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- Thread starter Yankel
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- Jan 26, 2012

- 4,193

\begin{align*}

zy&=x^{2}+y^{2} \\

0&=x^{2}+y^{2}-zy \\

\frac{z^{2}}{4} &=x^{2}+y^{2}-zy+ \frac{z^{2}}{4} \\

\frac{z^{2}}{4} &= x^{2}+ \left( y- \frac{z}{2} \right)^{ \! 2}.

\end{align*}

This is the equation of a circle centered at $(0,z/2)$ of radius $z/2$.

- Jan 29, 2012

- 1,151

- Feb 13, 2012

- 1,704

$\displaystyle \frac{x^{2} + y^{2}}{y} = a \implies r = a\ \sin \theta\ (1)$

... where $\theta$ must produce a value of $r \ge 0$...

Kind regards

$\chi$ $\sigma$