Use this A-Level Pure Mathematics September 2020 marking scheme for checking answers, revision, and study practice in Zimbabwe.
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Questions: Lockdown Pure Maths Test 51 Question Paper
This is the A-Level Pure Mathematics marking scheme from September 2020. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
Upper $6{ }^{\text {th }}$
LOCKDOWN PURE MATHS TEST 51
Thursday 24 September 2020
ANSWERS
30 MARKS
NB: If a numerical answer cannot be given exactly, and the accuracy required is not specified in the question, then in the case of an angle it should be given to the nearest degree, and in other cases it should be given correct to $\mathbf{2}$ significant figures.
1. The complex number $z$ is represented by the point $P(x, y)$ in the Argand diagram and
\[
|z+3|=k|z-i|,
\]
where $k$ is a real positive constant.
(a) When $k \neq 1$, the loci of $P$ is a circle. Find in terms of $k$,
(i) the equation of the circle,
\[
\begin{array}{l}
|z+3|=k|z-i|
\Rightarrow|x+i y+3|=k|x+i y-i|
\Rightarrow|(x+3)+i y|=k|x+i(y-1)|
\Rightarrow \sqrt{(x+3)^{2}+y^{2}}=k \sqrt{x^{2}+(y-1)^{2}}
\Rightarrow(x+3)^{2}+y^{2}=k^{2}\left[x^{2}+(y-1)^{2}\right]
\Rightarrow x^{2}+6 x+9+y^{2}=k^{2}\left[x^{2}+y^{2}-2 y+1\right]
\Rightarrow x^{2}+6 x+9+y^{2}=k^{2} x^{2}+k^{2} y^{2}-2 k^{2} y+k^{2}
\Rightarrow k^{2} x^{2}-x^{2}+k^{2} y^{2}-y^{2}-2 k^{2} y-6 x-9+k^{2}=0
\therefore\left(k^{2}-1\right) x^{2}+\left(k^{2}-1\right) y^{2}-6 x-2 k^{2} y+k^{2}-9=0
\end{array}
\]
(ii) the coordinates of the centre of the circle.
[7]
\[
\begin{array}{l}
\left(k^{2}-1\right) x^{2}+\left(k^{2}-1\right) y^{2}-6 x-2 k^{2} y+k^{2}-9=0
\Rightarrow x^{2}+y^{2}-\frac{6}{\left(k^{2}-1\right)} x-\frac{2 k^{2}}{\left(k^{2}-1\right)} y+\frac{\left(k^{2}-9\right)}{\left(k^{2}-1\right)}=0
\end{array}
\]
When mediocrity becomes the accepted norm, excellence dies a painful death. Endurance is
one of the most difficult disciplines, but victory comes to those who endure.
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