Let's place the buoys at the points (-1.5,0) and (1.5,0)
so they will be 3 units (miles) apart:



Let's draw an arbitrary variable point where the boat might be
at one instant and label it (x,y) that appears to be about twice 
as far from (-1.5,0) as it is from (1.5,0), and draw a line from 
this arbitrary variable point to each of the buoy points (-1.5,0)
and (1.5,0). Label the longer line d1 and the shorter 
one d2.



We set d1 equal to 2 times d2



Now we use the distance formula to substitute for the
two distances d1 and d2







Square both sides:



















Divide through by 3

x² - 5x + 2.25 + y² = 0

We can tell this is a circle.  We need to get it in
the standard form (x-h)² + (y-k)² = r²
Get the constant on the right:

x² - 5x + y² = -2.25

Complete the square on the x-terms.

Multiply the coefficient of x, which is -5 by , 
getting -2.5.  Square 2.5, getting 6.25

Add that to both sides:

x² - 5x + y² = -2.25
 

x² - 5x + 6.25 + y² = -2.25 + 6.25

(x - 2.5)(x - 2.5) + y² = 4

Write the (x - 25)(x - 2.5) as (x - 2.5)² 
write y² as (y - 0)²
Write 4 as 2²

(x - 2.5)² + (y - 0)² = 2²

So the boat travels in a circle with center (2.5,0)
and radius 2 

Here is the circle the boat travels in:



Edwin