| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - intersection or geometric properties |
| Difficulty | Standard +0.8 This is a multi-part coordinate geometry question requiring several techniques: distance formula, right-angle triangle geometry with tangents, chord of contact formula, and simultaneous equations. Part (b) requires recognizing the tangent-radius relationship and using trigonometry. Parts (c) and (d) involve finding the chord of contact equation and solving it with the circle equation. While systematic, it demands more problem-solving and integration of concepts than a routine question, placing it moderately above average difficulty. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((-6-8)^2 + (6-4)^2\) | M1 | OE |
| \(= 200\) | A1 | |
| \(\sqrt{200} > 10\), hence outside circle | A1 | AG ('Shown' not sufficient). Accept equivalents of \(\sqrt{200} > 10\) |
| Alternative method: | ||
| Radius \(= 10\) and \(C = (8, 4)\) | B1 | |
| \(\text{Min}(x)\) on circle \(= 8 - 10 = -2\) | M1 | |
| Hence outside circle | A1 | AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{angle} = \sin^{-1}\left(\frac{their\ 10}{their\ 10\sqrt{2}}\right)\) | M1 | Allow decimals for \(10\sqrt{2}\) at this stage. If cosine used, angle \(ACT\) or \(BCT\) must be identified, or implied by use of \(90°- 45°\) |
| \(\text{angle} = \sin^{-1}\left(\frac{1}{\sqrt{2}}\ \text{or}\ \frac{\sqrt{2}}{2}\ \text{or}\ \frac{10}{10\sqrt{2}}\ \text{or}\ \frac{10}{\sqrt{200}}\right) = 45°\) | A1 | AG Do not allow decimals |
| Alternative method: | ||
| \((10\sqrt{2})^2 = 10^2 + TA^2\) | M1 | |
| \(TA = 10 \to 45°\) | A1 | AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient, \(m\), of \(CT = -\frac{1}{7}\) | B1 | OE |
| Attempt to find mid-point (M) of \(CT\) | *M1 | Expect \((1, 5)\) |
| Equation of \(AB\) is \(y - 5 = 7(x-1)\) | DM1 | Through \(their\) \((1, 5)\) with gradient \(-\frac{1}{m}\) |
| \(y = 7x - 2\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-8)^2 + (7x-2-4)^2 = 100\) or equivalent in terms of \(y\) | M1 | Substitute \(their\) equation of \(AB\) into equation of circle |
| \(50x^2 - 100x = 0\) | A1 | |
| \(x = 0\) and \(2\) | A1 | WWW |
| Alternative method: | ||
| \(\mathbf{MC} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}\) | M1 | |
| \(\begin{pmatrix}1\\5\end{pmatrix} + \begin{pmatrix}-1\\-7\end{pmatrix} = \begin{pmatrix}0\\-2\end{pmatrix}\), \(\begin{pmatrix}1\\5\end{pmatrix} + \begin{pmatrix}1\\7\end{pmatrix} = \begin{pmatrix}2\\12\end{pmatrix}\) | A1 | |
| \(x = 0\) and \(2\) | A1 | |
| 3 |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(-6-8)^2 + (6-4)^2$ | M1 | OE |
| $= 200$ | A1 | |
| $\sqrt{200} > 10$, hence outside circle | A1 | AG ('Shown' not sufficient). Accept equivalents of $\sqrt{200} > 10$ |
| **Alternative method:** | | |
| Radius $= 10$ and $C = (8, 4)$ | B1 | |
| $\text{Min}(x)$ on circle $= 8 - 10 = -2$ | M1 | |
| Hence outside circle | A1 | AG |
| | **3** | |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{angle} = \sin^{-1}\left(\frac{their\ 10}{their\ 10\sqrt{2}}\right)$ | M1 | Allow decimals for $10\sqrt{2}$ at this stage. If cosine used, angle $ACT$ or $BCT$ must be identified, or implied by use of $90°- 45°$ |
| $\text{angle} = \sin^{-1}\left(\frac{1}{\sqrt{2}}\ \text{or}\ \frac{\sqrt{2}}{2}\ \text{or}\ \frac{10}{10\sqrt{2}}\ \text{or}\ \frac{10}{\sqrt{200}}\right) = 45°$ | A1 | AG Do not allow decimals |
| **Alternative method:** | | |
| $(10\sqrt{2})^2 = 10^2 + TA^2$ | M1 | |
| $TA = 10 \to 45°$ | A1 | AG |
| | **2** | |
---
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient, $m$, of $CT = -\frac{1}{7}$ | B1 | OE |
| Attempt to find mid-point (M) of $CT$ | *M1 | Expect $(1, 5)$ |
| Equation of $AB$ is $y - 5 = 7(x-1)$ | DM1 | Through $their$ $(1, 5)$ with gradient $-\frac{1}{m}$ |
| $y = 7x - 2$ | A1 | |
| | **4** | |
---
## Question 11(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-8)^2 + (7x-2-4)^2 = 100$ or equivalent in terms of $y$ | M1 | Substitute $their$ equation of $AB$ into equation of circle |
| $50x^2 - 100x = 0$ | A1 | |
| $x = 0$ and $2$ | A1 | WWW |
| **Alternative method:** | | |
| $\mathbf{MC} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}$ | M1 | |
| $\begin{pmatrix}1\\5\end{pmatrix} + \begin{pmatrix}-1\\-7\end{pmatrix} = \begin{pmatrix}0\\-2\end{pmatrix}$, $\begin{pmatrix}1\\5\end{pmatrix} + \begin{pmatrix}1\\7\end{pmatrix} = \begin{pmatrix}2\\12\end{pmatrix}$ | A1 | |
| $x = 0$ and $2$ | A1 | |
| | **3** | |11 A circle with centre $C$ has equation $( x - 8 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 100$.
\begin{enumerate}[label=(\alph*)]
\item Show that the point $T ( - 6,6 )$ is outside the circle.\\
Two tangents from $T$ to the circle are drawn.
\item Show that the angle between one of the tangents and $C T$ is exactly $45 ^ { \circ }$.\\
The two tangents touch the circle at $A$ and $B$.
\item Find the equation of the line $A B$, giving your answer in the form $y = m x + c$.
\item Find the $x$-coordinates of $A$ and $B$.\\
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 Q11 [12]}}