| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic coordinate geometry: midpoint formula (routine algebra), distance formula for radius, standard circle equation manipulation, and tangent perpendicularity. All parts are standard textbook exercises requiring only direct application of formulas with no problem-solving insight needed. Easier than average A-level content. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4+p}{2} = -1,\quad \frac{5+q}{2} = 3\) | M1 | Correct method (may be implied by one correct coordinate) |
| \(p = -6\) | A1 | |
| \(q = 1\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r^2 = (4-{-1})^2 + (5-3)^2\) | M1 | Use of \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) for either radius or diameter |
| \(r = \sqrt{29}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x+1)^2 + (y-3)^2 = 29\) | M1 | \((x+1)^2\) and \((y-3)^2\) seen |
| M1 | \((x \pm 1)^2 + (y \pm 3)^2 = \text{their } r^2\) | |
| \(x^2 + y^2 + 2x - 6y - 19 = 0\) | A1 | 3 Correct equation in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| gradient of radius \(= \frac{3-5}{-1-4} = \frac{2}{5}\) | M1 | Uses \(\frac{y_2 - y_1}{x_2 - x_1}\) |
| A1 | oe | |
| gradient of tangent \(= -\frac{5}{2}\) | B1\(\checkmark\) | oe |
| \(y - 5 = -\frac{5}{2}(x-4)\) | M1 | Correct equation of straight line through \((4,5)\), any non-zero gradient |
| \(y = -\frac{5}{2}x + 15\) | A1 | 5 oe 3 term equation e.g. \(5x + 2y = 30\) |
# Question 9:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4+p}{2} = -1,\quad \frac{5+q}{2} = 3$ | M1 | Correct method (may be implied by one correct coordinate) |
| $p = -6$ | A1 | |
| $q = 1$ | A1 | **3** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r^2 = (4-{-1})^2 + (5-3)^2$ | M1 | Use of $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ for either radius or diameter |
| $r = \sqrt{29}$ | A1 | **2** |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x+1)^2 + (y-3)^2 = 29$ | M1 | $(x+1)^2$ and $(y-3)^2$ seen |
| | M1 | $(x \pm 1)^2 + (y \pm 3)^2 = \text{their } r^2$ |
| $x^2 + y^2 + 2x - 6y - 19 = 0$ | A1 | **3** Correct equation in correct form |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| gradient of radius $= \frac{3-5}{-1-4} = \frac{2}{5}$ | M1 | Uses $\frac{y_2 - y_1}{x_2 - x_1}$ |
| | A1 | oe |
| gradient of tangent $= -\frac{5}{2}$ | B1$\checkmark$ | oe |
| $y - 5 = -\frac{5}{2}(x-4)$ | M1 | Correct equation of straight line through $(4,5)$, any non-zero gradient |
| $y = -\frac{5}{2}x + 15$ | A1 | **5** oe 3 term equation e.g. $5x + 2y = 30$ |
---9 (i) The line joining the points $A ( 4,5 )$ and $B ( p , q )$ has mid-point $M ( - 1,3 )$. Find $p$ and $q$.\\
$A B$ is the diameter of a circle.\\
(ii) Find the radius of the circle.\\
(iii) Find the equation of the circle, giving your answer in the form $x ^ { 2 } + y ^ { 2 } + a x + b y + c = 0$.\\
(iv) Find an equation of the tangent to the circle at the point $( 4,5 )$.
\hfill \mbox{\textit{OCR C1 2010 Q9 [13]}}