Jee Mains

Let the circle $\mathrm{C}_{1}:|\mathrm{z}|=\mathrm{r}$ and $\mathrm{C}_{2}:|\mathrm{z}-3-4 \mathrm{i}|=5, \mathrm{z} \in \mathrm{C}$, be such that $\mathrm{C}_{2}$ lies within $\mathrm{C}_{1}$. If $\mathrm{z}_{1}$ moves on $\mathrm{C}_{1},

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  • 30 October, 2026 at 8:11 am #12457
    UPLIFTJEE
    Keymaster

    Let the circle $\mathrm{C}_{1}:|\mathrm{z}|=\mathrm{r}$ and $\mathrm{C}_{2}:|\mathrm{z}-3-4 \mathrm{i}|=5, \mathrm{z} \in \mathrm{C}$, be such that $\mathrm{C}_{2}$ lies within $\mathrm{C}_{1}$. If $\mathrm{z}_{1}$ moves on $\mathrm{C}_{1}, \mathrm{z}_{2}$ moves on $\mathrm{C}_{2}$ and $\min \left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=2$, then $\max \left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|$ is equal to [JEE MAIN 02 April 2026 S II]
    (A) 12
    (B) 17
    (C) 22
    (D) 24

    30 October, 2026 at 8:11 am #12458
    UPLIFTJEE
    Keymaster

    Solution

    minimum \& maximum distance lie along their common normal (line joining centres)

    \[ \begin{aligned} & |z|=r \text { and } O A=r \\ & \text { minimum }\left|z_{1}-z_{2}\right|=2=A B \\ & \Rightarrow O A-O B=A B \\ & \Rightarrow r-10=2 \Rightarrow r=12 \end{aligned} \]

    $\operatorname{Max}\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=\mathrm{CB}=\mathrm{OC}+\mathrm{OB}=\mathrm{r}+10=12+10=22$

    30 October, 2026 at 8:11 am #12459
    UPLIFTJEE
    Keymaster

    Answer

    C

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