Expert: Jack Cheng - 11/23/2006

Question
i couldn't find anyone
else that could answer this:

"this question has been denied by other experts, this is not homework.
No professor in their right mind would even attempt to assign a problem
of this difficulty. This is for a project of mine, dealing with a
prediction algorithm.


Known : -the coords of the centers of a few circles, and their radii
I need to find the point whose average proximity to all the circles is
the best. For example, if I have 3 circles
Circle1 (5,5) radius:4
Circle2 (2,8) radius:6
Circle3 (6,1) radius:3
The formula for the average proximity would end up to be:
3  / ( |sqrt((x-5)^2+(y-5)^2) - 4| + |sqrt((x-2)^2+(y-8)^2) - 6| +
|sqrt((x-6)^2+(y-1)^2) - 3| )
which in plain english: the inverse of the average distance from a
point (x,y) to the point on each circle which is closest to the (x,y).
I realized that to find the maximum within an 2 dimensional interval, i
would have to set the partial derivative in terms of y and in terms of
x to be 0, and solve - giving me the extrema. Then using the 2nd deriv
test, i would find which point is a max and which is a min, and after
pluging it back into the orignal formula, i would see which is the
absolute max and min.
The problem is that with absolute values, i get too many piece wise
formula segments. ( i need to find the point with regards to about 20
circles). My friend at school suggested taking the squareroot of the
square
of the each function:
f(x,y) = |g(x,y)| + |h(x,y)| ...
f(x,y) = sqrt((g(x,y)^2) + sqrt((h(x,y)^2) ...

I tried this, but i reach a very problematic point.
I get this huge formula for the partial derivatives of 1 function such
as g(x,y). Which is a fraction.
In order to make the whole equation equal to 0, i would have to
multiply each of the 20 fractions' numerators, by each of the other
denominors.
If i avoid this and simply us a ^-1 notation for the denominator, i
still get stuck with something that looks like a quadratic equation,
but
with a degree much larger than 2 or 3 or even 4.
Is there another way of doing this?

Averaging the centers of the circles, or finding the center of the mass
of the whole structure, while seem to be good methods, don't work,
especially in a case where the point of intersection/ almost
intersection
exists outside the boundry created by the polygon whose points are the
centers of all outermost circles.
Could you help me?
-roman

"

Answer
Hi Roman,

I would hate to reject your question given what you've been through, but frankly, this question is way out of my expertise (have you tried the computer science or physics section? They might have a better idea).

Here's my little (perhaps useless) advise. Have you tried finding an algorithm in just one dimension, i.e. along a single line? Since the x and y are independent (maybe I am wrong on this), perhaps you can merge the result for one-dimension into two-dimension? It's similar to the center of mass, except you would have to take into account the radius of the circle.

Sorry that I am unable to help.

~ Jack