SOLUTION: a sphere of radius 5 cm and a right circular cone of base radius 5 cm and height 10 cm stand on a plane. find the position of a plane that cuts the two solids in equal circular cr

I think your professor wants you to calculate how high above the 
plane that they are standing on, must another plane, parallel to 
the plane they are standing on, that cuts the two solids, be so 
that the cross sections are two equal circles.  Think of the 
picture below as a mid-cross section of the sphere and the cone. 
The green line at the bottom represents the plane they are resting 
on and the red line is the plane that cuts them so that diameters 
AB and CD are equal, making the circles equal.  We want to find h 
such that AB = CD.



We redraw the figure:



ΔCEQ ∽ ΔFGQ







   <--equation 1

And by applying the Pythagorean theorem to right
triangle CEP,





  <--equation 2

Solve equation 1 for x,  

Substitute in equation 2:






5r = 0;  r-4 = 0
 r = 0;    r = 4

Only r = 4 makes sense.  And since




OS = 5
OD + DS = OS = 5
x + h = 5
3 + h = 5
    h = 2

So the plane which is 2 units above and parallel to the 
plane they are standing on, cuts the sphere and cone
so that the cross sections will be equal circles.  I'm 
pretty sure that's what you want to find.

Edwin