | A1 mark | rad.\(^2 = \frac{85}{4}\) o.e. e.g. in circle eqn |
| \((5-6)^2 + (0-4.5)^2 = 1 + \frac{81}{4} [= \frac{85}{4}]\) | 1 mark | or M1 for \((x-a)^2 + (y-b)^2 = r^2\) soi or for ths correct working shown; or 'angle in semicircle \([=90°]\)' |
| 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\vec{BE} = \vec{ED} = \begin{pmatrix} 1 \\ 4.5 \end{pmatrix}\) | M1 mark | o.e. ft their centre; or for \(\vec{BC} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\) |
| D has coords \((6 + 1, 4.5 + 4.5)\) ft or \((5 + 2, 0 + 9) = (7, 9)\) | M1 mark | or \((9 - 2, 8 + 1)\); condone mixtures of vectors and coords. throughout part iii allow B3 for \((7,9)\) |
| 3 marks |
**Part (i)**
grad $AB = \frac{8}{4}$ or $2$ or $y = 2x - 10$ | 1 mark | or M1 for $AB^2 = 4^2 + 8^2$ or $80$ and $BC^2 = 2^2 + 1^2$ or $5$ and $AC^2 = 6^2 + 7^2$ or $85$; M1 for $AC^2 = AB^2 + BC^2$ and 1 for [Pythag.] true so $AB$ perp to $BC$; if $0$, allow G1 for graph of A, B, C
grad $BC = \frac{1}{-2}$ or $-\frac{1}{2}$ or $y = -\frac{1}{2}x + 2.5$ | 1 mark |
product of grads $= -1$ [so perp] | 1 mark |
| | 3 marks |
**Part (ii)**
midpt E of $AC = (6, 4.5)$ | 1 mark | allow seen in (i) only if used in (ii); or $AE^2 = (9 - \text{their } 6)^2 + (8 - \text{ their } 4.5)^2$ or
$AC^2 = (9 - 3)^2 + (8 - 1)^2$ or $85$ | M1 mark |
rad $= \frac{1}{2}\sqrt{85}$ o.e. $(x - 6)^2 + (y - 4.5)^2 = \frac{85}{4}$ o.e. | A1 mark | rad.$^2 = \frac{85}{4}$ o.e. e.g. in circle eqn
$(5-6)^2 + (0-4.5)^2 = 1 + \frac{81}{4} [= \frac{85}{4}]$ | 1 mark | or M1 for $(x-a)^2 + (y-b)^2 = r^2$ soi or for ths correct working shown; or 'angle in semicircle $[=90°]$'
| | 6 marks |
**Part (iii)**
$\vec{BE} = \vec{ED} = \begin{pmatrix} 1 \\ 4.5 \end{pmatrix}$ | M1 mark | o.e. ft their centre; or for $\vec{BC} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$
D has coords $(6 + 1, 4.5 + 4.5)$ ft or $(5 + 2, 0 + 9) = (7, 9)$ | M1 mark | or $(9 - 2, 8 + 1)$; condone mixtures of vectors and coords. throughout part iii allow B3 for $(7,9)$
| | 3 marks |
---A$(9, 8)$, B$(5, 0)$ and C$(3, 1)$ are three points.
\begin{enumerate}[label=(\roman*)]
\item Show that AB and BC are perpendicular. [3]
\item Find the equation of the circle with AC as diameter. You need not simplify your answer.
Show that B lies on this circle. [6]
\item BD is a diameter of the circle. Find the coordinates of D. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 2006 Q11 [12]}}