| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Standard +0.8 This is a multi-part circle question requiring completion of the square, finding a specific point on the circle, determining the tangent equation using perpendicular gradients, and calculating a triangle area. While each individual step uses standard techniques, the question requires sustained problem-solving across multiple connected parts with some algebraic manipulation, placing it moderately above average difficulty. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0,-8)\) | B1 | \(x=0\) or \(y=-8\) (may be seen on sketch) |
| \((0,-8)\) | B1 | \(x=0\) and \(y=-8\) (may be seen on sketch) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses 64, 100 and \(k\) (not \(k^2\)) to obtain value for \(k\) | M1 | |
| \(k = -36\) | A1 | cao. \(k=-36\) scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=-16 \Rightarrow a=6\) | B1 | Correct \(x\)-coordinate. Allow \(x=6\) or just sight of 6. May be seen on sketch |
| \(m_N = \frac{-16+8}{6-0}\left(=-\frac{4}{3}\right)\) or \(m_N = \frac{-16+8}{a-0}\left(=-\frac{8}{a}\right)\) | M1 | Correct attempt at gradient using centre and their \(A\). Allow one sign slip. If they use \(O\) for centre, this is M0. Allow if in terms of \(a\) |
| \(m_T = -1 \div "-\frac{4}{3}" = \ldots\) or \(m_T = -1 \div "-\frac{8}{a}" = \ldots\) | M1 | Correct use of perpendicular gradient rule. Allow if in terms of \(a\) |
| \(y+16 = \frac{3}{4}(x-"6")\) or \(y+16 = \frac{a}{8}(x-"6")\) | M1 | Correct straight line method using gradient which is not the radius gradient and their \(A\) or \((a,-16)\). Allow gradient in terms of \(a\) |
| \(x=0 \Rightarrow y=-\frac{41}{2}\), \(y=0 \Rightarrow x=\frac{82}{3}\) | A1 | Correct values |
| Area \(= \frac{1}{2}\times\frac{41}{2}\times\frac{82}{3}\) | dM1 | Correct method for area using vertices of form \((0,0)\), \((X,0)\) and \((0,Y)\) where \(X\) and \(Y\) are numeric and have come from intersections of tangent with axes. Allow negative lengths. Dependent on previous M |
| \(= \frac{1681}{6}\) or \(280\frac{1}{6}\) or \(280.1\dot{6}\) | A1 | cao. Must be positive. May be recovered from sign errors on \(-\frac{41}{2}\) and/or \(\frac{82}{3}\) but must be from correct tangent equation |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0,-8)$ | B1 | $x=0$ **or** $y=-8$ (may be seen on sketch) |
| $(0,-8)$ | B1 | $x=0$ **and** $y=-8$ (may be seen on sketch) |
---
## Question 14(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses 64, 100 and $k$ (not $k^2$) to obtain value for $k$ | M1 | |
| $k = -36$ | A1 | cao. $k=-36$ scores both marks |
---
## Question 14(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=-16 \Rightarrow a=6$ | B1 | Correct $x$-coordinate. Allow $x=6$ or just sight of 6. May be seen on sketch |
| $m_N = \frac{-16+8}{6-0}\left(=-\frac{4}{3}\right)$ or $m_N = \frac{-16+8}{a-0}\left(=-\frac{8}{a}\right)$ | M1 | Correct attempt at gradient using centre and their $A$. Allow one sign slip. If they use $O$ for centre, this is M0. Allow if in terms of $a$ |
| $m_T = -1 \div "-\frac{4}{3}" = \ldots$ or $m_T = -1 \div "-\frac{8}{a}" = \ldots$ | M1 | Correct use of perpendicular gradient rule. Allow if in terms of $a$ |
| $y+16 = \frac{3}{4}(x-"6")$ or $y+16 = \frac{a}{8}(x-"6")$ | M1 | Correct straight line method using gradient which is **not** the radius gradient and their $A$ or $(a,-16)$. Allow gradient in terms of $a$ |
| $x=0 \Rightarrow y=-\frac{41}{2}$, $y=0 \Rightarrow x=\frac{82}{3}$ | A1 | Correct values |
| Area $= \frac{1}{2}\times\frac{41}{2}\times\frac{82}{3}$ | dM1 | Correct method for area using vertices of form $(0,0)$, $(X,0)$ and $(0,Y)$ where $X$ and $Y$ are numeric and have come from intersections of tangent with axes. Allow negative lengths. **Dependent on previous M** |
| $= \frac{1681}{6}$ or $280\frac{1}{6}$ or $280.1\dot{6}$ | A1 | cao. Must be **positive**. May be recovered from sign errors on $-\frac{41}{2}$ and/or $\frac{82}{3}$ **but must be from correct tangent equation** |
Alternative by implicit differentiation: $x^2+y^2+16y+k=0 \Rightarrow 2x+2y\frac{dy}{dx}+16\frac{dy}{dx}=0$, M1 for $\alpha x + \beta y\frac{dy}{dx}+c\frac{dy}{dx}=0$; $2(6)+2(-16)\frac{dy}{dx}+16\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{12}{16}$. Note no penalty for incorrect $k$.
---14. The circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } + 16 y + k = 0$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of $C$.
Given that the radius of $C$ is 10
\item find the value of $k$.
The point $A ( a , - 16 )$, where $a > 0$, lies on the circle $C$. The tangent to $C$ at the point $A$ crosses the $x$-axis at the point $D$ and crosses the $y$-axis at the point $E$.
\item Find the exact area of triangle $O D E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q14 [11]}}