How do you prove a radical is a circle on the Cartesian plane? I know something like $ \sqrt{x^{2}-1}$ could be a circle because that's just what I've always been told and it relates to the Pythagorean theorem. But, I looked at $ \sqrt{-(x-1)x}.$ I know that certainly looks like a circle, and it probably is, but how do I know for sure it's a circle in the Cartesian plane?
Please note that $y=\sqrt{x^2-1}$ is not a circle. It is a hyperbola because after squaring, you get $x^2-y^2=1$ which is the equation of a hyperbola. Plus, since $y$ has to be non-negative, it can't be all of the hyperbola given by that equation, but only the half of the two branches of the hyperbola that are in the upper half of the plane.

For the second equation, again, note that it can't be a complete circle because we know that $y$ has to be non-negative. By squaring, you get 

$$y^2 = -x^2 -x = -(x+\frac{1}{2})^2+\frac{1}{4}$$

Therefore, $$(x+\frac{1}{2})^2+y^2=\frac{1}{4}$$

which is the equation of a circle centered at $(-\frac{1}{2},0)$ with radius $\frac{1}{2}$, but since $y$ has to be positive, it gives you only the upper half of the circle, not the rest.

One other way to see that it can't be a complete circle is to note that your equation defines $y$ as a function of $x$ and that means that its graph must pass the vertical line test. Therefore, it can't be a complete circle.

Please note that a circle is defined as the set of points that are at a fixed distance away from a point called the center. Knowing that distance in the plane for the two points $(x,y)$ and $(a,b)$ is defined as $\sqrt{(x-a)^2+(y-b)^2}$, the equation of a circle centered at $(a,b)$ with radius $R$ is then translated to:

$$\sqrt{(x-a)^2+(y-b)^2}=R$$
$$(x-a)^2+(y-b)^2=R^2$$

That's why $(x+\frac{1}{2})^2+y^2=\frac{1}{4}$ is in the form of the equation of a circle centered at $(-\frac{1}{2},0)$ with radius $\frac{1}{2}$, and the set of our points would've been a circle had we not have any restricting conditions on $y$. But since we do know that $y$ is non-negative, we have to restrict it to where $y$ is non-negative, which is the upper half of the plane.
$-x^2-x=\frac{1}{4}-(\frac{1}{4}+x+x^2)=\frac{1}{4}-(x+\frac{1}{2})^2$. Therefore, your equation is a circle with center $(-\frac{1}{2},0)$ and radius $\frac{1}{2}$