| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a standard multi-part circle question requiring routine techniques: finding radius using distance formula, using diameter properties, finding perpendicular gradient for tangent, and applying Pythagoras. All parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level. |
| 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, tangents1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((x-5)^2 + (y+3)^2 = \ldots\) | M1 | Or \((x-5)^2+(y--3)^2=\ldots\) |
| \(7^2+4^2\) or \(49+16\) or \(65\) | B1 | Or seen under square root |
| \((x-5)^2+(y+3)^2=65\) | A1 | Or \((x-5)^2+(y--3)^2=65\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x_B = 12\) | B1 | |
| \(y_B = -7\) | B1 | \(B(12,-7)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Grad \(AC = \dfrac{1--3}{-2-5}\) | M1 | Condone one sign error; FT their \(B\) if grad \(AB\) or grad \(BC\) is used |
| \(= -\dfrac{4}{7}\) | A1 | |
| Grad tgt \(= \dfrac{7}{4}\) | B1F | |
| Equation of tgt: \(y-1 = \text{"their"}\dfrac{7}{4}(x--2)\) | m1 | Or \(y=\text{"their"}\dfrac{7}{4}x+c\) and attempt to find \(c\) using \(x=-2\) and \(y=1\) |
| \(7x - 4y + 18 = 0\) | A1 | Any multiple; must have integer coefficients and all terms on one side |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(CT^2 = AT^2 + AC^2\); \((CT^2=)\ 4^2 + \text{"their"}\ 65\) | M1 | Pythagoras with hyp\(=CT\); \(AC^2=\text{"their"}\ k\) or correct |
| \((CT^2=)\ 81\) | A1 | Or \((CT=)\sqrt{81}\) |
| \((CT=)9\) | A1 | All notation correct; must simplify \(\sqrt{81}\) |
# Question 5:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $(x-5)^2 + (y+3)^2 = \ldots$ | M1 | Or $(x-5)^2+(y--3)^2=\ldots$ |
| $7^2+4^2$ or $49+16$ or $65$ | B1 | Or seen under square root |
| $(x-5)^2+(y+3)^2=65$ | A1 | Or $(x-5)^2+(y--3)^2=65$ |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $x_B = 12$ | B1 | |
| $y_B = -7$ | B1 | $B(12,-7)$ |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| Grad $AC = \dfrac{1--3}{-2-5}$ | M1 | Condone one sign error; FT their $B$ if grad $AB$ or grad $BC$ is used |
| $= -\dfrac{4}{7}$ | A1 | |
| Grad tgt $= \dfrac{7}{4}$ | B1F | |
| Equation of tgt: $y-1 = \text{"their"}\dfrac{7}{4}(x--2)$ | m1 | Or $y=\text{"their"}\dfrac{7}{4}x+c$ and attempt to find $c$ using $x=-2$ and $y=1$ |
| $7x - 4y + 18 = 0$ | A1 | Any multiple; must have integer coefficients and all terms on one side |
## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| $CT^2 = AT^2 + AC^2$; $(CT^2=)\ 4^2 + \text{"their"}\ 65$ | M1 | Pythagoras with hyp$=CT$; $AC^2=\text{"their"}\ k$ or correct |
| $(CT^2=)\ 81$ | A1 | Or $(CT=)\sqrt{81}$ |
| $(CT=)9$ | A1 | All notation correct; must simplify $\sqrt{81}$ |
---5 A circle with centre $C ( 5 , - 3 )$ passes through the point $A ( - 2,1 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the circle in the form
$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
\item Given that $A B$ is a diameter of the circle, find the coordinates of the point $B$.
\item Find an equation of the tangent to the circle at the point $A$, giving your answer in the form $p x + q y + r = 0$, where $p , q$ and $r$ are integers.
\item The point $T$ lies on the tangent to the circle at $A$ such that $A T = 4$. Find the length of $C T$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2016 Q5 [13]}}