| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Standard +0.3 This is a standard C1 circles question requiring completing the square (routine), applying distance formula for tangent conditions, using perpendicular bisector properties for chords, and solving simultaneous equations. All techniques are straightforward applications of core methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.03d 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 | Mark | Guidance |
| Centre \((4, 1)\) | B1 | Correct centre |
| \((x-4)^2 + (y-1)^2 - 16 - 1 - 3 = 0\) | M1 | Correct method to find \(r^2\); \(r^2 = (\pm\text{their }4)^2 + (\pm\text{their }1)^2 + 3\) |
| \((x-4)^2 + (y-1)^2 = 20\) | ||
| Radius \(= \sqrt{20}\) | A1 | Correct radius; \(\pm\sqrt{20}\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = 1 \pm \sqrt{20}\) | M1, A1ft | \(y\) ordinate of centre \(\pm\) radius, both correct unsimplified values |
| \(k = 1 \pm 2\sqrt{5}\) | A1 | cao; alternatives: a) substitutes \(k\) for \(y\), uses \(b^2-4ac=0\); b) recognises \(x=4\) is equation of normal, substitutes into circle equation; SR \(k=1+\sqrt{20}\) or \(k=1-\sqrt{20}\) www B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(MT^2 = r^2 - 2^2\) | M1 | Correct use of Pythagoras involving \(MT\); SR: ST=8 from particular S and T coordinates |
| \(MT = 4\) | A1ft | Correct value of \(MT\) for their \(r\) |
| \(ST = 8\) | A1 | cao; justifies solution the same for all possible chords B2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2y + 12\) | M1* | Attempt to solve equations simultaneously; must reduce to one variable using line equation and circle equation |
| \((2y+8)^2 + (y-1)^2 = 20\) | A1 | Correct unsimplified expression; if \(y\) eliminated: \((x-4)^2 + \left(\frac{1}{2}x - 7\right)^2 = 20\) |
| \(4y^2 + 32y + 64 + y^2 - 2y + 1 = 20\) | ||
| \(5y^2 + 30y + 45 = 0\) | A1 | Obtain correct 3-term quadratic |
| \(y^2 + 6y + 9 = 0\) | ||
| \((y+3)^2 = 0\) | DM1 | Correct method to solve quadratic \(ax^2+bx+c=0\) \((b \neq 0)\) |
| \(y = -3\) | A1 | \(y\) value correct, no extra solutions |
| \(x = 6\) | A1 | \(x\) value correct ISW |
| OR | ||
| \(y - 1 = -2(x-4)\) | M1 | Attempt to find equation of radius/normal |
| A1 | Correct equation | |
| Solve simultaneously with \(y = \frac{1}{2}x - 6\) | M1 | |
| \(x = 6\), \(y = -3\) | A1, A1 | |
| States line is tangent as meets at one point or verifies \((6,-3)\) lies on circle | B1 | Allow showing distance between \((6,-3)\) and \((4,1) = \sqrt{20}\); SR correct coordinates spotted www B2 |
## Question 9(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre $(4, 1)$ | B1 | Correct centre |
| $(x-4)^2 + (y-1)^2 - 16 - 1 - 3 = 0$ | M1 | Correct method to find $r^2$; $r^2 = (\pm\text{their }4)^2 + (\pm\text{their }1)^2 + 3$ |
| $(x-4)^2 + (y-1)^2 = 20$ | | |
| Radius $= \sqrt{20}$ | A1 | Correct radius; $\pm\sqrt{20}$ is A0 |
---
## Question 9(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = 1 \pm \sqrt{20}$ | M1, A1ft | $y$ ordinate of centre $\pm$ radius, both correct unsimplified values |
| $k = 1 \pm 2\sqrt{5}$ | A1 | cao; alternatives: a) substitutes $k$ for $y$, uses $b^2-4ac=0$; b) recognises $x=4$ is equation of normal, substitutes into circle equation; SR $k=1+\sqrt{20}$ or $k=1-\sqrt{20}$ www B1 |
---
## Question 9(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $MT^2 = r^2 - 2^2$ | M1 | Correct use of Pythagoras involving $MT$; SR: ST=8 from particular S and T coordinates |
| $MT = 4$ | A1ft | Correct value of $MT$ for their $r$ |
| $ST = 8$ | A1 | cao; justifies solution the same for all possible chords B2 |
---
## Question 9(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2y + 12$ | M1* | Attempt to solve equations simultaneously; must reduce to one variable using line equation and circle equation |
| $(2y+8)^2 + (y-1)^2 = 20$ | A1 | Correct unsimplified expression; if $y$ eliminated: $(x-4)^2 + \left(\frac{1}{2}x - 7\right)^2 = 20$ |
| $4y^2 + 32y + 64 + y^2 - 2y + 1 = 20$ | | |
| $5y^2 + 30y + 45 = 0$ | A1 | Obtain correct 3-term quadratic |
| $y^2 + 6y + 9 = 0$ | | |
| $(y+3)^2 = 0$ | DM1 | Correct method to solve quadratic $ax^2+bx+c=0$ $(b \neq 0)$ |
| $y = -3$ | A1 | $y$ value correct, no extra solutions |
| $x = 6$ | A1 | $x$ value correct ISW |
| **OR** | | |
| $y - 1 = -2(x-4)$ | M1 | Attempt to find equation of radius/normal |
| | A1 | Correct equation |
| Solve simultaneously with $y = \frac{1}{2}x - 6$ | M1 | |
| $x = 6$, $y = -3$ | A1, A1 | |
| States line is tangent as meets at one point or verifies $(6,-3)$ lies on circle | B1 | Allow showing distance between $(6,-3)$ and $(4,1) = \sqrt{20}$; SR correct coordinates spotted www B2 |9 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } - 8 x - 2 y - 3 = 0$.\\
(i) Find the coordinates of $C$ and the radius of the circle.\\
(ii) Find the values of $k$ for which the line $y = k$ is a tangent to the circle, giving your answers in simplified surd form.\\
(iii) The points $S$ and $T$ lie on the circumference of the circle. $M$ is the mid-point of the chord $S T$. Given that the length of $C M$ is 2 , calculate the length of the chord $S T$.\\
(iv) Find the coordinates of the point where the circle meets the line $x - 2 y - 12 = 0$.
\hfill \mbox{\textit{OCR C1 2011 Q9 [15]}}