OCR H240/02 2020 November — Question 7 8 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2020
SessionNovember
Marks8
PaperDownload PDF ↗
TopicCircles
TypeCircle from diameter endpoints
DifficultyModerate -0.8 This is a straightforward vectors and circle question requiring only recognition of standard geometric interpretations (distance, midpoint, circle equation) and routine algebraic manipulation. All parts are direct applications of definitions with no problem-solving or novel insight required, making it easier than average.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement

\(A\) and \(B\) are fixed points in the \(x\)-\(y\) plane. The position vectors of \(A\) and \(B\) are \(\mathbf{a}\) and \(\mathbf{b}\) respectively. State, with reference to points \(A\) and \(B\), the geometrical significance of
  1. the quantity \(|\mathbf{a} - \mathbf{b}|\), [1]
  2. the vector \(\frac{1}{2}(\mathbf{a} + \mathbf{b})\). [1]
The circle \(P\) is the set of points with position vector \(\mathbf{p}\) in the \(x\)-\(y\) plane which satisfy $$\left|\mathbf{p} - \frac{1}{2}(\mathbf{a} + \mathbf{b})\right| = \frac{1}{2}|\mathbf{a} - \mathbf{b}|.$$
  1. State, in terms of \(\mathbf{a}\) and \(\mathbf{b}\),
    1. the position vector of the centre of \(P\), [1]
    2. the radius of \(P\). [1]
It is now given that \(\mathbf{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\) and \(\mathbf{p} = \begin{pmatrix} x \\ y \end{pmatrix}\).
  1. Find a cartesian equation of \(P\). [4]

$A$ and $B$ are fixed points in the $x$-$y$ plane. The position vectors of $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively.

State, with reference to points $A$ and $B$, the geometrical significance of
\begin{enumerate}[label=(\alph*)]
\item the quantity $|\mathbf{a} - \mathbf{b}|$, [1]
\item the vector $\frac{1}{2}(\mathbf{a} + \mathbf{b})$. [1]
\end{enumerate}

The circle $P$ is the set of points with position vector $\mathbf{p}$ in the $x$-$y$ plane which satisfy
$$\left|\mathbf{p} - \frac{1}{2}(\mathbf{a} + \mathbf{b})\right| = \frac{1}{2}|\mathbf{a} - \mathbf{b}|.$$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item State, in terms of $\mathbf{a}$ and $\mathbf{b}$,
\begin{enumerate}[label=(\roman*)]
\item the position vector of the centre of $P$, [1]
\item the radius of $P$. [1]
\end{enumerate}
\end{enumerate}

It is now given that $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$ and $\mathbf{p} = \begin{pmatrix} x \\ y \end{pmatrix}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find a cartesian equation of $P$. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/02 2020 Q7 [8]}}