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# mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of larger sphere?

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Answer 1 · 2018-06-28T17:27:59

The distance is #2\sqrt(2) + 3# or #\approx 5.828427125#

Let the radiuses of the mutually tangent spheres be # A(1,1)# and # B(3,1)#
respectively

Let # C(2,y)# be the radius of the sphere that lies on top of them

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#overline{AC} = 3 # and #overline{BC} = 3 #
Recall the distance formula:
# d = sqrt{ (x_1 - x_2)^2 + (y_1-y_2)^2 } #

Thus,
# 1^2 + (y-1)^2 = 3^2 #
#1 + (y-1)^2 = 9 #
# (y-1)^2 = 8 #
# y = 1 \pm 2sqrt{2} #

Since # y # must be greater than 1 because it is on top of the spheres, reject the negative solution
# y = 2sqrt{2} + 1#

Since the radius of the sphere is 2, # h = y+2 #
height #h = y + 2 = 2sqrt{2} + 1 + 2 #

#\therefore# The distance is #2\sqrt(2) + 3#