| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Standard +0.3 This is a slightly above-average AS-level question requiring completion of the square to find the centre, then using the midpoint theorem (that A'B' is parallel to AB with half the length). While it involves multiple steps, the techniques are standard and the geometric insight about midpoints is a common circle theorem application. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 + (y-8)^2 [= 25]\) | M1 | Attempt to complete the square for either \(x\) or \(y\) terms |
| C is \((3, 8)\) | A1 | cao Ignore arithmetic slips on RHS |
| Answer | Marks | Guidance |
|---|---|---|
| Intersects \(y = x - 2\) when \((x-3)^2 + (x-2-8)^2 = 25\) | M1 | Substituting for \(y\) in equation of circle or substitution for \(x\) |
| \(x^2 - 6x + 9 + x^2 - 20x + 100 - 25 = 0\), \(2x^2 - 26x + 84 = 0\) | M1 | Simplifies to 3 term quadratic in \(x\) or \(y\) (\(2y^2 - 18y + 40 = 0\)) |
| Giving \(x = 6, 7\) | A1 | Both correct values seen |
| So A and B are \((6, 4)\) and \((7, 5)\) | A1 | Correct \(y\)-coordinates FT their \(x\)-coordinates. Allow full credit for fully correct trial and improvement method |
| Midpoints \((4.5, 6)\) and \((5, 6.5)\) | M1, A1 | Uses the midpoint formula at least once soi. Both correct FT their A, B and C |
| \(A'B' = \sqrt{(5-4.5)^2 + (6.5-6)^2} = \frac{\sqrt{2}}{2}\) | M1, A1 | Uses the distance formula for their \(A'\) and \(B'\). Must be exact. ISW if 0.71 also given |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB = \sqrt{(7-6)^2 + (5-4)^2} = \sqrt{2}\) | M1, A1 | Uses the distance formula for their A and B. FT their A and B |
| The triangle \(CA'B'\) is an enlargement scale factor \(\frac{1}{2}\) of triangle CAB so \(A'B' = \frac{1}{2}\sqrt{2}\) | M1, A1 | Uses similar triangles or enlargement to find \(A'B'\). Must be exact ISW if 0.71 also given |
## Question 8:
### Part (a):
$(x-3)^2 + (y-8)^2 [= 25]$ | M1 | Attempt to complete the square for either $x$ or $y$ terms
C is $(3, 8)$ | A1 | cao Ignore arithmetic slips on RHS
**[2]**
### Part (b):
Intersects $y = x - 2$ when $(x-3)^2 + (x-2-8)^2 = 25$ | M1 | Substituting for $y$ in equation of circle or substitution for $x$
$x^2 - 6x + 9 + x^2 - 20x + 100 - 25 = 0$, $2x^2 - 26x + 84 = 0$ | M1 | Simplifies to 3 term quadratic in $x$ or $y$ ($2y^2 - 18y + 40 = 0$)
Giving $x = 6, 7$ | A1 | Both correct values seen
So A and B are $(6, 4)$ and $(7, 5)$ | A1 | Correct $y$-coordinates FT their $x$-coordinates. Allow full credit for fully correct trial and improvement method
Midpoints $(4.5, 6)$ and $(5, 6.5)$ | M1, A1 | Uses the midpoint formula at least once soi. Both correct FT their A, B and C
$A'B' = \sqrt{(5-4.5)^2 + (6.5-6)^2} = \frac{\sqrt{2}}{2}$ | M1, A1 | Uses the distance formula for their $A'$ and $B'$. Must be exact. ISW if 0.71 also given
**Alternative for last 4 marks:**
$AB = \sqrt{(7-6)^2 + (5-4)^2} = \sqrt{2}$ | M1, A1 | Uses the distance formula for their A and B. FT their A and B
The triangle $CA'B'$ is an enlargement scale factor $\frac{1}{2}$ of triangle CAB so $A'B' = \frac{1}{2}\sqrt{2}$ | M1, A1 | Uses similar triangles or enlargement to find $A'B'$. Must be exact ISW if 0.71 also given
**[8]**
---8 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } - 6 x - 16 y + 48 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of C .
A line has equation $\mathrm { y } = \mathrm { x } - 2$ and intersects the circle at the points A and B . The midpoints of AC and BC are $\mathrm { A } ^ { \prime }$ and $\mathrm { B } ^ { \prime }$ respectively.
\item Determine the exact distance $\mathrm { A } ^ { \prime } \mathrm { B } ^ { \prime }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q8 [10]}}