# Why one cannot equate a sphere and a plane but a sphere with a sphere? [closed]

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!

• I don't understad either why you don't obtain a circle. In what form would you want your circle to be described? – Hagen von Eitzen Jan 21 at 11:47
• 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? – Hume2 Jan 21 at 11:49
• @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? – MBCLA Jan 21 at 11:53
• 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? – Hume2 Jan 21 at 12:04
• 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. – ancientmathematician Jan 21 at 12:09

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).

• That makes sense! But then why one can do this with two spheres? – MBCLA Jan 21 at 12:08
• we can't ! As I said, it is pure coincidence that you recipe works in your particular case – GreginGre Jan 21 at 14:13

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.

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 $$\mathbb{R}^2$$.