| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using right angle in semicircle |
| Difficulty | Moderate -0.8 This is a structured multi-part question with clear signposting through each step. Parts (i)-(iv) involve routine C1 techniques: gradient calculation, parallel line equations, and distance formula. Part (v) requires recognizing that the hypotenuse of a right-angled triangle is the diameter of its circumcircle, then finding the circle equation—a standard result that's easier than deriving a circle equation from scratch. All steps are straightforward applications of basic coordinate geometry with no novel problem-solving required. |
| 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 |
|---|---|---|
| (i) | Marks | Guidance |
| Gradient \(DE = -\frac{1}{2}\) | B1 1 | \(-\frac{1}{2}\) (any working seen must be correct) |
| (ii) | Marks | Guidance |
| \(y - 3 = -\frac{1}{2}(x-2)\) | M1 | Correct equation for straight line, any gradient, passing through F |
| A1 | \(y - 3 = -\frac{1}{2}(x-2)\) aef | |
| \(x + 2y - 8 = 0\) | A1 3 | \(x + 2y - 8 = 0\) (this form but can have fractional coefficients e.g. \(\frac{1}{2}x + y - 4 = 0\)) |
| (iii) | Marks | Guidance |
| Gradient \(EF = \frac{4}{2} = 2\) | B1 | Correct supporting working must be seen |
| \(-\frac{1}{2} \times 2 = -1\) | B1 2 | Attempt to show that product of their gradients = -1 o.e. |
| (iv) | Marks | Guidance |
| \(DF = \sqrt{4^2 + 3^2} = 5\) | M1 | \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) used |
| A1 2 | \(5\) | |
| (v) | Marks | Guidance |
| \(DF\) is a diameter as angle \(DEF\) is a right angle. | B1 | Justification that \(DF\) is a diameter |
| Mid-point of \(DF\) or centre of circle is \((0, 1\frac{1}{2})\) | B1 | Mid-point of \(DF\) or centre of circle is \((0, 1\frac{1}{2})\) |
| Radius \(= 2.5\) | B1 | Radius \(= 2.5\) |
| \(x^2 + (y - \frac{3}{2})^2 = (\frac{5}{2})^2\) | B1 \(\sqrt{--}\) | \(x^2 + (y - \frac{3}{2})^2 = (\frac{5}{2})^2\) |
| \(x^2 + y^2 - 3y + \frac{9}{4} = \frac{25}{4}\) | ||
| \(x^2 + y^2 - 3y - 4 = 0\) | B1 5 | \(x^2 + y^2 - 3y - 4 = 0\) obtained correctly with at least one line of intermediate working. |
| SR For working that only shows \(x^2 + y^2 - 3y - 4 = 0\) is equation for a circle with centre \((0, 1\frac{1}{2})\) and radius \(2.5\): B1 | ||
| 13 |
**(i)** | **Marks** | **Guidance**
---|---|---
Gradient $DE = -\frac{1}{2}$ | B1 1 | $-\frac{1}{2}$ (any working seen must be correct)
**(ii)** | **Marks** | **Guidance**
---|---|---
$y - 3 = -\frac{1}{2}(x-2)$ | M1 | Correct equation for straight line, any gradient, passing through F
| A1 | $y - 3 = -\frac{1}{2}(x-2)$ aef
$x + 2y - 8 = 0$ | A1 3 | $x + 2y - 8 = 0$ (this form but can have fractional coefficients e.g. $\frac{1}{2}x + y - 4 = 0$)
**(iii)** | **Marks** | **Guidance**
---|---|---
Gradient $EF = \frac{4}{2} = 2$ | B1 | Correct supporting working must be seen
$-\frac{1}{2} \times 2 = -1$ | B1 2 | Attempt to show that product of their gradients = -1 o.e.
**(iv)** | **Marks** | **Guidance**
---|---|---
$DF = \sqrt{4^2 + 3^2} = 5$ | M1 | $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ used
| A1 2 | $5$
**(v)** | **Marks** | **Guidance**
---|---|---
$DF$ is a diameter as angle $DEF$ is a right angle. | B1 | Justification that $DF$ is a diameter
Mid-point of $DF$ or centre of circle is $(0, 1\frac{1}{2})$ | B1 | Mid-point of $DF$ or centre of circle is $(0, 1\frac{1}{2})$
Radius $= 2.5$ | B1 | Radius $= 2.5$
$x^2 + (y - \frac{3}{2})^2 = (\frac{5}{2})^2$ | B1 $\sqrt{--}$ | $x^2 + (y - \frac{3}{2})^2 = (\frac{5}{2})^2$
$x^2 + y^2 - 3y + \frac{9}{4} = \frac{25}{4}$ | |
$x^2 + y^2 - 3y - 4 = 0$ | B1 5 | $x^2 + y^2 - 3y - 4 = 0$ obtained correctly with at least one line of intermediate working.
| | **SR** For working that only shows $x^2 + y^2 - 3y - 4 = 0$ is equation for a circle with centre $(0, 1\frac{1}{2})$ and radius $2.5$: B1
| **13** |10 The points $D , E$ and $F$ have coordinates $( - 2,0 ) , ( 0 , - 1 )$ and $( 2,3 )$ respectively.\\
(i) Calculate the gradient of $D E$.\\
(ii) Find the equation of the line through $F$, parallel to $D E$, giving your answer in the form $a x + b y + c = 0$.\\
(iii) By calculating the gradient of $E F$, show that $D E F$ is a right-angled triangle.\\
(iv) Calculate the length of $D F$.\\
(v) Use the results of parts (iii) and (iv) to show that the circle which passes through $D , E$ and $F$ has equation $x ^ { 2 } + y ^ { 2 } - 3 y - 4 = 0$.
\hfill \mbox{\textit{OCR C1 2005 Q10 [13]}}