| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on circle geometry requiring standard techniques: finding circle equation from diameter endpoints (midpoint for centre, distance for radius), applying a translation vector, finding intersection by substitution, and algebraic manipulation. All steps are routine AS-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03d 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 |
| Centre \(= (2, -1)\) | B1 | |
| \(r^2 = [2-(-3)]^2 + [-1-(-5)]^2\) or \([2-7]^2 + [-1-3]^2\) OE | M1 | OR \(\frac{1}{2}[(-3-7)^2 + (-5-3)^2]\) OE |
| \((x-2)^2 + (y+1)^2 = 41\) | A1 | Must not involve surd form. SCB3 \((x+3)(x-7)+(y+5)(y-3)=0\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Centre \(= \textit{their}\ (2,-1) + \binom{8}{4} = (10, 3)\) | B1FT | SOI. FT on *their* \((2,-1)\) |
| \((x-10)^2 + (y-3)^2 = \textit{their}\ 41\) | B1FT | FT on *their* 41 even if in surd form. SCB2 \((x-5)(x-15)+(y+1)(y-7)=0\) |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient \(m\) of line joining centres \(= \frac{4}{8}\) OE | B1 | |
| Attempt to find mid-point of line | M1 | Expect \((6, 1)\) |
| Equation of \(RS\) is \(y - 1 = -2(x-6)\) | M1 | Through *their* \((6,1)\) with gradient \(\frac{-1}{m}\) |
| \(y = -2x + 13\) | A1 | AG |
| Alternative method: \((x-2)^2+(y+1)^2-41=(x-10)^2+(y-3)^2-41\) OE | M1 | |
| \(x^2-4x+4+y^2+2y+1=x^2-20x+100+y^2-6y+9\) OE | A1 | Condone 1 error or errors caused by 1 error in first line |
| \(16x + 8y = 104\) | A1 | |
| \(y = -2x + 13\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-10)^2 + (-2x+13-3)^2 = 41\) | M1 | Or eliminate \(y\) between \(C_1\) and \(C_2\) |
| \(x^2-20x+100+4x^2-40x+100=41 \rightarrow 5x^2-60x+159=0\) | A1 | AG |
| Total: 2 |
## Question 12(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre $= (2, -1)$ | B1 | |
| $r^2 = [2-(-3)]^2 + [-1-(-5)]^2$ or $[2-7]^2 + [-1-3]^2$ OE | M1 | OR $\frac{1}{2}[(-3-7)^2 + (-5-3)^2]$ OE |
| $(x-2)^2 + (y+1)^2 = 41$ | A1 | Must not involve surd form. **SCB3** $(x+3)(x-7)+(y+5)(y-3)=0$ |
| **Total: 3** | | |
## Question 12(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre $= \textit{their}\ (2,-1) + \binom{8}{4} = (10, 3)$ | B1FT | SOI. FT on *their* $(2,-1)$ |
| $(x-10)^2 + (y-3)^2 = \textit{their}\ 41$ | B1FT | FT on *their* 41 even if in surd form. **SCB2** $(x-5)(x-15)+(y+1)(y-7)=0$ |
| **Total: 2** | | |
## Question 12(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient $m$ of line joining centres $= \frac{4}{8}$ OE | B1 | |
| Attempt to find mid-point of line | M1 | Expect $(6, 1)$ |
| Equation of $RS$ is $y - 1 = -2(x-6)$ | M1 | Through *their* $(6,1)$ with gradient $\frac{-1}{m}$ |
| $y = -2x + 13$ | A1 | AG |
| **Alternative method:** $(x-2)^2+(y+1)^2-41=(x-10)^2+(y-3)^2-41$ OE | M1 | |
| $x^2-4x+4+y^2+2y+1=x^2-20x+100+y^2-6y+9$ OE | A1 | Condone 1 error or errors caused by 1 error in first line |
| $16x + 8y = 104$ | A1 | |
| $y = -2x + 13$ | A1 | AG |
| **Total: 4** | | |
## Question 12(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-10)^2 + (-2x+13-3)^2 = 41$ | M1 | Or eliminate $y$ between $C_1$ and $C_2$ |
| $x^2-20x+100+4x^2-40x+100=41 \rightarrow 5x^2-60x+159=0$ | A1 | AG |
| **Total: 2** | | |12 A diameter of a circle $C _ { 1 }$ has end-points at $( - 3 , - 5 )$ and $( 7,3 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the circle $C _ { 1 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{01b98496-a717-4c68-8489-42d2203b700f-16_618_846_1062_644}
The circle $C _ { 1 }$ is translated by $\binom { 8 } { 4 }$ to give circle $C _ { 2 }$, as shown in the diagram.
\item Find an equation of the circle $C _ { 2 }$.\\
The two circles intersect at points $R$ and $S$.
\item Show that the equation of the line $R S$ is $y = - 2 x + 13$.
\item Hence show that the $x$-coordinates of $R$ and $S$ satisfy the equation $5 x ^ { 2 } - 60 x + 159 = 0$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q12 [11]}}