| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question on completing the square to find circle centre/radius, then using basic coordinate geometry (midpoint, gradient, perpendicularity). Part (i) is routine manipulation, part (ii) requires finding a line through two points, and part (iii) is a simple verification using perpendicular gradient condition. All techniques are standard with no problem-solving insight required, making it easier than average. |
| 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 |
**(i)**
Obtain fully correct $(x-3)^2 - 9 + (y-2)^2 - 4 = 12$ [M1]
Obtain $(x-3)^2 + (y-2)^2 = 25$ [A1]
Obtain $r = 5$ [B1]
**Subtotal: [3]**
**(ii)**
State gradient $= \dfrac{3}{4}$ $\left(= \dfrac{2-(-1)}{3-(-1)} = \dfrac{5-(-1)}{7-(-1)}\right)$ [B1]
Obtain equation of straight line $(y - q) = \text{their } m(x-p)$ where $(p, q) = (-1,-1)$, $(3, 2)$ or $(7, 5)$ only [M1]
Obtain $1 = 3x - 4y$ [A1]
**Subtotal: [3]**
**(iii)**
Calculate $\dfrac{2-6}{3-0}$ [B1]
**SR** A diagram used to justify $-\dfrac{4}{3}$ B0B1B1.
Obtain gradient $= -\dfrac{4}{3}$ [depB1]
Clearly state $-\dfrac{4}{3} \times \dfrac{3}{4} = -1$ or "negative reciprocal" [depB1]
**Subtotal: [3]**
**Total: [9]**5 A circle has equation $x ^ { 2 } - 6 x + y ^ { 2 } - 4 y = 12$.\\
(i) Show that the centre of the circle is at the point $( 3,2 )$ and find the radius.\\
(ii) $P Q$ is a diameter of the circle where $P$ has coordinates $( - 1 , - 1 )$. Find the equation of $P Q$, giving your answer in the form $a x + b y = c$ where $a , b$ and $c$ are integers.\\
(iii) Another diameter of the circle passes through the point $( 0,6 )$. Show that this diameter is perpendicular to $P Q$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2015 Q5 [9]}}