Let the circles C_1 : |z| = r and C_2 : |z - | vedclass

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(C) The center of $C_1$ is $O(0, 0)$ and its radius is $R = r$.The center of $C_2$ is $C(3, 4)$ and its radius is $r' = 5$.The distance between the ce

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$22$

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(C) The center of $C_1$ is $O(0, 0)$ and its radius is $R = r$.The center of $C_2$ is $C(3, 4)$ and its radius is $r' = 5$.The distance between the centers $O$ and $C$ is $d = \sqrt{3^2 + 4^2} = 5$.Since $C_2$ lies within $C_1$,the condition $R \ge d + r'$ must be satisfied,which gives $R \ge 5 + 5 = 10$.The minimum distance between points on two circles where one is inside the other is given by $\min |z_1 - z_2| = R - (d + r')$.Given $\min |z_1 - z_2| = 2$,we have $R - (5 + 5) = 2$,which implies $R - 10 = 2$,so $R = 12$.The maximum distance between points on two circles where one is inside the other is given by $\max |z_1 - z_2| = R + d + r'$.Substituting the values,$\max |z_1 - z_2| = 12 + 5 + 5 = 22$.

Let the circles $C_1 : |z| = r$ and $C_2 : |z - 3 - 4i| = 5, z \in \mathbb{C}$,be such that $C_2$ lies within $C_1$. If $z_1$ moves on $C_1, z_2$ moves on $C_2$ and $\min |z_1 - z_2| = 2$,then $\max |z_1 - z_2|$ is equal to:

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