Key Concepts and Formulas

• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. A circle centered at the origin has the equation $x^2 + y^2 = r^2$.
• Triangle Inequality: For any three points A, B, and C, the triangle inequality states that $AB + BC \ge AC$ and $|AB - BC| \le AC$.
• Locus: The set of all points satisfying a given condition.

Step-by-Step Solution

1. Understanding the Given Information

We are given a set of circles, each with a radius of 3. The centers of these circles lie on the circle $x^2 + y^2 = 25$. We need to find the locus of any point in this set of circles.

2. Defining Variables and Geometric Interpretation

• Let the origin be $O(0, 0)$.
• Let $C(h, k)$ be the center of any circle in the set. Since $C$ lies on the circle $x^2 + y^2 = 25$, we have $h^2 + k^2 = 25$. This means the distance $OC = \sqrt{h^2 + k^2} = 5$.
• Let $P(x, y)$ be any point on or inside one of the circles in the set. Since the radius of each circle is 3, the distance $CP \le 3$.
• We want to find the locus of $P$, which is the set of all possible points $(x, y)$ that satisfy the given conditions. In other words, we want to find the possible range of $OP = \sqrt{x^2 + y^2}$.

3. Finding the Minimum Distance from the Origin ($OP_{min}$)

By the triangle inequality, we have $OP + PC \ge OC$, which can be rearranged to $OP \ge OC - PC$. Since we want to minimize $OP$, we want to maximize $PC$. The maximum value of $PC$ is the radius of the circle, which is 3. Therefore, $OP \ge OC - 3 = 5 - 3 = 2$. Thus, the minimum distance from the origin to any point $P$ is $OP_{min} = 2$.

4. Finding the Maximum Distance from the Origin ($OP_{max}$)

Again, by the triangle inequality, we have $OC + CP \ge OP$, which can be rearranged to $OP \le OC + CP$. To maximize $OP$, we want to maximize $CP$. The maximum value of $CP$ is the radius of the circle, which is 3. Therefore, $OP \le OC + 3 = 5 + 3 = 8$. Thus, the maximum distance from the origin to any point $P$ is $OP_{max} = 8$.

5. Determining the Locus of P

We have found that $2 \le OP \le 8$, which means $2 \le \sqrt{x^2 + y^2} \le 8$. Squaring all parts of the inequality gives us: $4 \le x^2 + y^2 \le 64$.

Common Mistakes & Tips

• Confusing the Locus: Understand that the locus is the region covered by all the circles, not just the circle of centers.
• Triangle Inequality is Key: Correctly applying the triangle inequality is crucial for finding the minimum and maximum distances. Be careful with the direction of the inequality.
• Visualization: Always visualize the problem to get a better understanding of the geometry.

Summary

The locus of any point in the set of circles is the region between two concentric circles centered at the origin. The inner boundary is a circle with radius 2, and the outer boundary is a circle with radius 8. Therefore, the locus is described by the inequality $4 \le x^2 + y^2 \le 64$.

The final answer is $\boxed{4\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,64}$, which corresponds to option (A).