| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard circle techniques: finding a tangent line using perpendicular gradients, writing a circle equation from center and radius, and solving simultaneous equations. All parts follow routine procedures with no novel insight required, making it easier than average for A-level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(m_{AB} = \frac{4-2}{-1-3} = -\frac{1}{2}\) | B1 | |
| Equation of tangent is \(y - 2 = 2(x-3)\) | B1 FT | \((3, 2)\) with their gradient \(-\frac{1}{m_{AB}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = 4^2 + 2^2 = 20\) or \(r^2 = 20\) or \(r = \sqrt{20}\) or \(AB = \sqrt{20}\) | B1 | |
| Equation of circle centre \(B\) is \((x-3)^2 + (y-2)^2 = 20\) | M1 A1 | FT their \(20\) for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 + (2x-6)^2 = \text{their } 20\) | M1 | Substitute their \(y - 2 = 2x - 6\) into their circle, centre \(B\) |
| \(5x^2 - 30x + 25 = 0\) or \(5(x-3)^2 = 20\) | A1 | |
| \([(5)(x-5)(x-1)\) or \(x - 3 = \pm 2]\) \(\quad x = 5, 1\) | A1 |
## Question 9(a):
| $m_{AB} = \frac{4-2}{-1-3} = -\frac{1}{2}$ | B1 | |
|---|---|---|
| Equation of tangent is $y - 2 = 2(x-3)$ | B1 FT | $(3, 2)$ with their gradient $-\frac{1}{m_{AB}}$ |
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## Question 9(b):
| $AB^2 = 4^2 + 2^2 = 20$ or $r^2 = 20$ or $r = \sqrt{20}$ or $AB = \sqrt{20}$ | B1 | |
|---|---|---|
| Equation of circle centre $B$ is $(x-3)^2 + (y-2)^2 = 20$ | M1 A1 | FT their $20$ for M1 |
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## Question 9(c):
| $(x-3)^2 + (2x-6)^2 = \text{their } 20$ | M1 | Substitute their $y - 2 = 2x - 6$ into their circle, centre $B$ |
|---|---|---|
| $5x^2 - 30x + 25 = 0$ or $5(x-3)^2 = 20$ | A1 | |
| $[(5)(x-5)(x-1)$ or $x - 3 = \pm 2]$ $\quad x = 5, 1$ | A1 | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{fdd6e942-b5bc-4369-8587-6de120459776-12_583_661_260_740}
The diagram shows a circle with centre $A$ passing through the point $B$. A second circle has centre $B$ and passes through $A$. The tangent at $B$ to the first circle intersects the second circle at $C$ and $D$.
The coordinates of $A$ are ( $- 1,4$ ) and the coordinates of $B$ are ( 3,2 ).
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent CBD.
\item Find an equation of the circle with centre $B$.
\item Find, by calculation, the $x$-coordinates of $C$ and $D$.\\
\includegraphics[max width=\textwidth, alt={}, center]{fdd6e942-b5bc-4369-8587-6de120459776-14_746_652_262_744}
The diagram shows a sector $C A B$ which is part of a circle with centre $C$. A circle with centre $O$ and radius $r$ lies within the sector and touches it at $D , E$ and $F$, where $C O D$ is a straight line and angle $A C D$ is $\theta$ radians.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q9 [8]}}