| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard circle properties (center, radius, tangent, diameter) with straightforward calculations. Part (i) is direct reading from the equation, parts (ii)-(iii) use routine techniques (substitution, perpendicular gradient), and part (iv) requires finding a point on a line at a given distance—all standard C1 circle work with no novel insight required, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Centre \(C' = (3, -2)\) | 1 | |
| Radius \(5\) | 1 | 0 for \(\pm 5\) or \(-5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Showing \((6-3)^2 + (-6+2)^2 = 25\) | B1 | Interim step needed |
| Showing that \(\overrightarrow{AC'} = \overrightarrow{C'B} = \begin{pmatrix}-3\\4\end{pmatrix}\) o.e. | B2 | Or B1 each for two of: showing midpoint of \(AB = (3, -2)\); showing \(B(0, 2)\) is on circle; showing \(AB = 10\); or B2 for showing midpoint of \(AB = (3, -2)\) and saying this is centre of circle; or B1 for finding equation of AB as \(y = -\frac{4}{3}x + 2\) o.e. and B1 for finding one of its intersections with the circle is \((0, 2)\); or B1 for showing \(C'B = 5\) and B1 for showing \(AB = 10\) or that \(AC'\) and \(BC'\) have the same gradient; or B1 for showing that \(AC'\) and \(BC'\) have the same gradient and B1 for showing that \(B(0, 2)\) is on the circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Grad \(AC'\) or \(AB = -\frac{4}{3}\) o.e. | M1 | Or ft from their \(C'\), must be evaluated |
| Grad tgt \(= -1/\text{their } AC'\) grad | M1 | May be seen in equation for tgt; allow M2 for grad tgt \(= \frac{3}{4}\) o.e. soi as first step |
| \(y -(-6) = \text{their } m(x - 6)\) o.e. | M1 | Or M1 for \(y = \text{their } m \times x + c\) then substitute \((6, -6)\) |
| \(y = 0.75x - 10.5\) o.e. isw | A1 | e.g. A1 for \(4y = 3x - 42\); allow B4 for correct equation www isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Centre C is at \((12, -14)\) cao | B2 | B1 for each coordinate |
| Circle is \((x - 12)^2 + (y + 14)^2 = 100\) | B1 | ft their C if at least one coordinate correct |
## Question 11(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre $C' = (3, -2)$ | 1 | |
| Radius $5$ | 1 | 0 for $\pm 5$ or $-5$ |
## Question 11(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Showing $(6-3)^2 + (-6+2)^2 = 25$ | B1 | Interim step needed |
| Showing that $\overrightarrow{AC'} = \overrightarrow{C'B} = \begin{pmatrix}-3\\4\end{pmatrix}$ o.e. | B2 | Or B1 each for two of: showing midpoint of $AB = (3, -2)$; showing $B(0, 2)$ is on circle; showing $AB = 10$; **or** B2 for showing midpoint of $AB = (3, -2)$ and saying this is centre of circle; **or** B1 for finding equation of AB as $y = -\frac{4}{3}x + 2$ o.e. and B1 for finding one of its intersections with the circle is $(0, 2)$; **or** B1 for showing $C'B = 5$ and B1 for showing $AB = 10$ or that $AC'$ and $BC'$ have the same gradient; **or** B1 for showing that $AC'$ and $BC'$ have the same gradient and B1 for showing that $B(0, 2)$ is on the circle |
## Question 11(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Grad $AC'$ or $AB = -\frac{4}{3}$ o.e. | M1 | Or ft from their $C'$, must be evaluated |
| Grad tgt $= -1/\text{their } AC'$ grad | M1 | May be seen in equation for tgt; allow M2 for grad tgt $= \frac{3}{4}$ o.e. soi as first step |
| $y -(-6) = \text{their } m(x - 6)$ o.e. | M1 | Or M1 for $y = \text{their } m \times x + c$ then substitute $(6, -6)$ |
| $y = 0.75x - 10.5$ o.e. isw | A1 | e.g. A1 for $4y = 3x - 42$; allow B4 for correct equation www isw |
## Question 11(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre C is at $(12, -14)$ cao | B2 | B1 for each coordinate |
| Circle is $(x - 12)^2 + (y + 14)^2 = 100$ | B1 | ft their C if at least one coordinate correct |11 A circle has equation $( x - 3 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 25$.\\
(i) State the coordinates of the centre of this circle and its radius.\\
(ii) Verify that the point A with coordinates $( 6 , - 6 )$ lies on this circle. Show also that the point B on the circle for which AB is a diameter has coordinates $( 0,2 )$.\\
(iii) Find the equation of the tangent to the circle at A .\\
(iv) A second circle touches the original circle at A . Its radius is 10 and its centre is at C , where BAC is a straight line. Find the coordinates of C and hence write down the equation of this second circle.
\hfill \mbox{\textit{OCR MEI C1 2010 Q11 [12]}}