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A level set of a real-valued function $f$ of the real variables $x,y$ is a set of the form: $$L_{x_1}(f)=({(x,y):f(x,y)=x_1})$$ that is, a set where the function takes on a given constant value $x_1$

I wante to find the common level set of the following functions

$$f(x,y)= (x-1)/((x-1)² +(y-2)²))$$

and

$$g(x,y)=(y-2)/((x-1)² +(y-2)²))$$

for some constants $x_1$ and $x_2$.

I have no idea to start.

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    So "level set" is simply the preimage of a point? So you are asking for which $x_1, x_2$ the preimages $f^{-1}(x_1)$ and $f^{-1}(x_2)$ are equal as sets? Or are you looking for solutions to a system of equation $f(x, y)=x_1$ and $g(x, y)=x_2$? – 2017-01-31
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    @freakish: It is the system of equations with respect to $x$ and $y$ in which the two level sets are equal. – 2017-01-31

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I think something is wrong.

On a common level set, we will have that $f(x, y)/g(x, y)$ is constant, so $(x-1)/(y-2)$ is constant, which already tells you that any common level set will be contained in a line $(x-1) = k(y-2)$.

Now along such a line $\ell$, we would have $g|_\ell = \frac{y-2}{(y-2)^2(1+k^2)} = \frac{1}{1+k^2}\frac{1}{y-2}$.

Similarly, $f|_\ell = \frac{k}{1+k^2}\frac{1}{y-2}$. We conclude that, unless $\ell$ has infinite slope, no subset of $\ell$ (other than a singleton) is a common level set. If $\ell$ has infinite slope, it has to be the line $x = 1$, which by inspection again has no subsets which are common level sets.

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    I don't know if you should accept this answer so quickly -- I might very well be making a mistake. – 2017-01-31
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    The exercice is true and your answer is also true. The common level set is a singleton or empty – 2017-01-31