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If the center and radius of the circle $\left|\frac{\mathrm{z}-2}{\mathrm{z}-3}\right|=2$ are respectively $(\alpha, \beta)$ and $\gamma$, then $3(\alpha, \beta, \gamma)$ is equal to

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  • 30 October, 2026 at 8:11 am #12535
    UPLIFTJEE
    Keymaster

    If the center and radius of the circle $\left|\frac{\mathrm{z}-2}{\mathrm{z}-3}\right|=2$ are respectively $(\alpha, \beta)$ and $\gamma$, then $3(\alpha, \beta, \gamma)$ is equal to [JEE (Main) 01 Feb 2023 S I]
    (A) 12
    (B) 11
    (C) 9
    (D) 10

    30 October, 2026 at 8:11 am #12536
    UPLIFTJEE
    Keymaster

    Solution

    $\sqrt{(\mathrm{x}-2)^{2}+\mathrm{y}^{2}}=2 \sqrt{(\mathrm{x}-3)^{2}+\mathrm{y}^{2}} \quad=\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}+4=4 \mathrm{x}^{2}+4 \mathrm{y}^{2}-24 \mathrm{x}+36$

    \[ \begin{aligned} & =3 x^{2}+3 y^{2}-20 x+32=0=x^{2}+y^{2}-\frac{20}{3} x+\frac{32}{3}=0 \\ & =(\alpha, \beta)=\left(\frac{10}{3}, 0\right) \Rightarrow \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \Rightarrow 3(\alpha, \beta, \gamma)=3\left(\frac{10}{3}, \frac{2}{3}\right)=12 \end{aligned} \]
    30 October, 2026 at 8:11 am #12537
    UPLIFTJEE
    Keymaster

    Answer

    27

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