Find the equation of the cone whose vertex is at the origin and base is the circle #x=a#, #y^2+z^2=b^2#?
First the sphere #S# which intersected with the plane
#Pi-> x = a# gives the circle
#C-> y^2+z^2= b^2#
#S-> norm(p-p_0)= r#
with
#r = b/a sqrt(b^2+a^2)#
#p = (x,y,z)#
#p_0 = (x_0,y_0,z_0)#
#x_0 = a+ sqrt(r^2-b^2), y_0 = z_0 = 0#
Now considering the line
#L->p=lambda vec v# which passes by #(0,0,0)#
the intersections #S nn L# are determined by solving
#norm(lambda vec v - p_0) = r# or
#lambda^2 norm(vec v)^2-2 lambda << vec v, p_0 >> + norm(p_0)^2 = r^2#
solving for #lambda# we get
#lambda = << vec v, p_0 >> pm sqrt(<< vec v, p_0 >> ^2-norm(vec v)^2(norm(p_0)^2-r^2))#
but the line #L# must be tangent to #S# hence
#<< vec v, p_0 >> ^2-norm(vec v)^2(norm(p_0)^2-r^2)=0#
This is the locus defining the cone surface. Putting #vec v = (x,y,z)# and applying the values for #p_0, r# we get
#b^4 x^2 - a^4 (y^2 + z^2) - 2 a^2 b^2 (y^2 + z^2) +
a^2 b^2 (x^2 + y^2 + z^2)=0#
Attached a plot shoving the conic surface for
#a = 1, b = 1/2#
The circle is shown in red.
