       Re: Cirlce in 3D?

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• Subject: [mg132723] Re: Cirlce in 3D?
• From: "djmpark" <djmpark at comcast.net>
• Date: Mon, 12 May 2014 22:25:54 -0400 (EDT)
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• References: <10319860.232311.1399870151373.JavaMail.root@m03>

```Using John Browne's Grassmann algebra application and with the three points
at Cartesian coordinates {ax, ay, az}, {bx, by, bz}, and {cx, cy, cz} I
calculated the center of the circle at:

{x -> -(4 (az (by - cy) + bz cy - by cz +
ay (-bz +
cz)) (-(ax^2 + ay^2 + az^2 - bx^2 - by^2 - bz^2) (-by cx +
ay (-bx + cx) + ax (by - cy) + bx cy) -
2 (az - bz) (az by cx - ay bz cx - az bx cy + ax bz cy +
ay bx cz - ax by cz)) - (2 (ax^2 + ay^2 + az^2 - bx^2 -
by^2 - bz^2) (bz - cz) -
2 (az - bz) (bx^2 + by^2 + bz^2 - cx^2 - cy^2 -
cz^2)) (-2 (ay - by) (ay (bx - cx) + by cx - bx cy +
ax (-by + cy)) -
2 (az - bz) (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz))))/(-4 (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz)) (-2 (ay - by) (ay (bx - cx) + by cx - bx cy +
ax (-by + cy)) -
2 (az - bz) (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz))) +
4 (az (by - cy) + bz cy - by cz +
ay (-bz + cz)) (2 (ax - bx) (-by cx + ay (-bx + cx) +
ax (by - cy) + bx cy) +
2 (az - bz) (az (by - cy) + bz cy - by cz + ay (-bz + cz)))),
y -> (-2 (ax^2 + ay^2 + az^2 - bx^2 - by^2 - bz^2) (bz - cz) +
2 (az - bz) (bx^2 + by^2 + bz^2 - cx^2 - cy^2 -
cz^2) + (4 (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz)) (4 (az (by - cy) + bz cy - by cz +
ay (-bz +
cz)) (-(ax^2 + ay^2 + az^2 - bx^2 - by^2 -
bz^2) (-by cx + ay (-bx + cx) + ax (by - cy) +
bx cy) -
2 (az - bz) (az by cx - ay bz cx - az bx cy + ax bz cy +
ay bx cz - ax by cz)) - (2 (ax^2 + ay^2 + az^2 -
bx^2 - by^2 - bz^2) (bz - cz) -
2 (az - bz) (bx^2 + by^2 + bz^2 - cx^2 - cy^2 -
cz^2)) (-2 (ay - by) (ay (bx - cx) + by cx - bx cy +
ax (-by + cy)) -
2 (az - bz) (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz)))))/(-4 (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz)) (-2 (ay - by) (ay (bx - cx) + by cx -
bx cy + ax (-by + cy)) -
2 (az - bz) (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz))) +
4 (az (by - cy) + bz cy - by cz +
ay (-bz + cz)) (2 (ax - bx) (-by cx + ay (-bx + cx) +
ax (by - cy) + bx cy) +
2 (az - bz) (az (by - cy) + bz cy - by cz +
ay (-bz + cz)))))/(4 (az (by - cy) + bz cy - by cz +
ay (-bz + cz))),
z -> (-az^2 by + by cx^2 + az^2 cy - bx^2 cy - by^2 cy - bz^2 cy +
by cy^2 + ax^2 (-by + cy) + ay^2 (-by + cy) + by cz^2 +
ay (bx^2 + by^2 + bz^2 - cx^2 - cy^2 - cz^2))/(2 (-bz cy +
az (-by + cy) + ay (bz - cz) + by cz)) + ((-by cx +
ay (-bx + cx) + ax (by - cy) +
bx cy) (4 (az (by - cy) + bz cy - by cz +
ay (-bz +
cz)) (-(ax^2 + ay^2 + az^2 - bx^2 - by^2 -
bz^2) (-by cx + ay (-bx + cx) + ax (by - cy) +
bx cy) -
2 (az - bz) (az by cx - ay bz cx - az bx cy + ax bz cy +
ay bx cz - ax by cz)) - (2 (ax^2 + ay^2 + az^2 - bx^2 -
by^2 - bz^2) (bz - cz) -
2 (az - bz) (bx^2 + by^2 + bz^2 - cx^2 - cy^2 -
cz^2)) (-2 (ay - by) (ay (bx - cx) + by cx - bx cy +
ax (-by + cy)) -
2 (az - bz) (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz)))))/((az (by - cy) + bz cy - by cz +
ay (-bz + cz)) (-4 (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz)) (-2 (ay - by) (ay (bx - cx) + by cx -
bx cy + ax (-by + cy)) -

2 (az - bz) (az (bx - cx) + bz cx - bx cz +
ax (-bz + cz))) +
4 (az (by - cy) + bz cy - by cz +
ay (-bz + cz)) (2 (ax - bx) (-by cx + ay (-bx + cx) +
ax (by - cy) + bx cy) +
2 (az - bz) (az (by - cy) + bz cy - by cz +
ay (-bz + cz)))))}

I have a notebook on it but it uses the GrassmannCalculus and Presentations
Applications. Presentations has Circle3D and Disk3D primitives. The notebook
also calculates and displays random cases.

David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/index.html

From: Ste[hen Gray [mailto:stevebg at roadrunner.com]

I'm looking for a neat formula to find the center of a circle in 3D through
3 points. I also need a good way to display it, preferably thickened so I
can show several and see whether they are linked, etc. To my surprise I did
not find anything on the Wolfram sites about these problems. (I have
Mathematica 7, if that matters.)

```

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