The centres of a set of circles, each of radius $3$, lie on the circle ${x}^{2}+{y}^{2}=25$. The lotus of any point in the set is ;

- The centers of the circles lie on the circle $x^{2}+y^{2}=25$. This means the centers are at a distance of 5 units (since the radius of this circle is $\sqrt{25}=5$ ) from the origin.

- Each of these circles has a radius of 3 .

- We need to find the locus of all points that lie on any of these circles.

Visualize the Situation

- The centers of the circles lie on a circle of radius 5 .

- Each of these circles has a radius of 3 .

- The locus of all points on these circles will form a ring (annulus) around the original circle.

Find the Locus

The locus of all points on these circles will be the set of points that are at a distance of $5 \pm 3$ from the origin. This is because:

- The farthest point from the origin will be at a distance of $5+3=8$.

- The closest point from the origin will be at a distance of $5-3=2$.

Thus, the locus of any point in the set is the region between two concentric circles with radii 2 and 8 , centered at the origin.

The locus is the set of all points $(x, y)$ such that:

$2^{2} \leq x^{2}+y^{2} \leq 8^{2}$

or

$4 \leq x^{2}+y^{2} \leq 64$

Final Answer:

The locus of any point in the set is the region between the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}=64$. In inequality form, this is:

$4 \leq x^{2}+y^{2} \leq 64$