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I have to find the level sets of the function $$f(x, y) = \frac{1}{\sqrt{36 - 25x^{2} - 25y^{2}}},$$ so I make $f(x, y) = k$ and after manipulating this equation come to $$x^{2} + y^{2} = \left( 36 - \frac{1}{k^{2}} \right) \frac{1}{25}.$$ Now, previoulsy I had found that $range f = \{ z : z \geq \frac{1}{6}\}$, so $$k \geq \frac{1}{6} \implies 36 \geq \frac{1}{k^{2}} \implies 36 - \frac{1}{k^{2}} \geq 0 \implies \left( 36 - \frac{1}{k^{2}} \right) \frac{1}{25} \geq 0.$$ Does that mean that the level sets of $f$ are circles with radius of any length?

UPDATE: I think I'm now sure this is the answer since the four possibilities are circles of radius $(1)$ $\leq \frac{6}{5}$ (counterexample $k = 1$), $(2)$ $< \frac{6}{5}$ (counterexample $k = 1$), $(3)$ $> \frac{6}{5}$ (counterexample $k = \frac{1 + \sqrt{6}}{12}$), and $(4)$ of any length.

Carl Langlois
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    It just means that the level set of $f$ with level $k$ is a circle of radius $\frac 15 \sqrt{36-\frac{1}{k^2}}$ centered at origin – Samrat Mukhopadhyay Feb 17 '14 at 19:55
  • @SamratMukhopadhyay I understand that, but want to know if there is some lower bound other than cero for the values of $\frac{1}{5} \left( \sqrt{36 - \frac{1}{k^{2}}}\right)$ (that is, if there exists some $c > 0$ such that $\frac{1}{5} \left( \sqrt{36 - \frac{1}{k^{2}}}\right) \geq c$). Knowing this is important for me since I've got four possible answers for the problem, with only one possible answer. – Carl Langlois Feb 17 '14 at 20:01
  • Given the bound on $k$, there does not seem to be any larger lower bound on the radius of the circle. – Samrat Mukhopadhyay Feb 17 '14 at 20:10

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