| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question requiring standard techniques: finding circle equation from diameter endpoints (formula recall), finding intersections by substitution (x=0), and finding tangent using perpendicular gradient. All methods are routine C1 procedures with no novel problem-solving 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/Working | Mark | Guidance |
| C, mid pt of \(AB = \left(\frac{11+(-1)}{2}, \frac{4}{2}\right) = (5, 2)\) | B1 | evidence of method required; may show equal steps, or start at A or B and go half the difference |
| \([AB^2 =] 12^2 + 4^2 [= 160]\) oe or \([CB^2 =] 6^2 + 2^2 [=40]\) oe with AC | B1 | or square root of these; accept unsimplified |
| Quote of \((x-a)^2 + (y-b)^2 = r^2\) o.e. with different letters | B1 | or \((5,2)\) clearly identified as centre and \(\sqrt{40}\) as \(r\) (or 40 as \(r^2\)) www or quote of \(gfc\) formula and finding \(c = -11\) |
| Completion (ans given) | B1 | dependent on centre (or midpt) and radius (or radius²) found independently and correctly |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct substitution of \(x = 0\) in circle eqn soi | M1 | |
| \((y-2)^2 = 15\) or \(y^2 - 4y - 11 [= 0]\) | M1 | condone one error |
| \(y - 2 = \pm\sqrt{15}\) or ft | M1 | or use of quad formula (condone one error in formula); ft only for 3 term quadratic in \(y\) |
| \([y =] 2 \pm \sqrt{15}\) cao | A1 | if \(y = 0\) subst, allow SC1 for \((11, 0)\) found; alt method: M1 for \(y\) values are \(2 \pm a\); M1 for \(a^2 + 5^2 = 40\) soi; M1 for \(a^2 = 40 - 5^2\) soi; A1 for \([y =] 2 \pm \sqrt{15}\) cao |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| grad \(AB = \frac{4}{11-(-1)}\) or \(1/3\) o.e. | M1 | or grad AC (or BC) |
| so grad tgt \(= -3\) | M1 | or ft \(-1\)/their gradient of AB |
| eqn of tgt is \(y - 4 = -3(x - 11)\) | M1 | or subst \((11, 4)\) in \(y = -3x + c\) or ft (no ft for their grad AB used) |
| \(y = -3x + 37\) or \(3x + y = 37\) | A1 | accept other simplified versions |
| \((0, 37)\) and \((37/3, 0)\) o.e. ft isw | B2 | B1 each, ft their tgt for grad \(\neq 1\) or \(1/3\); accept \(x = 0\), \(y = 37\) etc; NB alt method: intercepts may be found first by proportion then used to find eqn |
| Total: 6 marks |
# Question 11:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| C, mid pt of $AB = \left(\frac{11+(-1)}{2}, \frac{4}{2}\right) = (5, 2)$ | B1 | evidence of method required; may show equal steps, or start at A or B and go half the difference |
| $[AB^2 =] 12^2 + 4^2 [= 160]$ oe or $[CB^2 =] 6^2 + 2^2 [=40]$ oe with AC | B1 | or square root of these; accept unsimplified |
| Quote of $(x-a)^2 + (y-b)^2 = r^2$ o.e. with different letters | B1 | or $(5,2)$ clearly identified as centre and $\sqrt{40}$ as $r$ (or 40 as $r^2$) www or quote of $gfc$ formula and finding $c = -11$ |
| Completion (ans given) | B1 | dependent on centre (or midpt) and radius (or radius²) found independently and correctly |
| **Total: 4 marks** | | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct substitution of $x = 0$ in circle eqn soi | M1 | |
| $(y-2)^2 = 15$ or $y^2 - 4y - 11 [= 0]$ | M1 | condone one error |
| $y - 2 = \pm\sqrt{15}$ or ft | M1 | or use of quad formula (condone one error in formula); ft only for 3 term quadratic in $y$ |
| $[y =] 2 \pm \sqrt{15}$ cao | A1 | if $y = 0$ subst, allow SC1 for $(11, 0)$ found; alt method: M1 for $y$ values are $2 \pm a$; M1 for $a^2 + 5^2 = 40$ soi; M1 for $a^2 = 40 - 5^2$ soi; A1 for $[y =] 2 \pm \sqrt{15}$ cao |
| **Total: 4 marks** | | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| grad $AB = \frac{4}{11-(-1)}$ or $1/3$ o.e. | M1 | or grad AC (or BC) |
| so grad tgt $= -3$ | M1 | or ft $-1$/their gradient of AB |
| eqn of tgt is $y - 4 = -3(x - 11)$ | M1 | or subst $(11, 4)$ in $y = -3x + c$ or ft (no ft for their grad AB used) |
| $y = -3x + 37$ or $3x + y = 37$ | A1 | accept other simplified versions |
| $(0, 37)$ and $(37/3, 0)$ o.e. ft isw | B2 | B1 each, ft their tgt for grad $\neq 1$ or $1/3$; accept $x = 0$, $y = 37$ etc; NB alt method: intercepts may be found first by proportion then used to find eqn |
| **Total: 6 marks** | | |
---(i) Show that the equation of the circle with AB as diameter may be written as
$$( x - 5 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 40$$
(ii) Find the coordinates of the points of intersection of this circle with the $y$-axis. Give your answer in the form $a \pm \sqrt { b }$.\\
(iii) Find the equation of the tangent to the circle at B . Hence find the coordinates of the points of intersection of this tangent with the axes.
\hfill \mbox{\textit{OCR MEI C1 2009 Q11 [14]}}