| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Easy -1.2 This is a straightforward multi-part question testing basic coordinate geometry: midpoint formula, distance formula, circle equation, and perpendicular tangent. All parts use standard algorithms with no problem-solving insight required. The 12 marks reflect routine application of multiple techniques rather than difficulty. |
| 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 |
|---|---|---|
| \(\left(\frac{4+10}{2}, \frac{-2+6}{2}\right)\) | M1 | Uses \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) |
| \((7, 2)\) | A1 | 2 marks; (7, 2) (integers required) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{(7-4)^2 + (2-(-2))^2}\) | M1 | Uses \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) |
| \(= \sqrt{3^2 + 4^2}\) | A1 | |
| \(= 5\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-7)^2 + (y-2)^2 = 25\) | B1√ | \((x-7)^2\) and \((y-2)^2\) used (their centre) |
| B1√ | \(r^2 = 25\) used (their \(r^2\)) | |
| B1 | 3 marks; \((x-7)^2 + (y-2)^2 = 25\) cao | |
| Expanded form: -14x and -4y used | B1√ | |
| \(r = \sqrt{g^2 + f^2 - c}\) used | B1√ | |
| \(x^2 + y^2 - 14x - 4y + 28 = 0\) | B1 cao | |
| By using ends of diameter: | ||
| \((x-4)(x-10) + (y+2)(y-6) = 0\) | ||
| Both x brackets correct | B1 | |
| Both y brackets correct | B1 | |
| Final equation fully correct | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AB = \frac{6-(-2)}{10-4} = \frac{4}{3}\) | B1 | oe |
| Gradient of tangent = \(-\frac{3}{4}\) | B1√ | |
| Correct equation of straight line through A, any non-zero gradient | M1 | |
| \(y - (-2) = -\frac{3}{4}(x-4)\) | A1 | |
| \(3x + 4y = 4\) | A1 | 5 marks; \(a, b, c\) need not be integers |
## (i)
$\left(\frac{4+10}{2}, \frac{-2+6}{2}\right)$ | M1 | Uses $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
$(7, 2)$ | A1 | 2 marks; (7, 2) (integers required)
## (ii)
$\sqrt{(7-4)^2 + (2-(-2))^2}$ | M1 | Uses $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
$= \sqrt{3^2 + 4^2}$ | A1 |
$= 5$ | A1 | 2 marks
## (iii)
$(x-7)^2 + (y-2)^2 = 25$ | B1√ | $(x-7)^2$ and $(y-2)^2$ used (their centre)
| B1√ | $r^2 = 25$ used (their $r^2$)
| B1 | 3 marks; $(x-7)^2 + (y-2)^2 = 25$ cao
Expanded form: -14x and -4y used | B1√ |
$r = \sqrt{g^2 + f^2 - c}$ used | B1√ |
$x^2 + y^2 - 14x - 4y + 28 = 0$ | B1 cao |
By using ends of diameter: | |
$(x-4)(x-10) + (y+2)(y-6) = 0$ | |
Both x brackets correct | B1 |
Both y brackets correct | B1 |
Final equation fully correct | B1 |
## (iv)
Gradient of $AB = \frac{6-(-2)}{10-4} = \frac{4}{3}$ | B1 | oe
Gradient of tangent = $-\frac{3}{4}$ | B1√ |
Correct equation of straight line through A, any non-zero gradient | M1 |
$y - (-2) = -\frac{3}{4}(x-4)$ | A1 |
$3x + 4y = 4$ | A1 | 5 marks; $a, b, c$ need not be integers
---The points $A$ and $B$ have coordinates $(4, -2)$ and $(10, 6)$ respectively. $C$ is the mid-point of $AB$. Find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$, [2]
\item the length of $AC$, [2]
\item the equation of the circle that has $AB$ as a diameter, [3]
\item the equation of the tangent to the circle in part (iii) at the point $A$, giving your answer in the form $ax + by = c$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2006 Q9 [12]}}