<s>[INST] Intersection of two paraboloids Consider two paraboloids. The first one is given by $x^2 + y^2 = z+5$. So, it intersects the x-y plane in the circle $x^2+y^2=5$. The second paraboloid is exactly the same as the first one, only shifted in the x-y plane. It's equation becomes $(x-1)^2+(y-1)^2=z+5$. From the figure below, it seems clear that the two should intersect in a parabola. 

However, when we actually try and solve the two equations simultaneously, we get from the second equation
$$x^2 + y^2 -2x - 2y + 2 = z+5.$$
And substituting $x^2+y^2=z+5$ into this we get
$$2x+2y=2.$$
Now this is a linear function. However, the picture seems to suggest that it should be a parabola which is non-linear. What am I missing here?
 [/INST] Alternatively, exploiting  symmetry profitably, you can simply express the equations as
$$
\eqalign{
  & x^{\,2}  + y^{\,2}  = \left( {x - 1} \right)^{\,2}  + \left( {y - 1} \right)^{\,2}  = z + 5  \cr 
  & \left( {x - 1/2 + 1/2} \right)^{\,2}  + \left( {y - 1/2 + 1/2} \right)^{\,2}  = \left( {x - 1/2 - 1/2} \right)^{\,2}  + \left( {y - 1/2 - 1/2} \right)^{\,2}  = z + 5  \cr 
  & \left\{ \matrix{
  \left( {x - 1/2} \right)^{\,2}  + 1/4 + \left( {x - 1/2} \right) + \left( {y - 1/2} \right)^{\,2}  + 1/4 + \left( {y - 1/2} \right) = z + 5 \hfill \cr 
  \left( {x - 1/2} \right)^{\,2}  + 1/4 - \left( {x - 1/2} \right) + \left( {y - 1/2} \right)^{\,2}  + 1/4 - \left( {y - 1/2} \right) = z + 5 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  \left( {x - 1/2} \right)^{\,2}  + \left( {y - 1/2} \right)^{\,2}  = z + 9/2 \hfill \cr 
  \left( {x - 1/2} \right) + \left( {y - 1/2} \right) = 0 \hfill \cr}  \right. \cr} 
$$
That is, a paraboloid with vertex in $(1/2,\, 1/2,\, -9/2)$ cut
by the diametral plane through the vertex.
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