Parametrize a circle as a tube?

Question

I have this code:

Show[ContourPlot3D[{y == 0}, {x, -4, 6}, {y, -9, 1}, {z, -2, 8},
  Mesh -> {Range[-4, 6], Range[-9, 1], Range[-2, 8]},
  Axes -> True, AxesLabel -> {x, y, z},
  Ticks -> {Range[-4, 6, 1], Range[-9, 1, 1], Range[-2, 8, 1]}
  ],
 ParametricPlot3D[{3 Cos[φ] + 1, 0, 3 Sin[φ] + 3}, {φ, 
   0, 2 π},
  PlotStyle -> {Thickness[.02], Red}],
 Graphics3D[{Sphere[{1, -4, 3}, 5]}],
 ViewPoint -> {5, 5, 5}]

Which produces this image.

I have a couple of questions.

1. The intersection of the sphere with center at $(1,-4,3)$ and radius $5$ with the $xz$ plane is a circle with equation $(x-1)^2+(z-3)^2=9$. I parametrized the circle as $x=3\cos\phi+1$, $y=0$, and $z=3\sin\phi+3$, then used ParametricPlot3D to add it to my picture. Anybody have a simple way of plotting the circle without my parametrization?

2. How can I add my circle so it looks like a tube?

Answer 1

cf[x_, y_, z_] := Plus @@ (({x, y, z} - {1, -4, 3})^2);
opts = {ViewPoint -> {5, 5, 5}, Axes -> True, AxesLabel -> {x, y, z}, 
   Ticks -> {Range[-4, 6, 1], Range[-9, 1, 1], Range[-2, 8, 1]}, ImageSize -> 300};

BoundaryStyle

Use BoundaryStyle->{{1,2}->Directive[Red, Tube@@#&]} (or BoundaryStyle -> {{1, 2} -> Directive[Red, # /. Line -> Tube &]}) to make the boundary between the sphere and the plane rendered as a red tube.

cp = ContourPlot3D[{cf[x, y, z]== 25,  y == 0}, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
  ContourStyle -> {Directive[Blue, Opacity[0.5]], Directive[Pink, Opacity[0.5]]}, 
  BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Red, Tube @@ # &]}, 
  Mesh -> None, Evaluate@opts]

cp2 = ContourPlot3D[{y == 0}, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
  ContourStyle ->None, Mesh -> {Range[-4, 6], Range[-9, 1], Range[-2, 8]}];

Show[cp, cp2]

MeshFunctions

cp3 = ContourPlot3D[cf[x, y, z] == 25, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
  ContourStyle -> Opacity[.5, Blue], 
  MeshFunctions -> {Function[{x, y, z}, y]}, Mesh -> {{0}}, 
  MeshStyle -> Directive[Red, Tube @@ # &], Evaluate@opts]

or MeshStyle -> Directive[Red, # /. Line -> Tube &], gives

cp4 = ContourPlot3D[{y == 0}, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
  ContourStyle -> Opacity[.5, Red], 
  Mesh -> {Range[-4, 6], Range[-9, 1], Range[-2, 8]}];

Show[cp3, cp4]

Notes: Two undocumented tricks are used in the above methods. The BoundaryStyle trick first appeared on this site in this answer by Daniel Lichtblau linked by @Guesswhoitis in the comments. It also features a second undocumented trick: using functions as style directives. On this second trick see this page for relevant links.

Post-processing

Alternatively, you can post-process the ContourPlot3D output to change Line to Tube.

ContourPlot3D[{cf[x, y, z] == 25, y == 0}, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
  ContourStyle -> {Directive[Blue, Opacity[0.5]], Directive[Pink, Opacity[0.5]]}, 
  Mesh -> None, BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Red}, 
  Evaluate@opts] /. Line -> Tube

ContourPlot3D[cf[x, y, z] == 25, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
  ContourStyle -> Opacity[.5, Blue], MeshFunctions -> {Function[{x, y, z}, y]}, 
  Mesh -> {{0}}, MeshStyle -> Red, Evaluate@opts] /. Line -> Tube

to get the same pictures as above.

Update:

How would you use this code to make the tube larger?

This is easiest with the post-processing approach: simply change /. Line->Tube to /. Line -> (Tube[#, .3]&).

With the first and second approaches, I expected that Directive[Red, Tube[#, .3]& @@ # &] would work, but it doesn't. However, Directive[Red, Module[{tt = Tube[#, .3] &}, tt @@ # &]] works. It seems that, somehow, with another layer of function evaluation it gets too deep for Directive to process the primitive properly.

Alternatively, you can define a tube primitive function tubeF outside, and use it inside Directive:

ClearAll[tubeF]
tubeF[col_: Red, rad_: .1] := Module[{tf = Tube[#, rad] &}, Directive[col, tf @@ # &]];

Row[ContourPlot3D[{cf[x, y, z] == 25, y == 0}, {x, -4, 6}, {y, -9, 1}, {z, -2, 8}, 
    ContourStyle -> {Directive[Blue, Opacity[0.5]], Directive[Pink, Opacity[0.5]]},
    BoundaryStyle -> {1->None, 2->None, {1, 2}->#}, Mesh -> None, Evaluate@opts] & /@ 
 {tubeF[], tubeF[Orange, .3], tubeF[Purple, .5]}]

Row[ContourPlot3D[cf[x, y, z] == 25, {x, -4, 6}, {y, -9, 1}, {z, -2, 8},
    ContourStyle -> Opacity[.5, Blue], MeshFunctions -> {Function[{x, y, z}, y]}, 
    Mesh -> {{0}}, MeshStyle -> #, Evaluate@opts] & /@
  {tubeF[], tubeF[Orange, .3], tubeF[Purple, .5]}]

Answer 2

Just using a slightly different approach and using the derived circle and interpreting question as generating a torus from the circle of intersection between sphere and x-z plane:

s = ParametricPlot3D[
   5 {Sin[u] Cos[v], Sin[u] Sin[v], Cos[u]} + {1, -4, 3}, {u, 0, 
    Pi}, {v, 0, 2 Pi}, 
   RegionFunction -> Function[{x, y, z, u, v}, y <= 0], Mesh -> None];
t[r_, p_, b_] := 
  ParametricPlot3D[
   p + b {Cos[u], 0, Sin[u]} + 
    r Cos[v] {Cos[u], 0, Sin[u]} + {0, r Sin[v], 0}, {u, 0, 
    2 Pi}, {v, 0, 2 Pi}, Mesh -> False, PlotStyle -> Red];
Manipulate[
 Show[s, t[r, {1, 0, 3}, 3], Boxed -> False, Axes -> False, 
  Background -> Black, PlotRange -> All], {r, Range[0.05, 0.5, 0.05]}]