I have a circle that's formed by the intersection of the sphere $$x^2 + y^2 + z^2 = 1$$ and the plane $x + y + z = 0$. How can I express the equation of this circle? I think the equation can be expressed in spherical coordinates. I have a fair understanding of how to convert cartesian coordinates to spherical, but I'm not sure how to approach this.
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have you made a Image to the Problem? – Dr. Sonnhard Graubner Nov 04 '17 at 09:43
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see also here https://math.stackexchange.com/questions/943383/determine-circle-of-intersection-of-plane-and-sphere – Dr. Sonnhard Graubner Nov 04 '17 at 09:47
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Link may be of interest. https://math.stackexchange.com/questions/2283585/intersection-of-plane-and-sphere/2284352 – Peter Szilas Nov 04 '17 at 09:57
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The circle in question is a curve in $3$-space. Therefore it does not have "an equation". It is given either by two equations (as in the formulation of the problem), or by a parametric representation. In order to obtain such a representation we need two orthogonal unit vectors ${\bf a}$, ${\bf b}$ in the plane of the circle. We may (guess and) choose $${\bf a}={1\over\sqrt{2}}(1,-1,0)\ .$$ For ${\bf b}$ we take the vector product $${\bf n}\times{\bf a}=\lambda\ (1,1,1)\times(1,-1,0)=\lambda'(1,1,-2)$$ and normalize: $${\bf b}={1\over\sqrt{6}}(1,1,-2)\ .$$ The circle $\gamma$ can then be presented as $$\gamma:\quad t\mapsto {\bf x}(t)=\cos t\>{\bf a}+\sin t\>{\bf b}\ .$$
Christian Blatter
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HINT.- The center of the sphere is clearly $O=(0,0,0)$ and this point is in the plane $x+y+z=0$. Making, for example $z=0$ you have the easy system $$x+y=0\\x^2+y^2=1$$ This way and similarly making after $y=0$ and $x=0$ you get the three points determining the searched circle: $$\left(\frac{\sqrt2}{2},-\frac{\sqrt2}{2},0\right)\\\left(\frac{\sqrt2}{2},0,-\frac{\sqrt2}{2}\right)\\\left(0,\frac{\sqrt2}{2},-\frac{\sqrt2}{2}\right)$$ NOTE.-If you want to try with angles note that $\cos45^{\circ}=\dfrac{\sqrt2}{2}$.
Ataulfo
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