| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question testing basic concepts: reading centre/radius from standard form, distance comparison, parallel line equation, and tangent verification using discriminant. All parts are routine C1 techniques with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-step nature of part (iv). |
| Spec | 1.03a 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 |
| \((5,\ 2)\) | 1 | |
| \(\sqrt{20}\) or \(2\sqrt{5}\) | 1 | 0 for \(\pm\sqrt{20}\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No, since \(\sqrt{20}<5\) or showing roots of \(y^2-4y+9=0\) o.e. are not real | 2 | Or ft from their centre and radius; M1 for attempt (no and mentioning \(\sqrt{20}\) or 5) or sketch or solving by formula or completing the square \((-5)^2+(y-2)^2=20\) [condone one error]; or SC1 for fully comparing distance from \(x\)-axis with radius and saying yes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2x-8\) or simplified alternative | 2 | M1 for \(y-2=2(x-5)\) or ft from (i); or M1 for \(y=2x+c\) and substitute their (i); or M1 for answer \(y=2x+k\), \(k\neq 0\) or \(-8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \((x-5)^2 + (2x)^2 = 20\) o.e. | M1 | subst \(2x + 2\) for \(y\) [oe for \(x\)] |
| \(5x^2 - 10x + 5[ = 0]\) or better equiv. | M1 | expanding brackets and rearranging to 0; condone one error; dep on first M1 |
| obtaining \(x = 1\) (with no other roots) or showing roots equal | M1 | dep on first M1 |
| one intersection [so tangent] | A1 | o.e.; must be explicit; or showing line joining \((1,4)\) to centre is perp to \(y = 2x + 2\) |
| \((1, 4)\) cao | A1 | allow \(y = 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(y - 2 = -\frac{1}{2}(x - 5)\) o.e. | M1 | line through centre perp to \(y = 2x + 2\) |
| \(2x + 2 = -\frac{1}{2}(x - 5)\) o.e. | M1 | dep; subst to find intn with \(y = 2x + 2\) |
| \(x = 1\) | A1 | |
| \(y = 4\) cao | A1 | |
| showing \((1, 4)\) is on circle | B1 | by subst in circle eqn or finding dist from centre \(= \sqrt{20}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| perp dist between \(y = 2x - 8\) and \(y = 2x + 2 = 10\cos\theta\) where \(\tan\theta = 2\) | M1 | |
| showing this is \(\sqrt{20}\) so tgt | M1 | |
| \(x = 5 - \sqrt{20}\sin\theta\) | M1 | or other valid method for obtaining \(x\) |
| \(x = 1\) | A1 | |
| \((1, 4)\) cao | A1 | allow \(y = 4\) |
| Total: 5 marks *(part iv)* | 11 marks *(full question)* |
## Question 4 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(5,\ 2)$ | 1 | |
| $\sqrt{20}$ or $2\sqrt{5}$ | 1 | 0 for $\pm\sqrt{20}$ etc |
---
## Question 4 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| No, since $\sqrt{20}<5$ or showing roots of $y^2-4y+9=0$ o.e. are not real | 2 | Or ft from their centre and radius; M1 for attempt (no and mentioning $\sqrt{20}$ or 5) or sketch or solving by formula or completing the square $(-5)^2+(y-2)^2=20$ [condone one error]; or SC1 for fully comparing distance from $x$-axis with radius and saying yes |
---
## Question 4 (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2x-8$ or simplified alternative | 2 | M1 for $y-2=2(x-5)$ or ft from (i); or M1 for $y=2x+c$ and substitute their (i); or M1 for answer $y=2x+k$, $k\neq 0$ or $-8$ |
## Question iv:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $(x-5)^2 + (2x)^2 = 20$ o.e. | M1 | subst $2x + 2$ for $y$ [oe for $x$] |
| $5x^2 - 10x + 5[ = 0]$ or better equiv. | M1 | expanding brackets and rearranging to 0; condone one error; dep on first M1 |
| obtaining $x = 1$ (with no other roots) or showing roots equal | M1 | dep on first M1 |
| one intersection [so tangent] | A1 | o.e.; must be explicit; or showing line joining $(1,4)$ to centre is perp to $y = 2x + 2$ |
| $(1, 4)$ cao | A1 | allow $y = 4$ |
**Alt Method 1:**
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $y - 2 = -\frac{1}{2}(x - 5)$ o.e. | M1 | line through centre perp to $y = 2x + 2$ |
| $2x + 2 = -\frac{1}{2}(x - 5)$ o.e. | M1 | dep; subst to find intn with $y = 2x + 2$ |
| $x = 1$ | A1 | |
| $y = 4$ cao | A1 | |
| showing $(1, 4)$ is on circle | B1 | by subst in circle eqn or finding dist from centre $= \sqrt{20}$ |
*Note: A similar method earns first M1 for eqn of diameter, 2nd M1 for intn of diameter and circle, A1 each for $x$ and $y$ coords and last B1 for showing $(1, 4)$ on line – award only A1 if $(1, 4)$ and $(9, 0)$ found without $(1, 4)$ being identified as the soln*
**Alt Method 2:**
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| perp dist between $y = 2x - 8$ and $y = 2x + 2 = 10\cos\theta$ where $\tan\theta = 2$ | M1 | |
| showing this is $\sqrt{20}$ so tgt | M1 | |
| $x = 5 - \sqrt{20}\sin\theta$ | M1 | or other valid method for obtaining $x$ |
| $x = 1$ | A1 | |
| $(1, 4)$ cao | A1 | allow $y = 4$ |
**Total: 5 marks** *(part iv)* | **11 marks** *(full question)*4 A circle has equation $( x - 5 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 20$.\\
(i) State the coordinates of the centre and the radius of this circle.\\
(ii) State, with a reason, whether or not this circle intersects the $y$-axis.\\
(iii) Find the equation of the line parallel to the line $y = 2 x$ that passes through the centre of the circle.\\
(iv) Show that the line $y = 2 x + 2$ is a tangent to the circle. State the coordinates of the point of contact.
\hfill \mbox{\textit{OCR MEI C1 Q4 [11]}}