Where is the centre of mass of two isolated particles of equal mass located?

Four identical spheres, each of radius $10 \,cm$ and equal mass $1 \,kg$, are placed on a horizontal surface touching each other such that their centers are located at the vertices of a square of side $20 \,cm$. What is the distance of their center of mass from the center of any sphere?

Obtain the position of the centre of mass of a thin rod of uniform density.

The balls $A, B$ and $C$ of masses $50 \ g, 100 \ g$ and $150 \ g$,respectively,are placed at the vertices of an equilateral triangle. The length of each side is $1 \ m$. If $A$ is placed at $(0,0)$ and $B$ is placed at $(1,0) \ m$,find the coordinates $(x, y)$ for the centre of mass of this system of the balls.

The center of mass of a system of particles with masses $1 \, g, 2 \, g$,and $3 \, g$ is at the origin. When a particle of mass $4 \, g$ with position vector $\alpha(\hat{i} + 2\hat{j} + 3\hat{k})$ is added,the center of mass of the system becomes $(1, 2, 3)$. If $\alpha$ is a constant,its value must be:

Centre of mass $(C.M.)$ of three particles of masses $1 \,kg, 2 \,kg$ and $3 \,kg$ lies at the point $(1, 2, 3)$ and $C.M.$ of another system of particles of $3 \,kg$ and $2 \,kg$ lies at the point $(-1, 3, -2)$. Where should we put a particle of mass $5 \,kg$ so that the $C.M.$ of the entire system lies at the $C.M.$ of the first system?