| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find parameter values for tangency using discriminant |
| Difficulty | Moderate -0.3 This is a comprehensive but routine circle question covering completing the square, verifying points lie on circles, finding normals, and using the discriminant for tangency. All techniques are standard C1 procedures with clear scaffolding through multiple parts. While lengthy (7 parts), each step follows directly from the previous one with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a)(i) | \((x-3)^2 + (y+5)^2\) | B1 |
| \(= 25 - 9 + 9 = 25\) \((= 5^2)\) | B1 B1 | LHS correct with + and squares. Condone RHS = 25 |
| 7(b)(i) | \(C(3, -5)\) | B1 |
| 7(b)(ii) | Radius \(= 5\) | B1 |
| 7(c)(i) | \((7-3)^2 + (-2+5)^2 = 16 + 9 = 25 \Rightarrow D\) lies on circle. Must see statement | B1 |
| 7(c)(ii) | Attempt at gradient of \(CD\) as normal | M1 |
| grad \(CD = \frac{-2-(-5)}{7-3} = \frac{3}{4}\) | ||
| \(y + 2 = \frac{3}{4}(x-7)\) or \(y + 5 = \frac{3}{4}(x-3)\) | A1 | Correct equation in any form \(y = \frac{3}{4}x - \frac{29}{4}\) |
| \(\Rightarrow 3x - 4y = 29\) | A1 | CSO Integer coefficients. Condone \(4y - 3x + 29 = 0\) etc |
| 7(d)(i) | \(y = kx\) sub'd into original circle equation | M1 |
| \(x^2 + (kx)^2 - 6x + 10kx + 9 = 0\) | ||
| \(\Rightarrow (k^2 + 1)x^2 + 2(5k-3)x + 9 = 0\) AG | A1 | CSO. must see at least previous line for A1. any error such as \(kx^2 = ... = k^2x^2\) gets A0 |
| 7(d)(ii) | \(4(5k-3)^2 - 36(k^2+1)\) | M1 |
| \(= 64k^2 - 120k\) | A1 | or \(8k^2 - 15k = 0\) OE |
| Equal roots: \(4(5k-3)^2 - 36(k^2+1) = 0\) | \(b^2 - 4ac = 0\) clearly stated or evident by an equation in \(k\) with at most 2 slips. | |
| \(8k^2 - 15k = 0\) | m1 | Attempt to solve their quadratic or linear equation if \(k\) has been cancelled |
| \(\Rightarrow k = 0, k = \frac{15}{8}\) | A1 | OE but must have \(k=0\). If "=" not seen but correct values of \(k\) are found, candidate will lose B1 mark but may earn all other marks |
| 7(d)(iii) | (Line is a) tangent (to the circle) | E1 |
**7(a)(i)** | $(x-3)^2 + (y+5)^2$ | B1 | One term correct
| $= 25 - 9 + 9 = 25$ $(= 5^2)$ | B1 B1 | LHS correct with + and squares. Condone RHS = 25
**7(b)(i)** | $C(3, -5)$ | B1 | |
**7(b)(ii)** | Radius $= 5$ | B1 | Correct or ft their RHS provided $> 0$
**7(c)(i)** | $(7-3)^2 + (-2+5)^2 = 16 + 9 = 25 \Rightarrow D$ lies on circle. Must see statement | B1 | Or sub'n of $(7, -2)$ in original equation: $7^2 + (-2)^2 - 42 - 20 + 9 = 0$ or sub $x=7$ into eqn & showing $y = -2$ etc
**7(c)(ii)** | Attempt at gradient of $CD$ as normal | M1 | withhold if subsequently uses $m_1m_2 = -1$. $\Delta y/\Delta x$ (condone one slip) FT their centre $C$
| grad $CD = \frac{-2-(-5)}{7-3} = \frac{3}{4}$ | | |
| $y + 2 = \frac{3}{4}(x-7)$ or $y + 5 = \frac{3}{4}(x-3)$ | A1 | Correct equation in any form $y = \frac{3}{4}x - \frac{29}{4}$
| $\Rightarrow 3x - 4y = 29$ | A1 | CSO Integer coefficients. Condone $4y - 3x + 29 = 0$ etc
**7(d)(i)** | $y = kx$ sub'd into original circle equation | M1 | or using their completed square form and multiplying out
| $x^2 + (kx)^2 - 6x + 10kx + 9 = 0$ | | |
| $\Rightarrow (k^2 + 1)x^2 + 2(5k-3)x + 9 = 0$ AG | A1 | CSO. must see at least previous line for A1. any error such as $kx^2 = ... = k^2x^2$ gets A0
**7(d)(ii)** | $4(5k-3)^2 - 36(k^2+1)$ | M1 | Discriminant in $k$ (can be seen in quad formula). Condone one slip
| $= 64k^2 - 120k$ | A1 | or $8k^2 - 15k = 0$ OE
| Equal roots: $4(5k-3)^2 - 36(k^2+1) = 0$ | | $b^2 - 4ac = 0$ clearly stated or evident by an equation in $k$ with at most 2 slips.
| $8k^2 - 15k = 0$ | m1 | Attempt to solve their quadratic or linear equation if $k$ has been cancelled
| $\Rightarrow k = 0, k = \frac{15}{8}$ | A1 | OE but must have $k=0$. If "=" not seen but correct values of $k$ are found, candidate will lose B1 mark but may earn all other marks
**7(d)(iii)** | (Line is a) tangent (to the circle) | E1 | Line touches circle at one point
**Total: 17 marks**
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## TOTAL: 75 marks7 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } - 6 x + 10 y + 9 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Express this equation in the form
$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }$$
\item Write down:
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$;
\item the radius of the circle.
\end{enumerate}\item The point $D$ has coordinates (7, -2).
\begin{enumerate}[label=(\roman*)]
\item Verify that the point $D$ lies on the circle.
\item Find an equation of the normal to the circle at the point $D$, giving your answer in the form $m x + n y = p$, where $m , n$ and $p$ are integers.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item A line has equation $y = k x$. Show that the $x$-coordinates of any points of intersection of the line and the circle satisfy the equation
$$\left( k ^ { 2 } + 1 \right) x ^ { 2 } + 2 ( 5 k - 3 ) x + 9 = 0$$
\item Find the values of $k$ for which the equation
$$\left( k ^ { 2 } + 1 \right) x ^ { 2 } + 2 ( 5 k - 3 ) x + 9 = 0$$
has equal roots.
\item Describe the geometrical relationship between the line and the circle when $k$ takes either of the values found in part (d)(ii).
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q7 [17]}}