How do you prove a radical is a circle on the Cartesian plane?

Question

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?

Answer 1

$-x^2-x=\frac{1}{4}-(\frac{1}{4}+x+x^2)=\frac{1}{4}-(x+\frac{1}{2})^2$. Therefore, your equation is a circle with center $(-\frac{1}{2},0)$ and radius $\frac{1}{2}$

Answer 2

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.

Answer 3

By definition a circle is a set of points which are a fixed distance away from some fixed point called the center. In the language of coordinates: a set $S$, whose elements are ordered pairs, forms a circle only if there are constants $h, k, r$ such that every ordered pair $(x_0, y_0) \in S$ saitsfies

$$(x-h)^2+(y-k)^2=r^2$$

Note that the above is a neccesary but not sufficient for the point of $S$ to form a circle; the set $S = \{(0, -1), (0, 1)\}$ sastisfies the above condition with $h=0, k=0, r=1$, but it doe not describe a circle).

Now we wish to determine if the points of $S=\{(x, y)| \ $ $y= \sqrt{-x^2-x } \ \}$ form a circle; we have to answer: do there exist constants $h, k, r$ such that $(x-h)^2+(y-k)^2=r^2$ for every $(x, y)\in S$? Looking at the graph for inspiration, we can see that $h=-0.5, k=0, r=0.5$ might work; and indeed they do:

Take any $(x_0, y_0) \in S$. We want to show that this pair satisfies $(x_0+0.5)^2+ y_0^2=0.5^2$. And indeed it does;

$(x_0+0.5)^2+ y_0^2$

$=x_0^2+x_0+0.5^2+(\sqrt {-x^2-x})^2$

$=x_0^2+x_0+0.5^2+|-x^2-x|$

$=x_0^2+x_0+0.5^2-x_0^2-x_0$

$=0.5^2$

Thus $S$ is the subset of some circle. We already knew from the graph that it wasn't going to be a full circle; but we also haven't proved that it doesn't contain any holes.