Draw a circle on a plane

[markraz]

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]

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

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

 

[pka]

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]

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]

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.

 

[Dr.Peterson]

Well they are confidantes of points on the circle.

I think that's "coordinates".

 

[LCKurtz]

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

 

[markraz]

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

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 !!!!