| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.3 This is a straightforward two-part question testing standard circle techniques: completing the square to find centre/radius, then substituting a linear equation into the circle equation to find intersection points. Both parts are routine textbook exercises requiring no problem-solving insight, though the algebraic manipulation keeps it from being trivial. Slightly easier than average due to the mechanical nature of both parts. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-1)^2 - 1 + (y+2)^2 - 4 = 20\) | M1, A1 | Attempt to complete the square for at least one variable. Fully correct, need not be simplified |
| Centre \((1, -2)\) | A1 | FT their completed square form |
| Radius \(5\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Rewrite equation of the line \(x = 10 - 3y\) | B1 | soi |
| Substitute: \((10 - 3y - 1)^2 + (y+2)^2 = 25\) | M1 | Attempt to form quadratic in \(y\) only. Allow either form of equation used |
| \(10y^2 - 50y + 60 = 0\) | M1 | Attempt to simplify the quadratic to 3 terms |
| \(y = 2,\ 3\) | A1 | Both roots seen |
| Points of intersection at \((4, 2)\) and \((1, 3)\) | A1 | FT their \(y\)-values. No extra points. ISW \((2,4)\) and \((3,1)\) if \(x=4\) and \(x=1\) seen matched to their \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Rewrite \(y = \frac{10-x}{3}\) | B1 | soi |
| \(x^2 + \left(\frac{10-x}{3}\right)^2 - 2x + 4\left(\frac{10-x}{3}\right) - 20 = 0\) | M1 | Attempt to form quadratic in \(x\) only |
| \(\frac{10}{9}x^2 - \frac{50}{9}x + \frac{40}{9} = 0\) | M1 | Attempt to simplify to 3 terms |
| \(x = 1,\ 4\) | A1 | Both roots seen |
| Points of intersection at \((1, 3)\) and \((4, 2)\) | A1 | FT their \(x\)-values. No extra points. Do not allow \((2,4)\) and \((3,1)\) |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-1)^2 - 1 + (y+2)^2 - 4 = 20$ | M1, A1 | Attempt to complete the square for at least one variable. Fully correct, need not be simplified |
| Centre $(1, -2)$ | A1 | FT their completed square form |
| Radius $5$ | A1 | cao |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rewrite equation of the line $x = 10 - 3y$ | B1 | soi |
| Substitute: $(10 - 3y - 1)^2 + (y+2)^2 = 25$ | M1 | Attempt to form quadratic in $y$ only. Allow either form of equation used |
| $10y^2 - 50y + 60 = 0$ | M1 | Attempt to simplify the quadratic to 3 terms |
| $y = 2,\ 3$ | A1 | Both roots seen |
| Points of intersection at $(4, 2)$ and $(1, 3)$ | A1 | FT their $y$-values. No extra points. ISW $(2,4)$ and $(3,1)$ if $x=4$ and $x=1$ seen matched to their $y$ |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rewrite $y = \frac{10-x}{3}$ | B1 | soi |
| $x^2 + \left(\frac{10-x}{3}\right)^2 - 2x + 4\left(\frac{10-x}{3}\right) - 20 = 0$ | M1 | Attempt to form quadratic in $x$ only |
| $\frac{10}{9}x^2 - \frac{50}{9}x + \frac{40}{9} = 0$ | M1 | Attempt to simplify to 3 terms |
| $x = 1,\ 4$ | A1 | Both roots seen |
| Points of intersection at $(1, 3)$ and $(4, 2)$ | A1 | FT their $x$-values. No extra points. Do not allow $(2,4)$ and $(3,1)$ |
---8 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Find the centre and radius of the circle with equation $x ^ { 2 } + y ^ { 2 } - 2 x + 4 y - 20 = 0$.
\item Find the points of intersection of the circle with the line $x + 3 y - 10 = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2023 Q8 [9]}}