Scaling down a set of points into a smaller area A visibility graph $G(P) = (V,E)$ of a set $P = \{p_1, \dots, p_n\}$ of points is defined as follows.

Each vertex $u \in V$ corresponds to a point $p_u \in P$.
There exists an edge $uv \in E$ if, and only if the line segment $\overline{p_up_v}$ does not contain any other points in $P$.

Assume that the coordinates of every $p_i \in P$ has $O(n^c)$ digits, where $c$ is a constant.
My aim is to scale the point set, without changing the relative positions, into a smaller area.
For the sake of simplicity, assume that the area is a circle of raduis 1.
Consider three points: $p_1(100, 340)$, $p_2(500,150)$, $p_3(240, -600)$.
The new coordinates after scaling would be $p'_1(0.1, 0.34)$, $p'_2(0.5, 0.15)$, $p'_3(0.24, -0.6)$, and the relative positions stay the same, as well as the visibility relations.
My question is two folded:

In the very simple example above, the coordiantes are divided by $100$, and it is pretty easy to find that number. But what is the algorithm to find this number for any given point set?
Assuming that the coordinates are of $O(n^c)$ length, can there be a coordinate of $O(d^n)$ (exponential) length?

Regarding Discrete lizard's comment,
$100$ is an integer, which can be represented using $O(\log n)$ bits.
However, instead of $100$, the number was say $1998.12341234123422335$, then (i) it is not trivial to determine the number, and (ii) the division itself requires exponentially many digits.
The following procedure achieves your requirements:
Step 1: Find the smallest circle that encloses all points.  (There are standard algorithms for this.)
Step 2: Compute the area of that circle.
Step 3: Compute the scaling ratio $c$ needed to achieve the desired area.  (For instance if the area of that circle is 100 and you want area 5, then your scaling ratio is $c = \sqrt{100/5} = \sqrt{20}$.)
Step 4: Scale everything down by a factor of $c$, centered at the center of the circle found in Step 1.
This can be implemented in linear time, so it should be very efficient.