Calling
#S_1 -> x^2+y^2+z^2+alphax-y=0#
#S_2->x^2+y^2+z^2+x+2z+1=0#
or
#S_1->(x+alpha/2)^2+(y-1/2)^2+z^2=1/4(alpha^2+1)#
#S_2->(x+1/2)^2+y^2+(z+1)^2=1/2#
or
#S_1->norm(p-p_1)=r_1#
#S_2->norm(p-p_2)=r_2#
with
#p = (x,y,z)#
#p_1 = (-alpha/2,1/2,0)# with #r_1 = 1/2 sqrt(alpha^2+1)#
#p_2=(-1/2,0,-1)# with #r_2 = sqrt2/2#
The intersection plane is obtained as #S_1-S_2 = 0# or
#(1-alpha)x+y+2z+1=0#
The sphere's center distance is #d = norm(p_1-p_2)# and considering the intersection point as a vertice and the sphere centers as the remaining two triangle vertices,
#r_1^2+r_2^2-2r_1 r_2 cos alpha = d^2#
then
#cosalpha = (r_1^2+r_2^2-d^2)/(2 r_1 r_2) = sqrt2/2# or
#1/4(alpha^2+1)+1/2-(1/4(alpha-1)^2+1/4+1)=sqrt2 1/2sqrt(alpha^2+1)sqrt2/2 = sqrt(alpha^2+1)#
or
#alpha^2-(alpha-1)^2-2=4sqrt(alpha^2+1)#
note that #alpha^2# cancels to the left and we get a second degree polynomial in #alpha# to be solved. This is left as an exercise to the reader.
