| 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 -0.3 This is a standard multi-part circle question requiring routine techniques: finding tangent gradient (perpendicular to radius), writing line equation, solving simultaneous equations, and verifying tangency by substitution/discriminant. All methods are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| grad \(AC = \frac{7-3}{3-1}\) or \(4/2\) o.e. \([= 2]\) | M1 | not from using \(-\frac{1}{2}\) |
| so grad \(AT = -\frac{1}{2}\) | M1 | or ft their grad AC [for use of \(m_1 m_2 = -1\)] |
| eqn of \(AT\) is \(y - 7 = -\frac{1}{2}(x - 3)\) | M1 | or subst \((3, 7)\) in \(y = -\frac{1}{2}x + c\) or in \(2y + x = 17\); allow ft from their grad of AT, except 2 (may be AC not AT) |
| one correct constructive step towards \(x + 2y = 17\) [ans given] | M1 | or working back from given line to \(y = -\frac{1}{2}x + 8.5\) o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x + 2(2x - 9) = 17\) | M1 | attempt at subst for \(x\) or \(y\) or elimination |
| \(5x - 18 = 17\) or \(5x = 35\) o.e. | A1 | allow \(2.5x = 17.5\) etc |
| \(x = 7\) and \(y = 5\) [so \((7, 5)\)] | B1 | graphically: allow M2 for both lines correct or showing \((7, 5)\) fits both lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-1)^2 + (2x-12)^2 = 20\) | M1 | subst \(2x - 9\) for \(y\) [oe for \(x\)] |
| \(5x^2 - 50x + 125 [= 0]\) | M1 | rearranging to 0; condone one error |
| \((x-5)^2 = 0\) | A1 | showing 5 is root and only root |
| equal roots so tangent | B1 | explicit statement of condition needed (may be obtained earlier in part) or showing line is perp. to radius at point of contact |
| \((5, 1)\) | B1 | condone \(x = 5\), \(y = 1\) |
| or \(y - 3 = -\frac{1}{2}(x-1)\) o.e. seen | M1 | or if \(y = 2x - 9\) is tgt then line through C with gradient \(-\frac{1}{2}\) is radius |
| subst or elim. with \(y = 2x - 9\), \(x = 5\), \((5,1)\), showing \((5,1)\) on circle | M1, A1, B1, B1 | or showing distance between \((1, 3)\) and \((5, 1) = \sqrt{20}\) |
## Question 3:
**Part i:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| grad $AC = \frac{7-3}{3-1}$ or $4/2$ o.e. $[= 2]$ | M1 | not from using $-\frac{1}{2}$ |
| so grad $AT = -\frac{1}{2}$ | M1 | or ft their grad AC [for use of $m_1 m_2 = -1$] |
| eqn of $AT$ is $y - 7 = -\frac{1}{2}(x - 3)$ | M1 | or subst $(3, 7)$ in $y = -\frac{1}{2}x + c$ or in $2y + x = 17$; allow ft from their grad of AT, except 2 (may be AC not AT) |
| one correct constructive step towards $x + 2y = 17$ [ans given] | M1 | or working back from given line to $y = -\frac{1}{2}x + 8.5$ o.e. |
**Part ii:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + 2(2x - 9) = 17$ | M1 | attempt at subst for $x$ or $y$ or elimination |
| $5x - 18 = 17$ or $5x = 35$ o.e. | A1 | allow $2.5x = 17.5$ etc |
| $x = 7$ and $y = 5$ [so $(7, 5)$] | B1 | graphically: allow M2 for both lines correct or showing $(7, 5)$ fits both lines |
**Part iii:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-1)^2 + (2x-12)^2 = 20$ | M1 | subst $2x - 9$ for $y$ [oe for $x$] |
| $5x^2 - 50x + 125 [= 0]$ | M1 | rearranging to 0; condone one error |
| $(x-5)^2 = 0$ | A1 | showing 5 is root and only root |
| equal roots so tangent | B1 | explicit statement of condition needed (may be obtained earlier in part) or showing line is perp. to radius at point of contact |
| $(5, 1)$ | B1 | condone $x = 5$, $y = 1$ |
| **or** $y - 3 = -\frac{1}{2}(x-1)$ o.e. seen | M1 | or if $y = 2x - 9$ is tgt then line through C with gradient $-\frac{1}{2}$ is radius |
| subst or elim. with $y = 2x - 9$, $x = 5$, $(5,1)$, showing $(5,1)$ on circle | M1, A1, B1, B1 | or showing distance between $(1, 3)$ and $(5, 1) = \sqrt{20}$ |
---3
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{50cfc73d-850e-4a9b-b088-cc9741b66ffb-2_445_617_1008_741}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Not to scale
A circle has centre $\mathrm { C } ( 1,3 )$ and passes through the point $\mathrm { A } ( 3,7 )$ 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 Q3 [12]}}