| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Geometric properties with circles |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard coordinate geometry techniques: calculating gradients to verify perpendicularity, applying the angle-in-semicircle theorem, and comparing distances to a radius. All parts follow routine 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^21.03e Complete the square: find centre and radius of circle1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AB^2 = (8-15)^2 + (17-10)^2 = 98\) | M1 | Uses technique which could lead to showing two lines are perpendicular; obtains at least one correct distance (or distance²) or gradient |
| \(AC^2 = (8--2)^2 + (17--7)^2 = 676\) | A1 | Obtains three correct distances (or distance²) or two gradients; lengths \(7\sqrt{2}, 17\sqrt{2}, 26\); gradients \(AB = -\frac{7}{7}\), \(BC = \frac{17}{17}\) |
| \(CB^2 = (15--2)^2 + (10--7)^2 = 578\) | ||
| \(AB^2 + BC^2 = 98 + 578 = 676 = AC^2\) | R1 | Completes correct rigorous argument; uses Pythagoras OR multiplies gradients to show product is \(-1\); AND writes concluding statement |
| Angle \(ABC\) is a right angle. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The angle subtended by a diameter is \(90°\), \(\therefore AC\) must be a diameter of the circle. | E1 | Explains why \(AC\) is a diameter; must reference angle subtended by diameter (condone "angle in a semi-circle") or give full explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Radius \(\frac{\sqrt{676}}{2} = 13\) | B1 | Deduces correct radius (or radius²) |
| Centre \(\left(\frac{8-2}{2}, \frac{17-7}{2}\right) = (3, 5)\) | B1 | Obtains mid-point of diameter |
| Distance from centre to \(D\): \((3--8)^2 + (5--2)^2 = 11^2 + 7^2 = 170 > 169\) | M1, R1 | Uses \(D(-8,-2)\) to find distance (or distance²) from their centre \(OE\); completes rigorous argument by comparing \(\sqrt{170} > 13\) (or \(170 > 169\)) to show \(D\) lies outside the circle |
| So \(D\) lies outside the circle. |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB^2 = (8-15)^2 + (17-10)^2 = 98$ | M1 | Uses technique which could lead to showing two lines are perpendicular; obtains at least one correct distance (or distance²) or gradient |
| $AC^2 = (8--2)^2 + (17--7)^2 = 676$ | A1 | Obtains three correct distances (or distance²) or two gradients; lengths $7\sqrt{2}, 17\sqrt{2}, 26$; gradients $AB = -\frac{7}{7}$, $BC = \frac{17}{17}$ |
| $CB^2 = (15--2)^2 + (10--7)^2 = 578$ | | |
| $AB^2 + BC^2 = 98 + 578 = 676 = AC^2$ | R1 | Completes correct rigorous argument; uses Pythagoras OR multiplies gradients to show product is $-1$; AND writes concluding statement |
| Angle $ABC$ is a right angle. | | |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The angle subtended by a diameter is $90°$, $\therefore AC$ must be a diameter of the circle. | E1 | Explains why $AC$ is a diameter; must reference angle subtended by diameter (condone "angle in a semi-circle") or give full explanation |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Radius $\frac{\sqrt{676}}{2} = 13$ | B1 | Deduces correct radius (or radius²) |
| Centre $\left(\frac{8-2}{2}, \frac{17-7}{2}\right) = (3, 5)$ | B1 | Obtains mid-point of diameter |
| Distance from centre to $D$: $(3--8)^2 + (5--2)^2 = 11^2 + 7^2 = 170 > 169$ | M1, R1 | Uses $D(-8,-2)$ to find distance (or distance²) from their centre $OE$; completes rigorous argument by comparing $\sqrt{170} > 13$ (or $170 > 169$) to show $D$ lies outside the circle |
| So $D$ lies outside the circle. | | |
---7 Three points $A , B$ and $C$ have coordinates $A ( 8,17 ) , B ( 15,10 )$ and $C ( - 2 , - 7 )$
7
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A B C$ is a right angle.\\
7
\item $\quad A , B$ and $C$ lie on a circle.\\
7 (b) (i) Explain why $A C$ is a diameter of the circle.\\
7 (b) (ii) Determine whether the point $D ( - 8 , - 2 )$ lies inside the circle, on the circle or outside the circle.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q7 [8]}}