Bruno Thompson

2022-07-26

Show that ${x}^{2}+{y}^{2}=6$
is a level curve of $f\left(x,y\right)=\sqrt{{x}^{2}+{y}^{2}}-{x}^{2}-{y}^{2}+2$

Hassan Watkins

Expert

Write $f\left(x,y\right)=\sqrt{{x}^{2}+{y}^{2}}-{x}^{2}-{y}^{2}+2$ as a function of ${x}^{2}+{y}^{2}$
let ${x}^{2}+{y}^{2}=k$
then $\sqrt{k}+k+2=c$ repesent the level curves f(x,y)=c
solving the above for k,
$⇒\sqrt{k}=\frac{-1±\sqrt{1-4\left(2-c\right)}}{2}$
$⇒k=\left(\frac{-1±\sqrt{1-4}\left(2-c\right)}{2}{\right)}^{2}$
$⇒{x}^{2}+{y}^{2}=\left(\frac{-1±\sqrt{1-4\left(2-c\right)}}{2}{\right)}^{2}$
To get the level curve ${x}^{2}+{y}^{2}=6$, we simply solve for c
such that $\left(\frac{-1±\sqrt{1-4\left(2-c\right)}}{2}{\right)}^{2}=6$
$⇒-1±\sqrt{1-4\left(2-c\right)}=±2\sqrt{6}$
$⇒1-4\left(2-c\right)=\left(1±2\sqrt{6}{\right)}^{2}$
$⇒c=\frac{1}{4}\left(7+\left(1±2\sqrt{6}{\right)}^{2}\right)={c}_{1},{c}_{2}$
$\therefore f\left(x,y\right)={c}_{1}$ or $f\left(x,y\right)={c}_{2}$ repesent the curve ${x}^{2}+{y}^{2}=6$.

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