- Thread starter markraz
- Start date

- Joined
- Nov 12, 2017

- Messages
- 5,624

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

- Joined
- Jan 29, 2005

- Messages
- 9,015

As abstract as this question is, a detailed help is difficult.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 !!!!

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 \(\displaystyle R\).

- Joined
- Feb 19, 2014

- Messages
- 169

Thank you !!!! What do f, g, h vars represent?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 \(\displaystyle R\).

Thanks

- Joined
- Jan 29, 2005

- Messages
- 9,015

Well they are coordinates of points on the circle.Thank you !!!! What do f, g, h vars represent?

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.

Last edited by a moderator:

- Joined
- Nov 12, 2017

- Messages
- 5,624

I think that's "Well they areconfidantesof points on the circle.

- Joined
- May 3, 2019

- Messages
- 165

\(\displaystyle \langle x,y,z \rangle= \langle p,q,r \rangle + \vec u R\cos t + \vec v R\sin t,~0 \le t \le 2\pi\),

- Joined
- Feb 19, 2014

- Messages
- 169

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.

\(\displaystyle \langle x,y,z \rangle= \langle p,q,r \rangle + \vec u R\cos t + \vec v R\sin t,~0 \le t \le 2\pi\),

Thanks !!!!