Plane of intersection of two spheres What is the plane of intersection of spheres $$x^2+y^2+z^2+2x+2y+2z+2=0$$ and $$x^2+y^2+z^2+x+y+z-\frac{1}{4}=0$$
I am not sure of how to do this, i just subtracted the two equations and i got a plane equation $$4x+4y+4z=-9$$ But i think its not correct since when two spheres meet we get a curved surface or solid. please correct me
 A: Looks fine.

The red sphere is the one of your first equation, the green one the one of your second equation.
The yellow circle is the intersection of both sphere surfaces.
The light blue plane is the intersection plane you calculated.
A: Let $(x_1,y_1,z_1)$ be a point that belong to both spheres, then it satisfies both equations and any linear combination of them, in particular the linear combination
\begin{align}
x_1^2+y_1^2+z_1^2+2x_1+2y_1+2z_1+2-\left(\color{blue}{x_1^2+y_1^2+z_1^2+x_1+y_1+z_1-\frac{1}{4}}\right)&=0\\
x_1+y_1+z_1+\frac{9}{4}&=0
\end{align}
So the point belong to the plane $4x+4y+4z+9=0$.
A: Every sphere in 3-space can be described by the equation, 
$$ (\vec{x}-\vec{x}_0)^2 - R^2 = 0,  $$
In our case we have two spheres with different centers, call these $\vec{q}$ and $\vec{p}$. Let $r$ be the center of the sphere with center $\vec{q}$ and $R$ be the center of the sphere with center $\vec{p}$. 
The intersection of the two spheres satisfies the equation of each sphere. We will we subtract the two equations to obtain a new equation which must be satisfied by the intersection,
$$ (\vec{x}-\vec{q})^2 - (\vec{x}-\vec{p})^2 + R^2- r^2 = 0,  $$
simplifying this we get, 
$$ 2 \vec{x} \cdot ( \vec{p}-\vec{q} ) + \vec{q}^2 + \vec{p}^2 + R^2 - r^2 =0,$$
which is a plane with normal vector, $\vec{p}-\vec{q}$. The points on the intersection then must satisfy this equation, i.e., lie on this plane; and satisfy the equation for each of the spheres. 
This method works because the validity of the equation for the plane stems from the fact that the equation of each sphere is satisfied by $\vec{x}$. If $\vec{x}$ is not an element of either sphere, then we can't claim that the difference of the two equations is still $0$. In this sense the new equation is only valid if the original two are, which is exactly what we need. 
A: Rewriting into standard forms,
$S_{1}$: $\: (x+1)^{2}+(y+1)^{2}+(z+1)^{2}=1$
$S_{2}$: $\: (x+\frac{1}{2})^{2}+(y+\frac{1}{2})^{2}+(z+\frac{1}{2})^{2}=1$
Both have radii of $1$ and distance between their centres $d=\frac{\sqrt{3}}{2}$.
Therefore, $r_{1}=r_{2}<d<r_{1}+r_{2}$ implying they intersect.
In general, any two non-concentric spheres $(S=0,S'=0)$ have one radical plane $(S-S'=0)$ no matter there's an intersection or not. We can generate a family of spheres with same radical plane in the form of $S+kS'=0$.
