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This is a general question about the intersection of a sphere with a plane or sphere which is confusing me.

To find the intersection between two spheres K1 and K2, you can equate them, solve the equation K1=K2, and find the intersection plane. But the same procedure is not possible between a sphere (K) and a plane (P). One cannot just solve the equation K=P and gets the circle. But I don't understand why? Can anyone explain this?

Edit: As an example: Let K : (x-1)² +(y-9)² + (z-4)² -85=0 be the sphere in 3d and P: 6x-2y+3z-49=0 be a plane. Why can't I just solve the equation K=P and get the intersection circle?

Thank you!

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  • $\begingroup$ I don't understad either why you don't obtain a circle. In what form would you want your circle to be described? $\endgroup$ – Hagen von Eitzen Jan 21 at 11:47
  • $\begingroup$ What kind of equations are you using to describe a sphere or a plane? There are more ways to do that. Can you please show us an example? $\endgroup$ – Hume2 Jan 21 at 11:49
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    $\begingroup$ @Hume2 Let K : (x-1)² +(y-9)² + (z-4)² -85=0 be the sphere in 3d and P: 6x-2y+3z-49=0 be a plane. Why can't I just solve the equation K=P and get the intersection circle? $\endgroup$ – MBCLA Jan 21 at 11:53
  • $\begingroup$ Actually, this shouldn't work even for two spheres. Can you please give us also an example of two spheres for which is does work? $\endgroup$ – Hume2 Jan 21 at 12:04
  • $\begingroup$ You seem to me to be confusing the intersection, which is a circle, with the intersection plane which is the plane in which the intersection lives. $\endgroup$ – ancientmathematician Jan 21 at 12:09
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The problem has nothing to do with geometry. The real problem is that saying $K=0$ and $P=0$ is NOT equivalent to saying $K=P$.

Example: In the real plane, the system $x=0$ and $y=0$ yields a single point $(0,0)$, while $x=y$ yields a line.

The fact that your false recipe $K=P$ works in a particular case is a pure coincidence (the proof is that tomi gave you a counterexample where your method does not work).

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  • $\begingroup$ That makes sense! But then why one can do this with two spheres? $\endgroup$ – MBCLA Jan 21 at 12:08
  • $\begingroup$ we can't ! As I said, it is pure coincidence that you recipe works in your particular case $\endgroup$ – GreginGre Jan 21 at 14:13
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You have to be a little careful in your formulation.

First, as mentioned in comments, it isn't true that $K_1=K_2$ gives the [circle of] intersection of the two spheres. Rather, it gives the plane of the circle of intersection. (This plane is the analogue of a radical axis, the line containing the points of intersection of two circles.)

Second, it's not even necessarily true that $K_1=K_2$ gives the equation of that plane. That magic works when (and only when) $K_1$ and $K_2$ have matching coefficients on $x^2$, $y^2$, $z^2$, because then (and only then) do those terms cancel in $K_1=K_2$; what remains is a linear equation in $x$, $y$, $z$: the target plane.

A problem, though, is that there's no a priori reason for those coefficients to match. Both $x^2+y^2+z^2-3=0$ and $4x^2+4y^2+4z^2-12=0$ represent the same sphere; either could be "$K_1$". And neither will yield a sphere when simply set equal to a "$K_2$" of the form $3(x-4)^2+3(y+5)^2+3(z-6)^2-7=0$.

Almost-certainly, you're assuming $K_1$ and $K_2$ to be in "standard form", with necessarily-matching coefficients of $1$ on their second-degree terms; that's not an unreasonable assumption (standard form is common), but it is an assumption and so must be stated explicitly. An alternative approach is to restate things thusly:

If $K_1(x,y,z)=0$ and $K_2(x,y,z)=0$ represent two spheres, then there are non-zero values $k_1$ and $k_2$ such that $k_1 K_1 = k_2 K_2$ is the equation of the plane containing the circle of intersection (if any).

(Specifically, we take $k_1$ to be the coefficient of $x^2$ (and $y^2$ and $z^2$) in $K_2$, and $k_2$ to be the corresponding coefficient in $K_1$. This guarantees that those terms cancel.)

The advantage of the restatement is that it applies when, say, $K_1=0$ represents a plane instead of a sphere. Since we end up taking $k_2=0$ (as $K_1$ has no second-degree terms), the equation $k_1 K_1=k_2 K_2$ reverts to $K_1=0$, as expected; after all, the plane containing the circle of intersection is that original plane.

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Well, it's not true that the intersection between two spheres is a plane. E.g., consider the two spheres $$x^2+y^2+z^2=1$$ and $$(x-1)^2+y^2+z^2=1,$$ the intersection gives $$ \left\{\begin{array}{l} x^2+y^2+z^2=1\\ 2x=1 \end{array}\right. $$ which is not a plane, but the a circonference that lies in the plane $2x=1$ and is contained in the spheres $x^2+y^2+z^2=1$; sure, the system $$ \left\{\begin{array}{l} (x-1)^2+y^2+z^2=1\\ 2x=1 \end{array}\right. $$ defined the same circumference.

Now, if you consider the plane $2x=1$ and the sphere $x^2+y^2+z^2=1$, their intersection is again $$ \left\{\begin{array}{l} (x-1)^2+y^2+z^2=1\\ 2x=1 \end{array}\right. $$ i.e., the same circumference as above. You can write the same system as $$ \left\{\begin{array}{l} y^2+z^2=\frac{3}{4}\\ 2x=1 \end{array}\right. $$ namely, you can eliminate the $x$ from the first equation. Now, what represents the equation $y^2+z^2=\frac{3}{4}$? Not a circumference: indeed, it is a cylinder (remember that we are working in $\mathbb{R}^3$), to be precise the cylinder having as directrix the circumference $$ \left\{\begin{array}{l} (x-1)^2+y^2+z^2=1\\ 2x=1 \end{array}\right. $$ and axis perpendicular to the plane $2x=1$.

Summing up: an equation as $y^2+z^2=1$ is not a circumference in $\mathbb{R}^3$ but a cylinder, the equation of a circumference in $\mathbb{R}^3$ is always given by a system, the same holds for a line in $\mathbb{R}^3$, indeed $2x=1$ is a plane $\mathbb{R}^3$, and a line in $\mathbb{R}^2$.

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