AQA AS Paper 2 2019 June — Question 7 6 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeCircle from diameter endpoints
DifficultyModerate -0.3 This is a straightforward coordinate geometry question requiring students to use collinearity to find point A, then apply the midpoint formula for the diameter to find the centre, and finally calculate the radius. All steps are standard techniques with no novel problem-solving required, making it slightly easier than average.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^2

7 The points \(A ( a , 3 )\) and \(B ( 10,6 )\) lie on a circle. \(A B\) is a diameter of the circle and passes through the point ( 2,4 )
The circle has equation $$( x - c ) ^ { 2 } + ( y - d ) ^ { 2 } = e$$ where \(c , d\) and \(e\) are rational numbers. Find the values of \(a , c , d\) and \(e\).

Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Gradient \((2,4)\) to \(B = \frac{6-4}{10-2} = \frac{1}{4}\); uses gradient or equation of \(AB\) or vectors or proportionate division to find \(a\); \(\frac{6-3}{10-a} = \frac{1}{4}\)M1 (AO 3.1a) PI by obtaining \(a=-2\)
\(a = -2\)A1 (AO 1.1b)
Midpoint \(= \left(\frac{a+10}{2}, \frac{3+6}{2}\right) = (4, 4.5)\); \(c=4\), \(d=4.5\); finds midpoint of \(AB\)M1 (AO 1.1a) PI by either coordinate correct; NB knowledge of \(a\) not required
Deduces \(c=4\) and \(d=4.5\)A1 (AO 2.2a)
Radius\(^2 = 6^2 + 1.5^2 = 38.25\); \(e = 38.25\); uses appropriate distance formula to find length of radius or radius squaredM1 (AO 1.1a) NB must be fully numerical; PI by use of 'their' \((10-c)^2+(6-d)^2\) or 38.25 seen anywhere or \(\frac{1}{2}\sqrt{(10-a)^2+3^2}\) for 'their' \(a\)
\(e = 38.25\) or \(\frac{153}{4}\) or \(38\frac{1}{4}\) OE CAOA1 (AO 2.2a) Do not ISW if \(e\) is square rooted or squared
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient $(2,4)$ to $B = \frac{6-4}{10-2} = \frac{1}{4}$; uses gradient or equation of $AB$ or vectors or proportionate division to find $a$; $\frac{6-3}{10-a} = \frac{1}{4}$ | M1 (AO 3.1a) | PI by obtaining $a=-2$ |
| $a = -2$ | A1 (AO 1.1b) | |
| Midpoint $= \left(\frac{a+10}{2}, \frac{3+6}{2}\right) = (4, 4.5)$; $c=4$, $d=4.5$; finds midpoint of $AB$ | M1 (AO 1.1a) | PI by either coordinate correct; NB knowledge of $a$ not required |
| Deduces $c=4$ and $d=4.5$ | A1 (AO 2.2a) | |
| Radius$^2 = 6^2 + 1.5^2 = 38.25$; $e = 38.25$; uses appropriate distance formula to find length of radius or radius squared | M1 (AO 1.1a) | NB must be fully numerical; PI by use of 'their' $(10-c)^2+(6-d)^2$ or 38.25 seen anywhere or $\frac{1}{2}\sqrt{(10-a)^2+3^2}$ for 'their' $a$ |
| $e = 38.25$ or $\frac{153}{4}$ or $38\frac{1}{4}$ OE CAO | A1 (AO 2.2a) | Do not ISW if $e$ is square rooted or squared |

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7 The points $A ( a , 3 )$ and $B ( 10,6 )$ lie on a circle.\\
$A B$ is a diameter of the circle and passes through the point ( 2,4 )\\
The circle has equation

$$( x - c ) ^ { 2 } + ( y - d ) ^ { 2 } = e$$

where $c , d$ and $e$ are rational numbers.

Find the values of $a , c , d$ and $e$.\\

\hfill \mbox{\textit{AQA AS Paper 2 2019 Q7 [6]}}