Draw a circle on a plane

markraz

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Feb 19, 2014
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Hi, I need to draw a circle on plane. I have a center point, radius and a plane equitation ax+by+cz + d = 0
does anyone have any ideas to get me started? thanks !!!!
 

Dr.Peterson

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Do you mean literally "draw", or "find the equation" (in some particular form)?

(You might want to look up "equitation", just for fun.)
 

pka

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Hi, I need to draw a circle on plane. I have a center point, radius and a plane equitation ax+by+cz + d = 0does anyone have any ideas to get me started? thanks !!!!
As abstract as this question is, a detailed help is difficult.
First have you checked to see if the centre you have is actually on that plane?
If \(\displaystyle \mathfrak{C}(p,q,r)\) is the centre then \(\displaystyle ap+bq+cr+d=0\) the center is on the plane.
Lets say \(\displaystyle R\) is the radius.

The circle is \(\displaystyle \left\{(f,g,h) : (p-f)^2+(q-g)^2+(r-h)^2=R^2\right\}\) and the vectors \(\displaystyle \left<a,b,c\right>\cdot\left<p-f,q-g,r-h\right>=0.\).
The is a vector from the centre to a point of the circle is perpendicular to the normal of the plane and has length \(\displaystyle R\).
 

markraz

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As abstract as this question is, a detailed help is difficult.
First gave you checked to see if the centre you have is actually on that plane?
If \(\displaystyle \mathfrak{C}(p,q,r)\) is the centre then \(\displaystyle ap+bq+cr+d=0\) the center is on the plane.
Lets say \(\displaystyle R\) is the radius.

The circle is \(\displaystyle \left\{(f,g,h) : (p-f)^2+(q-g)^2+(r-h)^2=R^2\right\}\) and the vectors \(\displaystyle \left<a,b,c\right>\cdot\left<p-f,q-g,r-h\right>=0.\).
The is a vector from the centre to a point of the circle is perpendicular to the normal of the plane and has length \(\displaystyle R\).
Thank you !!!! What do f, g, h vars represent?

Thanks
 

pka

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Thank you !!!! What do f, g, h vars represent?
Well they are coordinates of points on the circle.
The circle is centred at \(\displaystyle (p,q,r)\) so if \(\displaystyle (f,g,h)\) is a point on the circle of radius \(\displaystyle R\)
then the vector \(\displaystyle \vec{v}=\left<f-p,g-q,r-h\right>\) is such that \(\displaystyle \|\vec{v}\|=R\).
In order for the circle to be in a plane that plane must contain the centre and that vector most be perpendicular to the normal.
 
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Dr.Peterson

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LCKurtz

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Start with your equation \(\displaystyle ax+by+cz = d\) and your given point \(\displaystyle (p,q,r)\) on the plane. A normal vector to your plane is \(\displaystyle \vec N = \langle a,b,c \rangle.\). Create any nonzero vector \(\displaystyle \vec U\) perpendicular to \(\displaystyle \vec N\). This is trivial. For example \(\displaystyle \vec U = \langle -b, a, 0\rangle\) will work if \(\displaystyle a\) and \(\displaystyle b\) aren't \(\displaystyle 0\). (If you can't find two nonzero coefficients your problem is trivial anyway). Create another vector \(\displaystyle \vec V = \vec N \times \vec U\). Now divide \(\displaystyle \vec U\) and \(\displaystyle \vec V\) by their lengths to make unit vectors \(\displaystyle \hat u\) and \(\displaystyle \hat v\). These are perpendicular unit vectors in your plane. Then a circle of radius \(\displaystyle R\) in your plane with \(\displaystyle (p,q,r)\) as center is
\(\displaystyle \langle x,y,z \rangle= \langle p,q,r \rangle + \vec u R\cos t + \vec v R\sin t,~0 \le t \le 2\pi\),
 

markraz

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Start with your equation \(\displaystyle ax+by+cz = d\) and your given point \(\displaystyle (p,q,r)\) on the plane. A normal vector to your plane is \(\displaystyle \vec N = \langle a,b,c \rangle.\). Create any nonzero vector \(\displaystyle \vec U\) perpendicular to \(\displaystyle \vec N\). This is trivial. For example \(\displaystyle \vec U = \langle -b, a, 0\rangle\) will work if \(\displaystyle a\) and \(\displaystyle b\) aren't \(\displaystyle 0\). (If you can't find two nonzero coefficients your problem is trivial anyway). Create another vector \(\displaystyle \vec V = \vec N \times \vec U\). Now divide \(\displaystyle \vec U\) and \(\displaystyle \vec V\) by their lengths to make unit vectors \(\displaystyle \hat u\) and \(\displaystyle \hat v\). These are perpendicular unit vectors in your plane. Then a circle of radius \(\displaystyle R\) in your plane with \(\displaystyle (p,q,r)\) as center is
\(\displaystyle \langle x,y,z \rangle= \langle p,q,r \rangle + \vec u R\cos t + \vec v R\sin t,~0 \le t \le 2\pi\),
Cool thank you, this is starting to make a little sense now. Let me work on this and I will get back here since I'm sure I will have more questions.
Thanks !!!!
 
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