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If I have an inequality that defines a domain of a function $y-1-x \geq 0$, how might I rewrite that, so I can plot the in equality and visualise it as a circle? I tried rewriting it as a normal equation to $y^2-x^2 \geq 1^2$, however I see that I clearly doesn't get the same thing, and it isn't of course a circle either. Should it be $y^2+x^2 \geq 1^2$, or how might I do this? The original function is $$f(x,y) = \sqrt{y-1-x}+ \ln(x-y^2+4y-3).$$

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    Isn't the inequality a halfspace? why would you want to visualise it as a circle? – 2017-02-20
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    And what is it you need to find for $f(x,y)$? – 2017-02-20
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    Domain and range is what i need to find. I just wanted to know, if it was possible because of the looks. – 2017-02-20
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    Relating with circles doesn't seem to be the right approach – 2017-02-20

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Domain for the function means both terms must be defined. For the square root, we have $$y-x-1 \ge 0\implies y\ge x+1 \space \space\space (1)$$ For the $\ln$ term, we have $$x-(y-1)(y-3) \ge 0 \implies x \ge (y-1)(y-3)$$ This will give us that for the function to be defined, we would need $$(y-1)(y-3) \le x \le (y-1) , \space\space\space y\ge 1$$ (Easy to see, just graph the two functions above)

Now, for range, we observe that both functions are increasing with their parameters.For the minimum value , we consider the critical point $y = 1$, where $x$ must be $0$, hence, minimum value is $-\infty$ $(\ln0)$. For a fixed $y$, we observe that $f(x,y)$ increases with $x$. Hence, we see that the maximum value is attained on the line, and that to at $y=\frac{5}{2}$. Hence, maximum value is $2\ln(1.5)$