Intersection of spheres seems to produce equation of plane

Question

I'm trying to find the equation describing the curve formed by the intersection of two spheres. One of radius $R$ is centered at $(0,0,0)$ and one of radius $r_v$ is at a point $(x_v,y_v,z_v)$. The equations describing the spheres are then

$$x^2 + y^2 + z^2 = R^2$$ $$(x-x_v)^2 + (y-y_v)^2 + (z-z_v)^2 = {r_v}^2$$

Expanding the second equation:

$$x^2 - 2xx_v + {x_v}^2 + y^2 - 2yy_v + {y_v}^2 + z^2 - 2zz_v + {z_v}^2 = {r_v}^2$$

Collecting terms:

$$x^2 + y^2 + z^2 - 2(xx_v + yy_v + zz_v) + {x_v}^2 + {y_v}^2 + {z_v}^2 = {r_v}^2$$

From the first equation, we have $x^2 + y^2 + z^2 = R^2$, so we can substitute $R^2$ for $x^2 + y^2 + z^2$ in the preceding equation:

$$R^2 - 2(xx_v + yy_v + zz_v) + {x_v}^2 + {y_v}^2 + {z_v}^2 = {r_v}^2$$

Which leads to

$$xx_v + yy_v + zz_v = -\frac{1}{2} \left( {r_v}^2 - R^2 - {x_v}^2 - {y_v}^2 - {z_v}^2 \right)$$

This equation describes a plane, but the intersection of two spheres should be a circle. $R$ and $r_v$ are independent of $x, y,$ and $z$. So this final equation is of the form

$$ax + by + cz = d$$

I can only assume I've made an error somewhere, but I can't figure out where.

Answer 1

By subtracting the two equations, you’ve not produced a solution to the system. What you’ve got is another equation whose solution set includes the common solutions to the original equations.

To illustrate, lets look at finding the intersection of the lines in the plane with equations $x+2y=1$ and $x-y=3$. Subtracting the second from the first yields the equation $3y=-2$. This is the equation of another line that passes through the intersection point. It’s simpler than the original equations, though, and you can easily find the $y$-coordinate of the intersection using it, but you then have to go back to one of the original equations to find the $x$-coordinate. In the same way, subtracting one sphere equation from the other produced a simpler equation, but this simpler equation overgenerates solutions—an entire plane’s worth instead of only a circle—just like the way the line $3y=-2$ includes many points we’re not interested in.

Observe, too, that no matter how we manipulate them, it take two implicit equations to specify a point on the plane. Similarly, a curve in $\mathbb R^3$ can’t be represented by a single implicit equation such as the one you derived.

Answer 2

You try to solve a system with two equations and two unknowns. You made one new equation from both. But the system stays a system, that is, you have a new system containing the new equation and one of the original ones.

Example: take $x+y=3$ and $2x-3y=6$. Now add both equations to get $3x-2y=9$, an equation of a line. It’s clear that this line is not the solution of the system. The solution is the intersection of that new line and one of the original ones.

In your case you found that the intersection of two spheres is the intersection of a plane and one of the spheres.

Answer 3

You made no other mistake than dropping one of the original equations, which allows more solution points than should.

The intersection of two spheres is indeed a circle, with is contained in a plane. Hence you can recast the problem as that of the intersection of a sphere and a plane.