| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| 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 circle question requiring completing the square to find centre/radius, solving a quadratic for x-intercepts, finding perpendicular gradients, and basic coordinate geometry. All techniques are standard C1 procedures with no novel problem-solving required, making it easier than average but not trivial due to the multiple parts. |
| 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 |
|---|---|---|
| \(x^2 + y^2 - 4x - 6y - 12 = 0\) | M1 | |
| \(\Rightarrow (x-2)^2 + (y-3)^2 = 4 + 9 + 12 = 25\) | A1 | |
| \(\Rightarrow\) Centre \((2, 3)\), \(r = \sqrt{25} = 5\) | A1 A1 | Centre; Radius. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(x = 6,\ y = 0 \Rightarrow 4^2 + 3^2 = 25\) ✓ | B1 | Alt: solve \(x^2-4x-12=0\), B1 |
| Substitute \(y = 0 \Rightarrow (x-2)^2 + 9 = 25\) | M1 | M1 factorise |
| \(\Rightarrow (x-2)^2 = 16 \Rightarrow x-2 = \pm 4 \Rightarrow x = 6\) or \(-2\) | A1 | A1 both answers |
| \(\Rightarrow B\) is \((-2, 0)\) | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Grad \(QC = \frac{3-0}{2-6} = -\frac{3}{4} \Rightarrow\) Grad Tangent \(= \frac{4}{3}\) | B1 | |
| \(\Rightarrow y - 0 = \frac{4}{3}(x-6) \Rightarrow 3y = 4x - 24\) | M1 A1 | Total: 3 |
| Answer | Marks |
|---|---|
| Mid-point of \(BC\) is \((2, 0)\). | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Then \(3y = 8 - 24 = -16\) i.e. \(\left(2, -\frac{16}{3}\right)\) | A1 | Total: 2 |
# Question 12:
## Part (i):
$x^2 + y^2 - 4x - 6y - 12 = 0$ | M1 |
$\Rightarrow (x-2)^2 + (y-3)^2 = 4 + 9 + 12 = 25$ | A1 |
$\Rightarrow$ Centre $(2, 3)$, $r = \sqrt{25} = 5$ | A1 A1 | Centre; Radius. **Total: 4**
## Part (ii):
Substitute $x = 6,\ y = 0 \Rightarrow 4^2 + 3^2 = 25$ ✓ | B1 | Alt: solve $x^2-4x-12=0$, B1
Substitute $y = 0 \Rightarrow (x-2)^2 + 9 = 25$ | M1 | M1 factorise
$\Rightarrow (x-2)^2 = 16 \Rightarrow x-2 = \pm 4 \Rightarrow x = 6$ or $-2$ | A1 | A1 both answers
$\Rightarrow B$ is $(-2, 0)$ | | **Total: 3**
## Part (iii):
Grad $QC = \frac{3-0}{2-6} = -\frac{3}{4} \Rightarrow$ Grad Tangent $= \frac{4}{3}$ | B1 |
$\Rightarrow y - 0 = \frac{4}{3}(x-6) \Rightarrow 3y = 4x - 24$ | M1 A1 | **Total: 3**
## Part (iv):
Mid-point of $BC$ is $(2, 0)$. | M1 |
So $QM$ meets the tangent at $C$ when $x = 2$
Then $3y = 8 - 24 = -16$ i.e. $\left(2, -\frac{16}{3}\right)$ | A1 | **Total: 2**12 You are given that the equation of the circle shown in Fig. 12 is
$$x ^ { 2 } + y ^ { 2 } - 4 x - 6 y - 12 = 0$$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d20d10e0-6965-4f89-8855-8c6d32f5da90-4_742_971_422_481}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
(i) Show that the centre, Q , of the circle is $( 2,3 )$ and find the radius.\\
(ii) The circle crosses the $x$-axis at B and C .
Show that the coordinates of C are $( 6,0 )$ and find the coordinates of B .\\
(iii) Find the gradient of the line QC and hence find the equation of the tangent to the circle at C.\\
(iv) Given that M is the mid-point of BC , find the coordinates of the point where QM meets the tangent at C .
\hfill \mbox{\textit{OCR MEI C1 Q12 [12]}}