If the center and radius of the circle |(z - 2)/(z - 3)| = 2 are...

Given:

$\left|\dfrac{z - 2}{z - 3}\right| = 2$

Let $z = x + iy$, then we have

\left|\dfrac{x - 2 + iy}{x - 3 + iy}\right| = 2$</p><p>$⇒ \sqrt{{\left(x - 2\right)}^{2} + {y}^{2}} = 2\sqrt{{\left(x - 3\right)}^{2} + {y}^{2}}$</p><p>$⇒ {x}^{2} + {y}^{2} - 4x + 4 = 4{x}^{2} + 4{y}^{2} - 24x + 36$</p><p>$⇒ 3{x}^{2} + 3{y}^{2} - 20x + 32 = 0$</p><p>$⇒ {x}^{2} + {y}^{2} - \dfrac{20}{3}x + \dfrac{32}{3} = 0$</p><p>So, \left(\alpha, \beta\right) ≡ \left(\dfrac{10}{3}, 0\right)$

And,

$\gamma = \sqrt{\dfrac{100}{9} - \dfrac{32}{3}} = \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$

Therefore,

$3\left(\alpha + \beta + \gamma\right) = 3\left(\dfrac{10}{3} + \dfrac{2}{3}\right) = 12$