JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 7)
If the center and radius of the circle $$\left| {{{z - 2} \over {z - 3}}} \right| = 2$$ are respectively $$(\alpha,\beta)$$ and $$\gamma$$, then $$3(\alpha+\beta+\gamma)$$ is equal to :
12
10
11
9
Explanation
$$\left| {{{z - 2} \over {z - 3}}} \right| = 2$$
$$ \begin{aligned} & \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\ & \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\ & \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\ & \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\ & \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\ & \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\ & 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\ & =12 \end{aligned} $$
$$ \begin{aligned} & \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\ & \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\ & \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\ & \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\ & \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\ & \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\ & 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\ & =12 \end{aligned} $$
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