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    Question: https://gyazo.com/c7db45f12d47aca8a58575fc7eaa22da
    Book's solution: https://gyazo.com/5f4827baf8d45e35aede1419089a5a43
    https://gyazo.com/9a3ea9ab090078f825204a2e6a0ed0da
    My question: For the solution of QB, with
    \displaystyle\begin{equation*}\ {x} = 4\end{equation*}
    why is \displaystyle\begin{equation*} \frac{u}{u^2+v^2} = 2 \end{equation*}
    Shouldn't it be \displaystyle\begin{equation*} \frac{u}{u^2+v^2} = 4? \end{equation*}
    Thanks in advance.
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    (Original post by Nerrad)
    ...
    At a glance, I seemingly agree with you, although it seems like a fairly bizarre error to make.
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    (Original post by Zacken)
    At a glance, I seemingly agree with you, although it seems like a fairly bizarre error to make.
    And yet their working out continues, and seems to have a final (assuming it's correct) answer?
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    (Original post by Nerrad)
    And yet their working out continues, and seems to have a final (assuming it's correct) answer?
    Yeah, it's defo wrong - just had a proper look.

    Should be this: \frac{u}{u^2 + v^2} = 4 \Rightarrow u = 4u^2 + 4v^2 \Rightarrow u^2 + v^2 - \frac{u}{4} =0

    So \left(u - \frac{1}{8}\right)^2 - \frac{1}{64} + v^2 = 0 \Rightarrow \left(u - \frac{1}{8}\right)^2 + v^2 = \left(\frac{1}{8}\right)^2.

    Circle centre \left(\frac{1}{8}, 0\right) and radius \frac{1}{8}.
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    (Original post by Zacken)
    Yeah, it's defo wrong - just had a proper look.

    Should be this: \frac{u}{u^2 + v^2} = 4 \Rightarrow u = 4u^2 + 4v^2 \Rightarrow u^2 + v^2 - \frac{u}{4} =0

    So \left(u - \frac{1}{8}\right)^2 - \frac{1}{16} + v^2 = 0 \Rightarrow \left(u - \frac{1}{8}\right)^2 + v^2 = \left(\frac{1}{4}\right)^2.

    Circle centre \left(\frac{1}{4}, 0\right) and radius \frac{1}{4}.
    Radius =1/8
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    (Original post by B_9710)
    Radius =1/8
    Oh gosh, thanks!! :lol:
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    (Original post by Zacken)
    Oh gosh, thanks!! :lol:
    :borat::hat2:
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    Thanks guys
 
 
 
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