| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - intersection or geometric properties |
| Difficulty | Challenging +1.2 This is a multi-step circle geometry problem requiring substitution of a line into a circle equation, using the discriminant condition for tangency, then finding the angle between two tangents. While it involves several techniques (discriminant, tangent conditions, angle between lines), these are standard A-level methods applied in a straightforward context. The structure is guided and the techniques are well-practiced, making it moderately above average difficulty. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.05o Trigonometric equations: solve in given intervals1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 + (mx+2)^2 - 10x - 14(mx+2) + 64 = 0\) | M1 | Substitute equation of tangent into equation of circle; could work backwards eliminating \(m\) to obtain equation of circle |
| \(x^2 + m^2x^2 + 4mx + 4 - 10x - 14mx - 28 + 64 = 0\) leading to \((m^2+1)x^2 - 10(m+1)x + 40 = 0\) A.G. | A1 | Expand and tidy to given answer, including \(= 0\) in final answer; AG so unsimplified expansion needs to be seen |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(100(m+1)^2 - 160(m^2+1) = 0\) | M1* | Use \(b^2 - 4ac = 0\); M1 only awarded when \(= 0\) soi |
| \(60m^2 - 200m + 60 = 0\) | A1 | Obtain correct equation; any correct 3 term equation |
| \((3m-1)(m-3) = 0\) | M1d* | Attempt to solve quadratic; DR so method for solving must be shown |
| \(m = 3,\ m = \frac{1}{3}\) | ||
| \(y = 3x + 2\) | A1 | Obtain correct equation; SC B1 for correct equation if roots not justified; A0 if second equation also given |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| radius \(= \sqrt{10}\), \(PC = 5\sqrt{2}\), \(PA = PB = 2\sqrt{10}\), \(AB = 4\sqrt{2}\) | M1 | Attempt (at least 2) useful lengths; NB points of intersection are \((2,8)\) and \((6,4)\) |
| \(\tan(\frac{1}{2}APB) = \frac{1}{2}\) | A1 | Obtain a correct related trig ratio; \(\cos APB = \frac{3}{5}\) from cosine rule |
| \(\tan APB = \frac{1}{1 - \frac{1}{4}}\) | M1 | Attempt \(\tan APB\); DR so need to see use of identity or relevant triangle to find \(\tan APB\) |
| \(\tan APB = \frac{4}{3}\) | A1 | Obtain \(\frac{4}{3}\); from explicit, exact working |
| [4] |
## Question 11 - Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + (mx+2)^2 - 10x - 14(mx+2) + 64 = 0$ | M1 | Substitute equation of tangent into equation of circle; could work backwards eliminating $m$ to obtain equation of circle |
| $x^2 + m^2x^2 + 4mx + 4 - 10x - 14mx - 28 + 64 = 0$ leading to $(m^2+1)x^2 - 10(m+1)x + 40 = 0$ **A.G.** | A1 | Expand and tidy to given answer, including $= 0$ in final answer; AG so unsimplified expansion needs to be seen |
| | **[2]** | |
## Question 11 - Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $100(m+1)^2 - 160(m^2+1) = 0$ | M1* | Use $b^2 - 4ac = 0$; M1 only awarded when $= 0$ soi |
| $60m^2 - 200m + 60 = 0$ | A1 | Obtain correct equation; any correct 3 term equation |
| $(3m-1)(m-3) = 0$ | M1d* | Attempt to solve quadratic; DR so method for solving must be shown |
| $m = 3,\ m = \frac{1}{3}$ | | |
| $y = 3x + 2$ | A1 | Obtain correct equation; SC B1 for correct equation if roots not justified; A0 if second equation also given |
| | **[4]** | |
## Question 11 - Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| radius $= \sqrt{10}$, $PC = 5\sqrt{2}$, $PA = PB = 2\sqrt{10}$, $AB = 4\sqrt{2}$ | M1 | Attempt (at least 2) useful lengths; NB points of intersection are $(2,8)$ and $(6,4)$ |
| $\tan(\frac{1}{2}APB) = \frac{1}{2}$ | A1 | Obtain a correct related trig ratio; $\cos APB = \frac{3}{5}$ from cosine rule |
| $\tan APB = \frac{1}{1 - \frac{1}{4}}$ | M1 | Attempt $\tan APB$; DR so need to see use of identity or relevant triangle to find $\tan APB$ |
| $\tan APB = \frac{4}{3}$ | A1 | Obtain $\frac{4}{3}$; from explicit, exact working |
| | **[4]** | |
---\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of $A$ satisfies the equation $\left( m ^ { 2 } + 1 \right) x ^ { 2 } - 10 ( m + 1 ) x + 40 = 0$.
\item Hence determine the equation of the tangent to the circle at $A$ which passes through $P$. [4]
A second tangent is drawn from $P$ to meet the circle at a second point $B$. The equation of this tangent is of the form $y = n x + 2$, where $n$ is a constant less than 1 .
\end{enumerate}\item Determine the exact value of $\tan A P B$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2020 Q11 [10]}}