Standard +0.3 This is a straightforward conversion from Cartesian to polar form using standard substitutions (x = r cos θ, y = r sin θ). While it requires expanding brackets and collecting terms, it's a routine Further Maths exercise with a prescribed answer format that guides the solution. The 4-mark allocation reflects mechanical algebraic manipulation rather than conceptual insight.
2 The Cartesian equation of a circle is \(( x + 8 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 100\).
Using the origin \(O\) as the pole and the positive \(x\)-axis as the initial line, find the polar equation of this circle, giving your answer in the form \(r = p \sin \theta + q \cos \theta\).
(4 marks)
2 The Cartesian equation of a circle is $( x + 8 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 100$.\\
Using the origin $O$ as the pole and the positive $x$-axis as the initial line, find the polar equation of this circle, giving your answer in the form $r = p \sin \theta + q \cos \theta$.\\
(4 marks)
\hfill \mbox{\textit{AQA FP3 2013 Q2 [4]}}