| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Geometric properties with circles |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring coordinate geometry of circles, tangent properties, distance calculations, and sector area. Part (iii) requiring proof that ADBC is a square demands geometric insight and systematic verification of properties (equal sides, right angles), while part (iv) involves angle calculation in sectors. More demanding than standard circle questions but uses familiar techniques throughout. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(C\) is \((1, -1)\) | B1 [1] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(y = \frac{3}{4}x - 8\) into equation of circle: \((x-1)^2 + \left(\frac{3}{4}x - 8 + 1\right)^2 = 25\) | M1 | AG — Attempt to eliminate one variable |
| \(x^2 - 8x + 16 = 0\) | M1 | Attempt to expand and collect terms to obtain 3-term quadratic; a correct 3-term quadratic |
| Either \((x-4)^2 = 0\), or Discriminant \(= (-8)^2 - 4\times1\times16 = 0\) | A1 | |
| So equation has a repeated root, so the line is a tangent | A1 [4] | Clearly argued |
| \(x = 4\) and \(y = -5\), so \(B\) is \((4, -5)\) | B1 [1] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\angle CAD = \angle CBD = 90°\) (radius perpendicular to tangent) | B1 | Allow for one or other of these angles |
| Gradient of \(AC = \frac{2-(-1)}{5-1} = \frac{3}{4}\); Gradient of \(BC = \frac{(-1)-(-5)}{1-4} = -\frac{4}{3}\); so \(AC \perp BC\), \(\angle ACB = 90°\); ADBC is a rectangle | B1 | |
| Either \(AC = BC =\) radius \([=5]\), or \(AD = BD\) equal tangents, so ADBC is a square | B1 [3] | Complete proof AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| E is the point \((1, -6)\) | B1 (AO 2.1) | May be implied |
| EITHER: Right-angled triangle with \(C(1,-1)\), \(B(4,-5)\); \(\theta = \arctan\left(\frac{3}{4}\right) = 0.6435\) | M1 (AO 3.1a) | Right-angled triangle formed and use of arctan |
| \(\theta = 0.6435\) | A1 (AO 3.1a) | oe |
| OR: \(BE = \sqrt{(4-1)^2 + (-5-(-6))^2} = \sqrt{10}\); Cosine rule in triangle BCE: \(\cos BCE = \frac{5^2 + 5^2 - 10}{2 \times 5 \times 5} = \frac{40}{50}\) | (M1) | Using distance BC and the cosine rule |
| \(\angle BCE = 0.6435...\) | (A1) | oe |
| OR: M is midpoint of BE, \(M = (2.5, -5.5)\); \(BM = \sqrt{(4-2.5)^2 + (-5-(-5.5))^2} = \frac{1}{2}\sqrt{10}\); \(\angle BCM = \arcsin\left(\frac{\frac{1}{2}\sqrt{10}}{5}\right) = 0.32175...\) | (M1) | Using trig in triangle BCM or ECM; Allow for \(\angle BCM\) |
| \(\angle BCE = 0.6435...\) | (A1) | oe. Must be \(\angle BCE\) |
| Area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2} \times 25 \times 2 \times 0.32175... = 8.04376\) | M1dep (AO 1.1a) | Using the sector area formula |
| \(= 8.04376\) | A1 [5] (AO 1.1b) | FT their \(\angle BCM\) |
# Question 12:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ is $(1, -1)$ | B1 [1] | cao |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $y = \frac{3}{4}x - 8$ into equation of circle: $(x-1)^2 + \left(\frac{3}{4}x - 8 + 1\right)^2 = 25$ | M1 | AG — Attempt to eliminate one variable |
| $x^2 - 8x + 16 = 0$ | M1 | Attempt to expand and collect terms to obtain 3-term quadratic; a correct 3-term quadratic |
| Either $(x-4)^2 = 0$, or Discriminant $= (-8)^2 - 4\times1\times16 = 0$ | A1 | |
| So equation has a repeated root, so the line is a tangent | A1 [4] | Clearly argued |
| $x = 4$ and $y = -5$, so $B$ is $(4, -5)$ | B1 [1] | cao |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\angle CAD = \angle CBD = 90°$ (radius perpendicular to tangent) | B1 | Allow for one or other of these angles |
| Gradient of $AC = \frac{2-(-1)}{5-1} = \frac{3}{4}$; Gradient of $BC = \frac{(-1)-(-5)}{1-4} = -\frac{4}{3}$; so $AC \perp BC$, $\angle ACB = 90°$; ADBC is a rectangle | B1 | |
| Either $AC = BC =$ radius $[=5]$, or $AD = BD$ equal tangents, so ADBC is a square | B1 [3] | Complete proof AG |
# Question (iv) [Sector Area]:
| Answer | Marks | Guidance |
|--------|-------|----------|
| E is the point $(1, -6)$ | **B1** (AO 2.1) | May be implied |
| **EITHER:** Right-angled triangle with $C(1,-1)$, $B(4,-5)$; $\theta = \arctan\left(\frac{3}{4}\right) = 0.6435$ | **M1** (AO 3.1a) | Right-angled triangle formed and use of arctan |
| $\theta = 0.6435$ | **A1** (AO 3.1a) | oe |
| **OR:** $BE = \sqrt{(4-1)^2 + (-5-(-6))^2} = \sqrt{10}$; Cosine rule in triangle BCE: $\cos BCE = \frac{5^2 + 5^2 - 10}{2 \times 5 \times 5} = \frac{40}{50}$ | **(M1)** | Using distance BC and the cosine rule |
| $\angle BCE = 0.6435...$ | **(A1)** | oe |
| **OR:** M is midpoint of BE, $M = (2.5, -5.5)$; $BM = \sqrt{(4-2.5)^2 + (-5-(-5.5))^2} = \frac{1}{2}\sqrt{10}$; $\angle BCM = \arcsin\left(\frac{\frac{1}{2}\sqrt{10}}{5}\right) = 0.32175...$ | **(M1)** | Using trig in triangle BCM or ECM; Allow for $\angle BCM$ |
| $\angle BCE = 0.6435...$ | **(A1)** | oe. Must be $\angle BCE$ |
| Area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 25 \times 2 \times 0.32175... = 8.04376$ | **M1dep** (AO 1.1a) | Using the sector area formula |
| $= 8.04376$ | **A1** [5] (AO 1.1b) | FT their $\angle BCM$ |
---12 Fig. 12 shows the circle $( x - 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 25$, the line $4 y = 3 x - 32$ and the tangent to the circle at the point $\mathrm { A } ( 5,2 )$. D is the point of intersection of the line $4 y = 3 x - 32$ and the tangent at A .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{904025c9-6d68-4344-bd41-8c0fccfcf92f-07_750_773_1311_632}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Write down the coordinates of C , the centre of the circle.
\item (A) Show that the line $4 y = 3 x - 32$ is a tangent to the circle.\\
(B) Find the coordinates of B , the point where the line $4 y = 3 x - 32$ touches the circle.
\item Prove that ADBC is a square.
\item The point E is the lowest point on the circle. Find the area of the sector ECB .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q12 [14]}}