Level Curves Problem

Question

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

I know that the first equation is a circle but I do not know how to find out if the second one is it too.

Thanks for the help. (Sorry my English is not good).

Answer 1

Observe that when $x^2+y^2=6$,

$$f(x,y)=\sqrt6-6+2,$$ so that the circle belongs to a level curve.

We still have to show that it is a whole level curve, by solving

$$\sqrt{x^2+y^2}-x^2-y^2+2=\sqrt6-6+2.$$

As the quadratic equation $$r-r^2=\sqrt6-6$$ has a single positive root $r=\sqrt6$, $x^2+y^2=6$ is indeed a level curve.

Answer 2

With

$x^2 + y^2 = c, \; \text{ a constant}, \tag 1$

for any $c$, we have

$f(x, y) = \sqrt{x^2 + y^2} - x^2 - y^2 + 2 = \sqrt{x^2 + y^2} - (x^2 + y^2) + 2 = \sqrt c - c + 2; \tag 2$

thus the circle (1) lies in the $\sqrt c - c +2$-level set of $f(x, y)$; note we needn't prove

$f(x, y) = c_0, \; \text{a constant} \tag 3$

describes a circle to establish this.