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If you have for example a plan ($\mathcal{P}$) and a sphere ($\mathcal{S}$), let say :

$$(\mathcal{P}) \enspace \enspace \enspace \enspace z= \frac{1}{2} $$

$$(\mathcal{S}) \enspace \enspace \enspace \enspace x^2 + y^2 + z^2 = 1 $$

When you want to get the intersection, injecting ($\mathcal{P}$) in the expression of ($\mathcal{S}$), you get :

$$(\mathcal{P} \cap \mathcal{S}) \enspace \enspace \enspace \enspace x^2 + y^2 =\frac{3}{4} $$

The equation $(\mathcal{P} \cap \mathcal{S}) $ is supposed to be the equation a circle, but because we are still in 3D, this equation gives an infinite cylinder in space, having now a free parameter $z$, able to take any value.

In other words, how to only represent the circle in 3D space.

Thus, how to get a general expression of this 2D intersection, taking in consideration that we are still in 3D ?

Thank you in advance for your perspective.

jozinho22
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    Keep the other equation $z=\frac12.$ Just as two planes are needed to define the intersection line, you need two equations to get the circle from the cylinder and plane. – Jan-Magnus Økland Apr 02 '24 at 10:22
  • Because the intersection is a curve in $3D$, which is known to be a circle, then you need to find the center of the this circle, call it $C$, and two orthogonal vectors $v_1$ and $v_2$ such that the circle of intersection is parameterized as $P(t) = C + \cos(t) v_1 + \sin(t) v_2 $ –  Apr 02 '24 at 13:58
  • Not marking as a duplicate, but does this answer? https://math.stackexchange.com/questions/943383/determine-circle-of-intersection-of-plane-and-sphere/1218805#1218805 – Andrew D. Hwang Apr 02 '24 at 14:09

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The radius of the circle is, as you found out, is $R = \dfrac{\sqrt{3}}{2} $

And the center of the circle is $ C = (0, 0, \dfrac{1}{2}) $

Two vectors of length equal to $R$, that are mutually orthogonally, and perpendicular to the given plane are

$ V_1 = (\dfrac{\sqrt{3}}{2} , 0, 0 ) $

$ V_2 = ( 0, \dfrac{\sqrt{3}}{2}, 0) $

Therefore, the equation (parametric) of the circle is

$ p(t) = C + V_1 \cos t + V_2 \sin t = (\dfrac{\sqrt{3}}{2} \cos t , \dfrac{\sqrt{3}}{2} \sin t, \dfrac{1}{2} ) $

  • Can you be more explicit about your equation ? Is there no $(x,y)$ anymore ? – jozinho22 Apr 02 '24 at 17:22
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    @jozinho22 Just as you can have a parametrized line depending on 1 parameter like $(x,y,z)=(1+t,2-3t,5+7t)$, this is $(x,y,z)=(\frac{\sqrt3}2 \cos \theta , \frac{\sqrt3}2 \sin \theta, \frac12 ).$ If you don't mind to skip one point of the circle, you can even use a rational parametrization $(x,y,z)=(\frac{\sqrt3}2 \frac{1-t^2}{1+t^2} , \frac{\sqrt3}2 \frac{2t}{1+t^2}, \frac12 ),$ where $t=\tan{\frac{\theta}2}.$ – Jan-Magnus Økland Apr 02 '24 at 17:40
  • This solution is working well. I only did not understand why did you write the equation $p(t)$ through this form, because it can not be computed like this. Is it ? – jozinho22 Apr 02 '24 at 18:10
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    The circle of intersection is a one-dimensional curve that lies in a three-dimensional space. There is not a single algebraic equation to describe this circle, that's why I used a parametric equation. –  Apr 02 '24 at 19:03