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.