The circle x² + y² = 4 intersects the x axis respectively at the points E and F. A different circle with variable radius with its center located at F cuts the former circle at point G located above x axis and intersects the line segment EF at the point H.

Determine the maximum area of the triangle FHG.

First circle,

   x^2 + y^2 = 2^2

Second circle,

   (x - 2)^2 + y^2 = r^2, where r < 4.

Solving these for point G(x,y),

        8 - r^2
   x = --------- 
           4

        r*sqrt(16 - r^2)
   y = ------------------
               4

Area of triangle FGH,

   A = (1/2)(base)(altitude) = (1/2)(r)(y)

        r^2*sqrt(16 - r^2) 
     = --------------------
                8

Setting dA/dr = 0 and solving for r^2 gives,

   r^2 = 32/3

Plugging this into the area formula gives.

   A = 16*sqrt(3)/9 ~= 3.0792

This agrees with Geometer's Sketchpad.

Posted by Bractals on 2008-01-11 12:33:02