You draw a square. Then, you draw the largest possible circle, tangent to all four sides. If the lower left corner of the square is at (0,0), and the sides of the square are parallel to the coordinate axis, the point (2,1) is on the circle, within the square.
What is the radius of the circle?
Semantics aside, there are arguably two different locations for the center of the circle and the square: (1,1) and (5,5).
But what if we relax the constraint that the square must be parallel to
the coordinate axis? This turns out to be relatively simple to
solve.
If the center is at (x,y), then
(x^2 + y^2)/2 = (x-2)^2 + (y-1)^2.
Rearranging gives (x-4)^2 + (y-2)^2 = 10.
In other words, the set of all possible centers is a circle, centered
around (4,2) with a radius of sqrt(10). Obviously, (1,1) and
(5,5) are on this circle.
Just thought I'd share ...
Relaxing a constraint
