| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard techniques: finding a circle equation from centre and a point (using distance formula for radius), then substituting a linear equation into the circle equation to find intersection points and calculating chord length using the distance formula. While it involves multiple steps and algebraic manipulation, these are routine procedures for A-level students with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(r^2 [= (5-2)^2 + (7-5)^2] = 13\) | B1 | \(r^2 = 13\) or \(r = \sqrt{13}\) |
| Equation of circle is \((x-5)^2 + (y-2)^2 = 13\) | B1 FT | OE. FT on *their* 13 but LHS must be correct. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-5)^2 + (5x-10-2)^2 = 13\) | M1 | Substitute \(y = 5x - 10\) into *their* equation. |
| \(26x^2 - 130x + 156\ [=0]\) | A1 FT | OE 3-term quadratic with all terms on one side. FT on *their* circle equation. |
| \([26](x-2)(x-3)\ [=0]\) | M1 | Solve 3-term quadratic in \(x\) by factorising, using formula or completing the square. Factors must expand to give *their* coefficient of \(x^2\). |
| \((2, 0),\ (3, 5)\) | A1 A1 | Coordinates must be clearly paired; A1 for each correct point. A1 A0 available if two \(x\) or \(y\) values only. If M0 for solving quadratic, SC B2 can be awarded for correct coordinates, SC B1 if two \(x\) or \(y\) values only. |
| \((AB)^2 = (3-2)^2 + (5-0)^2\) | M1 | SOI. Using *their* points to find length of \(AB\). |
| \(AB = \sqrt{26}\) | A1 | ISW. Dependent on final M1 only. |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{y+10}{5}-5\right)^2 + (y-2)^2 = 13\) | M1 | Substitute \(x = \frac{y+10}{5}\) into *their* equation. |
| \(\frac{26y^2}{25} - \frac{26y}{5}\ [=0]\) | A1 FT | OE 2-term quadratic with all terms on one side. FT on *their* circle equation. |
| \([26]y(y-5)\ [=0]\) | M1 | Solve 2-term quadratic in \(y\) by factorising, using formula or completing the square. Factors must expand to give *their* coefficient of \(y^2\). |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r^2 [= (5-2)^2 + (7-5)^2] = 13$ | B1 | $r^2 = 13$ or $r = \sqrt{13}$ |
| Equation of circle is $(x-5)^2 + (y-2)^2 = 13$ | B1 FT | OE. FT on *their* 13 but LHS must be correct. |
| **Total: 2** | | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-5)^2 + (5x-10-2)^2 = 13$ | M1 | Substitute $y = 5x - 10$ into *their* equation. |
| $26x^2 - 130x + 156\ [=0]$ | A1 FT | OE 3-term quadratic with all terms on one side. FT on *their* circle equation. |
| $[26](x-2)(x-3)\ [=0]$ | M1 | Solve 3-term quadratic in $x$ by factorising, using formula or completing the square. Factors must expand to give *their* coefficient of $x^2$. |
| $(2, 0),\ (3, 5)$ | A1 A1 | Coordinates must be clearly paired; A1 for each correct point. A1 A0 available if two $x$ or $y$ values only. If M0 for solving quadratic, **SC B2** can be awarded for correct coordinates, **SC B1** if two $x$ or $y$ values only. |
| $(AB)^2 = (3-2)^2 + (5-0)^2$ | M1 | SOI. Using *their* points to find length of $AB$. |
| $AB = \sqrt{26}$ | A1 | ISW. Dependent on final M1 only. |
| **Total: 7** | | |
**Alternative method for 7(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{y+10}{5}-5\right)^2 + (y-2)^2 = 13$ | M1 | Substitute $x = \frac{y+10}{5}$ into *their* equation. |
| $\frac{26y^2}{25} - \frac{26y}{5}\ [=0]$ | A1 FT | OE 2-term quadratic with all terms on one side. FT on *their* circle equation. |
| $[26]y(y-5)\ [=0]$ | M1 | Solve 2-term quadratic in $y$ by factorising, using formula or completing the square. Factors must expand to give *their* coefficient of $y^2$. |
---7 A circle with centre $( 5,2 )$ passes through the point $( 7,5 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the circle.\\
The line $y = 5 x - 10$ intersects the circle at $A$ and $B$.
\item Find the exact length of the chord $A B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q7 [9]}}