# Draw a circle on a plane

#### markraz

##### Junior Member
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

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

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

#### pka

##### Elite Member
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

##### Junior Member
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

##### Elite Member
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

##### Elite Member
Well they are confidantes of points on the circle.
I think that's "coordinates".

• pka

#### LCKurtz

##### Full Member
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

##### Junior Member
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 !!!!