Question 2 - A-Level Maths

OCR MEI C1 — Question 2 11 marks

Exam Board	OCR MEI
Module	C1 (Core Mathematics 1)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Circle from diameter endpoints
Difficulty	Easy -1.2 This is a straightforward C1 circle question requiring only basic coordinate geometry: symmetry to find B, distance formula for radius, diameter endpoint calculation, and perpendicular gradient for tangent. All steps are routine applications of standard techniques with no problem-solving insight needed.
Spec	1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle

2 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]

\includegraphics[alt={},max width=\textwidth]{55e2d4f5-c84d-4577-988e-96071a220d60-2_689_811_430_662} \captionsetup{labelformat=empty} \caption{Fig. 10}

\end{figure}

Write down the coordinates of B .
Find the radius of the circle and hence write down the equation of the circle.
AD is a diameter of the circle. Find the coordinates of D .
Find the equation of the tangent to the circle at D . Give your answer in the form \(y = a x + b\).

Show mark scheme Show mark scheme source

Question 2(i):

Answer	Marks	Guidance
Answer	Mark	Guidance
\((7, 0)\)	1	accept \(x = 7\), \(y = 0\); condone \(7, 0\)

Question 2(ii):

Answer	Marks	Guidance
Answer	Mark	Guidance
\(\sqrt{13}\)	2	M1 for Pythagoras used correctly eg \([r^2 =] 3^2 + 2^2\) or for subst A or their B in \((x-4)^2 + (y-2)^2 [= r^2]\); or B1 for \([r =] \pm\sqrt{13}\); annotate if partially correct; allow recovery if some confusion between squares and roots but correct answer found
\((x-4)^2 + (y-2)^2 = 13\) or ft their evaluated \(r^2\), isw	2	M1 for one side correct, as part of an equation with \(x\) and \(y\) terms; do not accept \((\sqrt{13})^2\) instead of 13

Question 2(iii):

Answer	Marks	Guidance
Answer	Mark	Guidance
\((7, 4)\)	2	B1 each coord; accept \(x = 7\), \(y = 4\); if B0, then M1 for a vector or coordinates approach such as '3 along and 2 up' to get from A to C oe; or M1 for \(\frac{x_D + 1}{2} = 4\) and \(\frac{y_D + 0}{2} = 2\)

Question 2(iv):

Answer	Marks	Guidance
Answer	Mark	Guidance
grad tgt \(= -3/2\) oe	M2	correctly obtained or ft their D if used; M1 for grad \(AD = \frac{4-0}{7-1}\) oe isw or \(2/3\) oe seen or used in this part or for their grad tgt \(= -1/\) their grad AD; may use AD, CD or AC
\(y - \text{their } 4 = \text{their } (-3/2)(x - \text{their } 7)\)	M1	or subst \((7, 4)\) into \(y = \text{their } (-3/2)x + b\); M0 if grad AD oe used or if a wrong gradient appears with no previous 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 \(b = 14.5\) oe

## Question 2(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $(7, 0)$ | 1 | accept $x = 7$, $y = 0$; condone $7, 0$ |

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## Question 2(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{13}$ | 2 | M1 for Pythagoras used correctly eg $[r^2 =] 3^2 + 2^2$ or for subst A or their B in $(x-4)^2 + (y-2)^2 [= r^2]$; or B1 for $[r =] \pm\sqrt{13}$; annotate if partially correct; allow recovery if some confusion between squares and roots but correct answer found |
| $(x-4)^2 + (y-2)^2 = 13$ or ft their evaluated $r^2$, isw | 2 | M1 for one side correct, as part of an equation with $x$ and $y$ terms; do not accept $(\sqrt{13})^2$ instead of 13 |

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## Question 2(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $(7, 4)$ | 2 | B1 each coord; accept $x = 7$, $y = 4$; if B0, then M1 for a vector or coordinates approach such as '3 along and 2 up' to get from A to C oe; or M1 for $\frac{x_D + 1}{2} = 4$ and $\frac{y_D + 0}{2} = 2$ |

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## Question 2(iv):

| Answer | Mark | Guidance |
|--------|------|----------|
| grad tgt $= -3/2$ oe | M2 | correctly obtained or ft their D if used; M1 for grad $AD = \frac{4-0}{7-1}$ oe isw or $2/3$ oe seen or used in this part or for their grad tgt $= -1/$ their grad AD; may use AD, CD or AC |
| $y - \text{their } 4 = \text{their } (-3/2)(x - \text{their } 7)$ | M1 | or subst $(7, 4)$ into $y = \text{their } (-3/2)x + b$; M0 if grad AD oe used or if a wrong gradient appears with no previous 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 $b = 14.5$ oe |

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Show LaTeX source

2 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]{55e2d4f5-c84d-4577-988e-96071a220d60-2_689_811_430_662}
\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 $y = a x + b$.

\hfill \mbox{\textit{OCR MEI C1  Q2 [11]}}

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