Why are there two versions of a polar equation for a circle from geometric form In class today we learned that a rectangular/geometric equation for a circle such as $x^2+(y-5)^2 = 9$ can be converted into a polar equation by reducing it to the quadratic equation $r^2-10r\sin \theta+16=0$ applying the quadratic formula to this equation yields two equations 
$$
r=5\sin\theta + \sqrt {25\sin^2\theta-16}
$$ 
and
$$
r=5\sin\theta - \sqrt {25\sin^2\theta-16}
$$
However when these equations are graphed in polar mode on a calculator they are the exact same circle. How is this possible?
 A: Polar coordinates are both a blessing and a curse. A blessing, because they make it easier to describe many geometric objects. A curse, because they are not unique: the same point can be represented in different ways.
In particular, the coordinates $(r,\theta)$ and $(-r,\theta+\pi)$ refer to the same point. This should be clear from a drawing.
What does this have to do with your question? If you substitute $\theta+\pi$ into your second equation, you get $$r_2=-5\sin\theta-\sqrt{25\sin^2\theta-16}=-\left(5\sin\theta+\sqrt{25\sin^2\theta-16}\right)=-r_1$$ because $\sin(\theta+\pi)=-\sin\theta$.
A: I think it is a good observation, and I would like to explain it like this:
You are looking at the curve that is the circle showed in the picture below.

First of all, if you draw a straight line from the origin, it is not so strange that it generically intersects the circle either in no point or in two points (the exception being when the line is tangent).
Making a polar plot of this circle, a small calculation shows that the angles determined by the circles are 
$$\pi/2-\arctan(3/4)\leq\theta\leq\pi/2+\arctan(3/4)\quad\text{(modulus $2\pi$)}
$$ 
 (this is also the condition for having non-negative expression inside your square roots). Thus, we use your correct calculation, and plot
$$
r_1=5\sin\theta+\sqrt{25\sin^2\theta-16}\quad \text{(orange)}
$$
and
$$
r_2=5\sin\theta-\sqrt{25\sin^2\theta-16}\quad \text{(green)}
$$
in the domain mentioned above. Animated, it looks like this:

Update
If this does not answer why one gets the full circle looking at only one of the $r_1$ or $r_2$, that is answered by what happens in the other valid $\theta$-range (looking at what is inside the square root),
$$
3\pi/2-\arctan(3/4)\leq\theta\leq3\pi/2+\arctan(3/4).
$$
The same animation in this domain is

If you consider both these animations you see that both the green part and the orange part has filled the full circle.
This is not so strange, if you consider that $x=r\cos\theta$ and $y=r\sin\theta$ and the fact that $\cos(\theta+\pi)=-\cos\theta$ and $\sin(\theta+\pi)=-\sin(\theta)$.
A: If you substitute
$$\theta\leftrightarrow-\theta,\\r\leftrightarrow-r$$ one equation becomes the other.
And as
$$r\cos(\theta)=-(-r)\cos(-\theta),\\r\sin(\theta)=(-r)\sin(-\theta),$$
this substitution amounts to a reflection around the vertical axis ($x\leftrightarrow-x$), which you cannot see.
