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

9 \leq x^{2} + y^{2} \leq 49

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x^{2} + y^{2} \leq 25

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x^{2} + y^{2} \geq 25

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3 \leq x^{2} + y^{2} \leq 9

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Solution

The correct option is A $9 \leq x^{2} + y^{2} \leq 49$
Centre and radius of the circle $x^{2} + y^{2} = 25$ are $C (0, 0)$ and $5$ respectively.
Let $(h, k)$ be any point on the circle $x^{2} + y^{2} = 25$
$⟹ h^{2} + k^{2} = 25$
$r_{1} - r_{2} \leq$ Distance between two circles $\leq r_{1} + r_{2}$
$(5 - 2) \leq \sqrt{h^{2} + k^{2}} \leq (5 + 2)$
$⟹ 9 \leq h^{2} + k^{2} \leq 49$
Thus, the locus point $(h, k)$ is $9 \leq x^{2} + y^{2} \leq 49$
Hence, option A is correct.

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