Answers to many questions
is a rectangle a plane?
No, it is not. A plane is infinite in two dimensions, a rectangle has finite size.
I thought planes were defined by 3 points
That is correct but not useful for your application.
my rectangle is defined by four.
As you claim that your rectangle is aligned to the horizontal and vertical directions, two points suffice. For a general rectangle, a third point would be required, or in fact half of a third point (5 real degrees of freedom in total). But this, too, is irrelevant to your question.
the equation of a rectangle
There is no equation for a rectangle, there is only a set of inequalities. Which don't integrate well into the whole setup of dealing with multiple equations.
Is either solution #1 or solution #2 valid in solving collision detection?
Each one will work for a particular use case.
Which solution (#1 or #2) is preferable for what reasons?
As your rectangle is formed by four line segments, solution #2 is the one to take, as your edit in response to the comment by jshin47 already indicates.
The core question
How can I calculate the intersection between a 2D rectangle and a 2D circle (without splitting up the rectangle into four separate lines)?
For the general case of an arbitrary circle and rectangle, I'd go for the intersections between circle and line segments. Avoiding that will only make things more complicated.
Solution #2 sounds like a good starting point. It isn't too hard: If the line segment starts at $(x_1,y_1)$ and ends at $(x_2,y_2)$ and your circle is centered at $(x_C, y_C)$ and has radius $r$, then you have the single non-linear equation
$$ ((1-t)\cdot x_1 + t\cdot x_2 - x_C)^2 +
((1-t)\cdot y_1 + t\cdot y_2 - y_C)^2 = r^2 $$
Solving this equation for $t$, you might find up to two solutions. If these satisfy $0\le t\le1$ then there is an intersection along the line segment. Do this for all four line segments, and if any intersects the circle, then the rectangle intersects the circle.
The above equation originated from the equation of the circle, with a generic point on the line (segment) substituted for $x$ and $y$.
\begin{align*}
x &= (1-t)\cdot x_1 + t\cdot x_2 \\
y &= (1-t)\cdot y_1 + t\cdot y_2 \\
r^2 &= (x-x_C)^2 + (y-y_C)^2
\end{align*}
In contrast to this solution, I use $t=0$ to represent the preceived starting point $(x_1,y_1)$ of the line, and $t=1$ for the end point $(x_2,y_2)$. This choice has no impact on your problem, as the direction of your line segments is irrelevant. But it is more useful or common for other applications.
If you have trouble understanding these parametric equations for $x$ and $y$, I suggest you simply choose end points with simple coordinates, and plug in some values of $t$, both from within and outside the specified range. You will see how all of these lie on the connecting line, and the range $0\le t\le 1$ corresponds to the line segment.
Note that all of the above won't deal with the scenario where the circle is fully contained within the rectangle, or vice versa. So you might want to check whether the center of the circle lies within the rectangle, or whether any rectangle corner lies within the circle.
Aligned rectangle
If your rectangle is aligned to the horizontal and vertical axes, you can simplify the intersection checks somewhat. I'd still examine every edge separately, but using a simpler decision making process.
As an example, let's take a horizontal edge, from $(x_1, y)$ to $(x_2, y)$. The circle is still centered at $(x_C, y_C)$ with radius $r$.
- If $x_1 \le x_C \le x_2$, then the circle intersects the line if $\lvert y - y_C\rvert \le r$. This means that the circle center and the line are no more than $r$ apart.
- Otherwise, they intersect if $(x_1-x_C)^2+(y-y_C)^2\le r^2$ or $(x_2-x_C)^2+(y-y_C)^2\le r^2$. This means that either endpoint must lie within the circle.
At $z=0$, $2x+y=4 \implies y=4-2x$ this will help for the integration.
Note that the $x$, $y$ and $z$ intercepts are $2$, $4$ and $2$ respectively.
\begin{align*}
\boldsymbol{r} &= \left( x,y,2-x-\frac{y}{2} \right) \\
\boldsymbol{r}_{x} &= (1,0,-1) \\
\boldsymbol{r}_{y} &= \left( 0,1,-\frac{1}{2} \right) \\
\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} &=
\left( 1,\frac{1}{2}, 1 \right) \\
dA &= |\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}| \, dx \, dy \\
&= \frac{3}{2} \, dx \, dy \\
A &= \int_{0}^{2} \int_{0}^{4-2x} \frac{3}{2} \, dy \, dx \\
&= \int_{0}^{2} \frac{3}{2} (4-2x) dx \\
&= \left[ 6x-\frac{3x^{2}}{2} \right]_{0}^{2} \\
&= 6
\end{align*}
Best Answer
Not all "curves" in $\mathbb R^2$ admits a normal vector, and not all "curves" in $\mathbb R^2$ admits a "parametrization". For example, pictorially it is not possible to find a normal vector at the four corners of your rectangle $R$. Similarly:
To prove this, for simplicity assume that the rectangle $R$ has the four corners $(0,0), (1,0),(0,1),(1,1)$. I claim that if $\gamma : (-\epsilon, \epsilon) \to R$ is differentiable, one to one and $\gamma (0) = (0,0)$, then $\gamma'(0) = 0$.
Write $\gamma (t) = (\gamma_1(t), \gamma_2(t))$. We need to show $$\gamma_1'(0) = \gamma_2'(0) = 0.$$
Since $\gamma$ is one to one, we may assume that $\gamma(s)$ lies in the horizontal sides for $s>0$ and $\gamma(u)$ lies in the vertical sides when $u<0$. So $\gamma_2(s) = 0$ for $s>0$ and $\gamma_1(u) = 0$ for $u<0$. This implies
$$ \gamma_2(0) = \lim_{s\to 0^+} \frac{\gamma_2(s) - \gamma_2(0)}{s-0} = 0$$
and similarly $\gamma_1'(0) = 0$.