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 :
Solution
$\left| {{{z - 2} \over {z - 3}}} \right| = 2$
<br/><br/>$$
\begin{aligned}
& \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\
& \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\
& \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\
& \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\
& \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\
& \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\
& 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\
& =12
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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