For example, how to find the Cartesian equation for those two spheres?

$$(x-1)^2+(y-2)^2+(z-3)^2=64$$ and $$(x-2)^2+(y-3)^2+(z-5)^2=36$$?


It seems that you don't have the proper idea of what's going on here. "How to find the Cartesian equations of these two spheres?" Heavens, you write them down in the next two lines yourself!

Now it can be that these two spheres intersect. If this is the case the intersection is (a point or) a circle $\gamma$. This circle is lying in a certain plane. But as $1$-dimensional manifold in $3$-space it does not have "an equation". It can be given by two equations in many ways, e.g., by the two equations you wrote down yourself, or by the equation of said plane and one of the two given equations.

The most useful way to present $\gamma$ is a parametric representation. To obtain it you have to determine the center ${\bf c}$ of $\gamma$, the radius $r>0$, and two orthogonal unit vectors ${\bf u}$, ${\bf v}$ in the plane of $\gamma$. The parametric representation would then be $$\gamma:\quad \phi\mapsto{\bf x}(\phi)={\bf c}+r \>\cos\phi\>{\bf u}+r\>\sin\phi\>{\bf v}\qquad(0\leq \phi\leq 2\pi)\ .$$

  • $\begingroup$ Thank you for helping me! $\endgroup$ – Siwei Dec 25 '17 at 14:57

Substract second equation from the first one:

$$2x-3+2y-5+4z-16=28\implies x+y+2z=26$$

and we get a plane with normal $\;(1,1,2)\;$ , and now you only have to find out what circle this plae cuts on the first (or second: it doesn't matter) sphere.

For example, with the first sphere and substituting $\;x=26-y-2z\;$ :



The last one is a very ugly equation of a circle in three dimensions...Try to simplify things here.

Observe that this is basically the same method used by Watson in the other answer...but I used other parameters for the circle. It's up to you to show these two parametrizations (Watson's and mine) yields the same object.

  • $\begingroup$ Thank you for helping me. $\endgroup$ – Siwei Dec 25 '17 at 14:53

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