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(A) Let $(h, k)$ be the center of a circle with radius $r = 3$. The equation of such a circle is $(x - h)^2 + (y - k)^2 = 3^2 = 9$.Since the center $(

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

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(A) Let $(h, k)$ be the center of a circle with radius $r = 3$. The equation of such a circle is $(x - h)^2 + (y - k)^2 = 3^2 = 9$.Since the center $(h, k)$ lies on the circle $x^2 + y^2 = 25$,the distance of the center from the origin is $\sqrt{h^2 + k^2} = 5$.Any point $(x, y)$ on a circle with center $(h, k)$ and radius $3$ satisfies the condition that its distance from $(h, k)$ is $3$.By the triangle inequality,the distance $d$ of any point $(x, y)$ from the origin satisfies $|\sqrt{h^2 + k^2} - 3| \le \sqrt{x^2 + y^2} \le \sqrt{h^2 + k^2} + 3$.Substituting $\sqrt{h^2 + k^2} = 5$,we get $|5 - 3| \le \sqrt{x^2 + y^2} \le 5 + 3$.This simplifies to $2 \le \sqrt{x^2 + y^2} \le 8$.Squaring the inequality,we get $4 \le x^2 + y^2 \le 64$.

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

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