Moderate -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.
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 \(e\).
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 \(a\); \(\frac{6-3}{10-a} = \frac{1}{4}\)
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
## 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]}}