| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find parameter values for tangency using discriminant |
| Difficulty | Standard +0.3 This is a standard two-part circle question: (a) completing the square to find centre and radius is routine bookwork, while (b) using the discriminant condition for tangency (substituting the line into the circle equation and setting Δ=0) is a well-practiced technique. The algebra is straightforward with no conceptual surprises, making this slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-5)^2+(y+2)^2 = \ldots\) | M1 | Attempts to complete the square, halving both \(x\) and \(y\) terms. Award for sight of \((x\pm5)^2\), \((y\pm2)^2\) |
| Centre \((5,-2)\) | A1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = \sqrt{``5"^2 + ``{-2}"{^2}-11}\) | M1 | Correct strategy for radius or radius² |
| \(r = 3\sqrt{2}\) | A1 | Do not accept \(r = \pm3\sqrt{2}\) or \(\sqrt{18}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=3x+k \Rightarrow x^2+(3x+k)^2-10x+4(3x+k)+11=0 \Rightarrow x^2+9x^2+6kx+k^2-10x+12x+4k+11=0\) | M1 | Substitutes \(y=3x+k\) and expands. Condone lack of \(=0\) |
| \(\Rightarrow 10x^2+(6k+2)x+k^2+4k+11=0\) | A1 | Correct 3-term quadratic (terms collected) |
| \(b^2-4ac=0 \Rightarrow (6k+2)^2-4\times10\times(k^2+4k+11)=0\) | M1 | Uses \(b^2-4ac=0\) where \(b\) and \(c\) are expressions in \(k\) |
| \(\Rightarrow 4k^2+136k+436=0 \Rightarrow k=\ldots\) | M1 | Solves resulting 3TQ in \(k\) |
| \(k = -17\pm6\sqrt{5}\) | A1 | Correct simplified values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2+y^2-10x+4y+11=0 \Rightarrow 2x+2y\frac{dy}{dx}-10+4\frac{dy}{dx}=0\) | M1, A1 | Differentiates implicitly; correct differentiation |
| Sets \(\frac{dy}{dx}=3 \Rightarrow x+3y+1=0\), combines with \(C\) to solve quadratic in \(x\) or \(y\) | M1 | Sets gradient \(=3\), combines with circle equation |
| \(x = \frac{25\pm9\sqrt{5}}{5}\), \(y = \frac{-10\pm3\sqrt{5}}{5}\), \(k=y-3x \Rightarrow k=\ldots\) | M1 | Uses at least one pair of coordinates to find \(k\) |
| \(k=-17\pm6\sqrt{5}\) | A1 | Correct simplified values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=3x+k \Rightarrow m=3 \Rightarrow m_r = -\frac{1}{3}\) | M1 | Applies negative reciprocal rule |
| \(y+2 = -\frac{1}{3}(x-5)\) | A1 | Correct equation of radial line through centre |
| \((x-5)^2+(y+2)^2=18\), \(y+2=-\frac{1}{3}(x-5) \Rightarrow \frac{10}{9}(x-5)^2=18 \Rightarrow x=\ldots\) | M1 | Solves simultaneously |
| \(x=\frac{25\pm9\sqrt{5}}{5}\), \(y=\frac{-10\pm3\sqrt{5}}{5}\), \(k=y-3x \Rightarrow k=\ldots\) | M1 | Applies \(k=y-3x\) |
| \(k=-17\pm6\sqrt{5}\) | A1 | Correct simplified values |
# Question 7:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-5)^2+(y+2)^2 = \ldots$ | M1 | Attempts to complete the square, halving both $x$ and $y$ terms. Award for sight of $(x\pm5)^2$, $(y\pm2)^2$ |
| Centre $(5,-2)$ | A1 | Correct coordinates |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \sqrt{``5"^2 + ``{-2}"{^2}-11}$ | M1 | Correct strategy for radius or radius² |
| $r = 3\sqrt{2}$ | A1 | Do not accept $r = \pm3\sqrt{2}$ or $\sqrt{18}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=3x+k \Rightarrow x^2+(3x+k)^2-10x+4(3x+k)+11=0 \Rightarrow x^2+9x^2+6kx+k^2-10x+12x+4k+11=0$ | M1 | Substitutes $y=3x+k$ and expands. Condone lack of $=0$ |
| $\Rightarrow 10x^2+(6k+2)x+k^2+4k+11=0$ | A1 | Correct 3-term quadratic (terms collected) |
| $b^2-4ac=0 \Rightarrow (6k+2)^2-4\times10\times(k^2+4k+11)=0$ | M1 | Uses $b^2-4ac=0$ where $b$ and $c$ are expressions in $k$ |
| $\Rightarrow 4k^2+136k+436=0 \Rightarrow k=\ldots$ | M1 | Solves resulting 3TQ in $k$ |
| $k = -17\pm6\sqrt{5}$ | A1 | Correct simplified values |
**Alternative (b) Alt 1 - Gradient method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2+y^2-10x+4y+11=0 \Rightarrow 2x+2y\frac{dy}{dx}-10+4\frac{dy}{dx}=0$ | M1, A1 | Differentiates implicitly; correct differentiation |
| Sets $\frac{dy}{dx}=3 \Rightarrow x+3y+1=0$, combines with $C$ to solve quadratic in $x$ or $y$ | M1 | Sets gradient $=3$, combines with circle equation |
| $x = \frac{25\pm9\sqrt{5}}{5}$, $y = \frac{-10\pm3\sqrt{5}}{5}$, $k=y-3x \Rightarrow k=\ldots$ | M1 | Uses at least one pair of coordinates to find $k$ |
| $k=-17\pm6\sqrt{5}$ | A1 | Correct simplified values |
**Alternative (b) Alt 3 - Radius perpendicular method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=3x+k \Rightarrow m=3 \Rightarrow m_r = -\frac{1}{3}$ | M1 | Applies negative reciprocal rule |
| $y+2 = -\frac{1}{3}(x-5)$ | A1 | Correct equation of radial line through centre |
| $(x-5)^2+(y+2)^2=18$, $y+2=-\frac{1}{3}(x-5) \Rightarrow \frac{10}{9}(x-5)^2=18 \Rightarrow x=\ldots$ | M1 | Solves simultaneously |
| $x=\frac{25\pm9\sqrt{5}}{5}$, $y=\frac{-10\pm3\sqrt{5}}{5}$, $k=y-3x \Rightarrow k=\ldots$ | M1 | Applies $k=y-3x$ |
| $k=-17\pm6\sqrt{5}$ | A1 | Correct simplified values |\begin{enumerate}
\item The circle $C$ has equation
\end{enumerate}
$$x ^ { 2 } + y ^ { 2 } - 10 x + 4 y + 11 = 0$$
(a) Find\\
(i) the coordinates of the centre of $C$,\\
(ii) the exact radius of $C$, giving your answer as a simplified surd.
The line $l$ has equation $y = 3 x + k$ where $k$ is a constant.\\
Given that $l$ is a tangent to $C$,\\
(b) find the possible values of $k$, giving your answers as simplified surds.
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q7 [9]}}