Is it possible to find the domain and range of a circle function without graphing it?

Question

If I am given a circle, such as, $(x + 1)^2 + y^2 = 9$, is it possible to determine the domain and range without having to graph it up? I know the answer, but I don't see any connection with that and the equation. I know for functions such as $x^2 + y^2 = 9$ the domain and range is simply -3, 3 but the equation above is not in that form.

Answer 1

Defining $x'=x+1$, the equation becomes $x'^2+y^2=9$, that you know how to handle.

The domain of $x'$ being $-3, 3$, that of $x$ is $-4, 2$.

More generally, to find the domain of such an implicit function, you have to check for what values of $x$ is has solutions.

In this case, $(x+1)^2+y^2=9$ can be written as $y^2=9-(x+1)^2$, which makes sense only if $0\le9-(x+1)^2$ (because $y^2$ cannot be negative), or $(x+1)^2\le9$. $$-3\le x+1\le3.$$$$-4\le x\le2.$$ Very similarly for the range, $(x+1)^2=9-y^2$, so that $0\le9-y^2$. $$-3\le y\le3.$$ Let us take the case of a general conic, $ax^2+bxy+cy^2+dx+ey+f=0$. You can see that as a second degree equation in $y$: $$cy^2+(bx+e)y+(ax^2+dx+f)=0.$$ It has real solutions when the discriminant is positive, i.e. $$(bx+e)^2-4c(ax^2+dx+f)\ge 0,$$ $$(b^2-4ca)x^2+2(be-2cd)x+(e^2-4cf)\ge 0.$$ You now have to discuss the roots of this second degree equation in $x$, again based on its discriminant $$4(be-2cd)^2-4(b^2-4ca)(e^2-4cf).$$

Answer 2

$(x-a)^2 + (y-b)^2 = r^2$ is a circle with center at $(a, b)$ and radius $r$, so domain is $[a-r, a+r]$ and range is $[b-r, b+r]$.