| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question requiring standard techniques: finding a tangent equation using perpendicular gradients (part i), solving simultaneous linear equations (part ii), and using the discriminant method to verify tangency (part iii). All methods are routine C1 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 circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Radius \(CA\): gradient \(= \frac{7-3}{3-1} = 2\) | M1 | Gradient of radius |
| Tangent gradient \(= -\frac{1}{2}\) | M1 | Negative reciprocal |
| Tangent: \(y - 7 = -\frac{1}{2}(x-3)\) | M1 | Line through \(A\) |
| \(2y - 14 = -(x-3)\), \(x + 2y = 17\) | A1 | Correct equation shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2x - 9\) and \(x + 2y = 17\) | M1 | Simultaneous equations |
| \(x + 2(2x-9) = 17\) | M1 | Substitute |
| \(5x = 35\), \(x = 7\), \(y = 5\) | A1 | \(T = (7, 5)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(y = 2x-9\) into circle equation: | M1 | Substitution |
| \((x-1)^2 + (2x-9-3)^2 = 20\) | ||
| \((x-1)^2 + (2x-12)^2 = 20\) | M1 | Expand |
| \(x^2 - 2x + 1 + 4x^2 - 48x + 144 = 20\) | ||
| \(5x^2 - 50x + 125 = 0\) | A1 | Correct quadratic |
| \(x^2 - 10x + 25 = 0\) | M1 | Simplify |
| \((x-5)^2 = 0\) | A1 | Repeated root \(\Rightarrow\) tangent |
| \(x = 5\), \(y = 1\); point \((5, 1)\) | A1 | Correct coordinates |
## Question 11:
**(i)**
Radius $CA$: gradient $= \frac{7-3}{3-1} = 2$ | M1 | Gradient of radius
Tangent gradient $= -\frac{1}{2}$ | M1 | Negative reciprocal
Tangent: $y - 7 = -\frac{1}{2}(x-3)$ | M1 | Line through $A$
$2y - 14 = -(x-3)$, $x + 2y = 17$ | A1 | Correct equation shown
**(ii)**
$y = 2x - 9$ and $x + 2y = 17$ | M1 | Simultaneous equations
$x + 2(2x-9) = 17$ | M1 | Substitute
$5x = 35$, $x = 7$, $y = 5$ | A1 | $T = (7, 5)$
**(iii)**
Substitute $y = 2x-9$ into circle equation: | M1 | Substitution
$(x-1)^2 + (2x-9-3)^2 = 20$ | |
$(x-1)^2 + (2x-12)^2 = 20$ | M1 | Expand
$x^2 - 2x + 1 + 4x^2 - 48x + 144 = 20$ | |
$5x^2 - 50x + 125 = 0$ | A1 | Correct quadratic
$x^2 - 10x + 25 = 0$ | M1 | Simplify
$(x-5)^2 = 0$ | A1 | Repeated root $\Rightarrow$ tangent
$x = 5$, $y = 1$; point $(5, 1)$ | A1 | Correct coordinates
---11
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4d8caf0f-7594-42cb-bd40-e6c11e2b6832-3_442_1102_1384_717}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
A circle has centre $C ( 1,3 )$ and passes through the point $A ( 3,7 )$ as shown in Fig. 11.\\
(i) Show that the equation of the tangent at A is $x + 2 y = 17$.\\
(ii) The line with equation $y = 2 x - 9$ intersects this tangent at the point T .
Find the coordinates of T .\\
(iii) The equation of the circle is $( x - 1 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 20$.
Show that the line with equation $y = 2 x - 9$ is a tangent to the circle. Give the coordinates of the point where this tangent touches the circle.
\hfill \mbox{\textit{OCR MEI C1 2007 Q11 [12]}}