| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Standard +0.3 This is a standard circle tangent problem requiring perpendicular gradient for part (a), distance formula for radius in part (b), and perpendicular distance from centre to tangent line for part (c). All techniques are routine A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4y - x = 7\) | A1 | For expressing the equation of the tangent in the form \(4y - x = 7\) |
| M1 | Finding the equation of a line with gradient \(\frac{1}{4}\) and point \((-3,1)\) | |
| M1 | Correctly deduces the gradient of \(AQ\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((x + 3)^2 + (y - 1)^2 = 17\) | A1 | Correctly states the equation of the circle |
| M1 | An attempt to use the Pythagoras Theorem to find the radius or (radius)\(^2\) using their \(Q(1,2)\) and \(A(-3,1)\) | |
| A1 | Finds \(Q\) correctly | |
| M1 | For an attempt at solving \(4y - x = 7\) and \(y = -4x + 6\) simultaneously to find the coordinates of point \(Q\) |
| Answer | Marks |
|---|---|
| \(k = -28\) | A1 |
| M1 | For substituting \((-7,0)\) into \(y = -4x + k\) and proceeds to \(k = \ldots\) |
| M1 | For substituting \(2k\) and \(k\) into their \(\frac{B}{(2x-k)}\) and subtracting |
| M1 | For finding the end point of the diameter using mid-point formula \(\frac{x+1}{2} = -3 \Rightarrow x = -7\) and \(\frac{y+2}{2} = 1 \Rightarrow y = 0\) |
| M1 | Uses \(b^2 - 4ac = 0\) and finds a value of \(k\) |
| M1 | For solving \(l_2\) with their equation of the circle |
### Part a:
$4y - x = 7$ | A1 | For expressing the equation of the tangent in the form $4y - x = 7$
| M1 | Finding the equation of a line with gradient $\frac{1}{4}$ and point $(-3,1)$
| M1 | Correctly deduces the gradient of $AQ$
### Part b:
$(x + 3)^2 + (y - 1)^2 = 17$ | A1 | Correctly states the equation of the circle
| M1 | An attempt to use the Pythagoras Theorem to find the radius or (radius)$^2$ using their $Q(1,2)$ and $A(-3,1)$
| A1 | Finds $Q$ correctly
| M1 | For an attempt at solving $4y - x = 7$ and $y = -4x + 6$ simultaneously to find the coordinates of point $Q$
### Part c:
$k = -28$ | A1 |
| M1 | For substituting $(-7,0)$ into $y = -4x + k$ and proceeds to $k = \ldots$
| M1 | For substituting $2k$ and $k$ into their $\frac{B}{(2x-k)}$ and subtracting
| M1 | For finding the end point of the diameter using mid-point formula $\frac{x+1}{2} = -3 \Rightarrow x = -7$ and $\frac{y+2}{2} = 1 \Rightarrow y = 0$
| M1 | Uses $b^2 - 4ac = 0$ and finds a value of $k$
| M1 | For solving $l_2$ with their equation of the circle6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{63d85737-99d4-4916-a479-fe44f77b1505-10_951_1022_306_488}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The circle $C$ has centre $A$ with coordinates $( - 3,1 )$.\\
The line $l _ { 1 }$ with equation $y = - 4 x + 6$, is the tangent to $C$ at the point $Q$, as shown in Figure 3.\\
a. Find the equation of the line $A Q$ in the form $a x + b y = c$.\\
b. Show that the equation of the circle $C$ is $( x + 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 17$
The line $l _ { 2 }$ with equation $y = - 4 x + k , k \neq 6$, is also a tangent to $C$.\\
c. Find the value of the constant $k$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q6 [10]}}