| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.8 This is a multi-part C1 question covering standard circle properties (radius, equation, diameter, tangent) and coordinate geometry. All parts use routine techniques: symmetry for finding B, distance formula for radius, midpoint for diameter endpoint, perpendicular gradient for tangent. While multi-step, each component is straightforward recall and application of basic formulas with no novel problem-solving required. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02q Use intersection points: of graphs to solve equations1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| \((7, 0)\) | B1 | Accept \(x = 7, y = 0\); condone 7, 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{13}\) | M1 | Pythagoras used correctly e.g. \(r^2 = 3^2 + 2^2\) or substituting A into \((x-4)^2 + (y-2)^2 = r^2\) |
| \(\sqrt{13}\) | B1 | For \(r = \pm\sqrt{13}\); do not accept \((\sqrt{13})^2\) instead of 13 |
| \((x-4)^2 + (y-2)^2 = 13\) or ft their evaluated \(r^2\) | M1 | For one side correct as part of equation with \(x\) and \(y\) terms; allow M1 for LHS \((x-4)^2 + (y-2)^2 = r^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((7, 4)\) | B1 | Each coord; accept \(x=7, y=4\); condone 7, 4 |
| M1 | If B0, then M1 for vector approach such as '3 along and 2 up' from A to C, or M1 for \(\frac{x_D+1}{2}=4\) and \(\frac{y_D+0}{2}=2\) | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| grad tgt \(= -\frac{3}{2}\) oe | M2 | Correctly obtained or ft their D; M1 for grad \(AD = \frac{4-0}{7-1}\) oe isw or \(\frac{2}{3}\) seen/used; grad tgt \(= -1/\) their grad \(AD\) |
| \(y - \text{their } 4 = \text{their } (-\frac{3}{2})(x - \text{their } 7)\) | M1 | Or substitute \((7,4)\) into \(y = \text{their}(-\frac{3}{2})x + b\); M0 if grad \(AD\) used or wrong gradient with no working |
| \(y = -1.5x + 14.5\) oe isw | A1 | Must be in form \(y = ax + b\); condone \(y = \frac{-3x+29}{2}\); condone \(y = -1.5x + b\) and \(b = 14.5\) |
## Question 10:
**Part (i):**
$(7, 0)$ | B1 | Accept $x = 7, y = 0$; condone 7, 0
**Part (ii):**
$\sqrt{13}$ | M1 | Pythagoras used correctly e.g. $r^2 = 3^2 + 2^2$ or substituting A into $(x-4)^2 + (y-2)^2 = r^2$
$\sqrt{13}$ | B1 | For $r = \pm\sqrt{13}$; do not accept $(\sqrt{13})^2$ instead of 13
$(x-4)^2 + (y-2)^2 = 13$ or ft their evaluated $r^2$ | M1 | For one side correct as part of equation with $x$ and $y$ terms; allow M1 for LHS $(x-4)^2 + (y-2)^2 = r^2$ | **[4]**
**Part (iii):**
$(7, 4)$ | B1 | Each coord; accept $x=7, y=4$; condone 7, 4
| M1 | If B0, then M1 for vector approach such as '3 along and 2 up' from A to C, or M1 for $\frac{x_D+1}{2}=4$ and $\frac{y_D+0}{2}=2$ | **[2]**
**Part (iv):**
grad tgt $= -\frac{3}{2}$ oe | M2 | Correctly obtained or ft their D; M1 for grad $AD = \frac{4-0}{7-1}$ oe isw or $\frac{2}{3}$ seen/used; grad tgt $= -1/$ their grad $AD$
$y - \text{their } 4 = \text{their } (-\frac{3}{2})(x - \text{their } 7)$ | M1 | Or substitute $(7,4)$ into $y = \text{their}(-\frac{3}{2})x + b$; M0 if grad $AD$ used or wrong gradient with no working
$y = -1.5x + 14.5$ oe isw | A1 | Must be in form $y = ax + b$; condone $y = \frac{-3x+29}{2}$; condone $y = -1.5x + b$ and $b = 14.5$ | **[4]**
---10 Fig. 10 shows a sketch of a circle with centre $\mathrm { C } ( 4,2 )$. The circle intersects the $x$-axis at $\mathrm { A } ( 1,0 )$ and at B .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e8f2d63-8a25-4da2-8c3e-9e75ea1b7c08-3_680_800_1146_628}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
(i) Write down the coordinates of B .\\
(ii) Find the radius of the circle and hence write down the equation of the circle.\\
(iii) AD is a diameter of the circle. Find the coordinates of D .\\
(iv) Find the equation of the tangent to the circle at D . Give your answer in the form $y = a x + b$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e8f2d63-8a25-4da2-8c3e-9e75ea1b7c08-4_643_853_269_589}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows a sketch of the curve with equation $y = ( x - 4 ) ^ { 2 } - 3$.\\
(i) Write down the equation of the line of symmetry of the curve and the coordinates of the minimum point.\\
(ii) Find the coordinates of the points of intersection of the curve with the $x$-axis and the $y$-axis, using surds where necessary.\\
(iii) The curve is translated by $\binom { 2 } { 0 }$. Show that the equation of the translated curve may be written as $y = x ^ { 2 } - 12 x + 33$.\\
(iv) Show that the line $y = 8 - 2 x$ meets the curve $y = x ^ { 2 } - 12 x + 33$ at just one point, and find the coordinates of this point.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e8f2d63-8a25-4da2-8c3e-9e75ea1b7c08-5_775_1461_317_296}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
Fig. 12 shows the graph of a cubic curve. It intersects the axes at $( - 5,0 ) , ( - 2,0 ) , ( 1.5,0 )$ and $( 0 , - 30 )$.\\
(i) Use the intersections with both axes to express the equation of the curve in a factorised form.\\
(ii) Hence show that the equation of the curve may be written as $y = 2 x ^ { 3 } + 11 x ^ { 2 } - x - 30$.\\
(iii) Draw the line $y = 5 x + 10$ accurately on the graph. The curve and this line intersect at ( $- 2,0$ ); find graphically the $x$-coordinates of the other points of intersection.\\
(iv) Show algebraically that the $x$-coordinates of the other points of intersection satisfy the equation
$$2 x ^ { 2 } + 7 x - 20 = 0 .$$
Hence find the exact values of the $x$-coordinates of the other points of intersection.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR MEI C1 2014 Q10 [11]}}